How to add an object to an array
How can I add an object to an array (in javascript or jquery)?
For example, what is the problem with this code?
function()
var a = new array();
var b = new object();
a[0]=b;
I would like to use this code to save many objects in the array of function1 and call function2 to use the object in the array.
- How can I save an object in an array?
- How can I put an object in an array and save it to a variable?
javascript arrays object
add a comment |
How can I add an object to an array (in javascript or jquery)?
For example, what is the problem with this code?
function()
var a = new array();
var b = new object();
a[0]=b;
I would like to use this code to save many objects in the array of function1 and call function2 to use the object in the array.
- How can I save an object in an array?
- How can I put an object in an array and save it to a variable?
javascript arrays object
10
Please don't correct errors in posted code, as edit 5 did. If you correct a simple error in the question, often there is no reason for the answer to exist.
– Paul
Aug 6 '15 at 3:01
Also,push
has been suggested multiple times here already. Please, do not pollute this thread anymore with anotherpush
suggestion.
– Boghyon Hoffmann
Apr 14 '18 at 14:17
add a comment |
How can I add an object to an array (in javascript or jquery)?
For example, what is the problem with this code?
function()
var a = new array();
var b = new object();
a[0]=b;
I would like to use this code to save many objects in the array of function1 and call function2 to use the object in the array.
- How can I save an object in an array?
- How can I put an object in an array and save it to a variable?
javascript arrays object
How can I add an object to an array (in javascript or jquery)?
For example, what is the problem with this code?
function()
var a = new array();
var b = new object();
a[0]=b;
I would like to use this code to save many objects in the array of function1 and call function2 to use the object in the array.
- How can I save an object in an array?
- How can I put an object in an array and save it to a variable?
javascript arrays object
javascript arrays object
edited Feb 18 '18 at 14:29
Hassan Imam
11.7k31330
11.7k31330
asked Jun 6 '11 at 15:05
nasernaser
1,40231112
1,40231112
10
Please don't correct errors in posted code, as edit 5 did. If you correct a simple error in the question, often there is no reason for the answer to exist.
– Paul
Aug 6 '15 at 3:01
Also,push
has been suggested multiple times here already. Please, do not pollute this thread anymore with anotherpush
suggestion.
– Boghyon Hoffmann
Apr 14 '18 at 14:17
add a comment |
10
Please don't correct errors in posted code, as edit 5 did. If you correct a simple error in the question, often there is no reason for the answer to exist.
– Paul
Aug 6 '15 at 3:01
Also,push
has been suggested multiple times here already. Please, do not pollute this thread anymore with anotherpush
suggestion.
– Boghyon Hoffmann
Apr 14 '18 at 14:17
10
10
Please don't correct errors in posted code, as edit 5 did. If you correct a simple error in the question, often there is no reason for the answer to exist.
– Paul
Aug 6 '15 at 3:01
Please don't correct errors in posted code, as edit 5 did. If you correct a simple error in the question, often there is no reason for the answer to exist.
– Paul
Aug 6 '15 at 3:01
Also,
push
has been suggested multiple times here already. Please, do not pollute this thread anymore with another push
suggestion.– Boghyon Hoffmann
Apr 14 '18 at 14:17
Also,
push
has been suggested multiple times here already. Please, do not pollute this thread anymore with another push
suggestion.– Boghyon Hoffmann
Apr 14 '18 at 14:17
add a comment |
13 Answers
13
active
oldest
votes
Put anything into an array using Array.push().
var a=, b=;
a.push(b);
// a[0] === b;
Extra information on Arrays
Add more than one item at a time
var x = ['a'];
x.push('b', 'c');
// x = ['a', 'b', 'c']
Add items to the beginning of an array
var x = ['c', 'd'];
x.unshift('a', 'b');
// x = ['a', 'b', 'c', 'd']
Add the contents of one array to another
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
x.push.apply(x, y);
// x = ['a', 'b', 'c', 'd', 'e', 'f']
// y = ['d', 'e', 'f'] (remains unchanged)
Create a new array from the contents of two arrays
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
var z = x.concat(y);
// x = ['a', 'b', 'c'] (remains unchanged)
// y = ['d', 'e', 'f'] (remains unchanged)
// z = ['a', 'b', 'c', 'd', 'e', 'f']
I also have a question: myArray = ; myArray.push('text': 'some text', 'id' : 13); and now myArray is empty. So if we try get the value from myArray[0]['text'] it will be empty, why? take.ms/jSvbn
– fdrv
Mar 16 '16 at 14:55
Something doesn't seem right. Make sure there isn't a scoping problem. Use let/var to declare myArray. Because myArray.push will always mutate myArray.
– John Strickler
Mar 16 '16 at 19:14
add a comment |
First of all, there is no object
or array
. There are Object
and Array
. Secondly, you can do that:
a = new Array();
b = new Object();
a[0] = b;
Now a
will be an array with b
as its only element.
3
+1 for the least overly-complicated answer. I've expanded it below to include an answer to OP's question 2.
– Sam
Jun 6 '11 at 15:34
add a comment |
var years = ;
for (i= 2015;i<=2030;i=i+1)
years.push(operator : i)
here array years is having values like
years[0]=operator:2015
years[1]=operator:2016
it continues like this.
add a comment |
- JavaScript is case-sensitive. Calling
new array()
andnew object()
will throw aReferenceError
since they don't exist. - It's better to avoid
new Array()
due to its error-prone behavior.
Instead, assign the new array with= [val1, val2, val_n]
. For objects, use=
. - There are many ways when it comes to extending an array (as shown in John's answer) but the safest way would be just to use
concat
instead ofpush
.concat
returns a new array, leaving the original array untouched.push
mutates the calling array which should be avoided, especially if the array is globally defined. - It's also a good practice to freeze the object as well as the new array in order to avoid unintended mutations. A frozen object is neither mutable nor extensible (shallowly).
Applying those points and to answer your two questions, you could define a function like this:
function appendObjTo(thatArray, newObj)
const frozenObj = Object.freeze(newObj);
return Object.freeze(thatArray.concat(frozenObj));
Usage:
// Given
const myArray = ["A", "B"];
// "save it to a variable"
const newArray = appendObjTo(myArray, hello: "world!");
// returns: ["A", "B", hello: "world!"]. myArray did not change.
"Usage" is what made this my personal preferred answer.
– HPWD
Oct 3 '17 at 20:59
@boghyon, I'm curious about the freeze and mutations. Are you talking about changes to the array that could happen directly because of the concat method, or because of changes that might be made to the array from elsewhere in the script while in the process of performing a '.concat()'? If the second, shouldn't this be automatically prevented due to javascript being single-threaded, even in the case of asynchronous code since control should, at least theoretically, not be returned until the method is completed?
– jdmayfield
Feb 1 '18 at 19:30
@jdmayfield: JS's evaluation strategy is call by object-sharing. When objects are passed around, they're not copied. The mutations in those objects, while being passed around, are always immediately visible to the callers regardless whether the mutator was called asynchronously, completed, or not. Since such side effects often led to bugs, we're returning new object, decoupled from the original one. Making the objects immutable (e.g. viafreeze
) is just an additional step to prevent unintended mutations.
– Boghyon Hoffmann
Feb 1 '18 at 22:36
@jdmayfield: But as you might have guessed already, creating new objects every time might become inefficient quickly when the objects get huge. For such cases, there are helpful libs, which make it efficient by leveraging persistent data structures, such as Mori (kinda dead) and Immutable.js. Here is a short introduction to "why immutability": youtu.be/e-5obm1G_FY
– Boghyon Hoffmann
Feb 1 '18 at 22:45
add a comment |
Using ES6
notation, you can do something like this:
For appending you can use the spread operator like this:
var arr1 = [1,2,3]
var obj = 4
var newData = [...arr1, obj] // [1,2,3,4]
console.log(newData);
add a comment |
Expanding Gabi Purcaru's answer to include an answer to number 2.
a = new Array();
b = new Object();
a[0] = b;
var c = a[0]; // c is now the object we inserted into a...
add a comment |
obejct
is clearly a typo. But both object
and array
need capital letters.
You can use short hands for new Array
and new Object
these are and
You can push data into the array using .push
. This adds it to the end of the array. or you can set an index to contain the data.
function saveToArray()
var o = ;
o.foo = 42;
var arr = ;
arr.push(o);
return arr;
function other()
var arr = saveToArray();
alert(arr[0]);
other();
Educational post with example, good
– GoldBishop
Mar 29 '17 at 13:27
add a comment |
/* array literal */
var aData = ;
/* object constructur */
function Data(firstname, lastname)
this.firstname = firstname;
this.lastname = lastname;
this.fullname = function()
return (this.firstname + " " + this.lastname);
;
/* store object into array */
aData.push(new Data("Jhon", "Doe"));
aData.push(new Data("Anna", "Smith"));
aData.push(new Data("Black", "Pearl"));
/* convert array of object into string json */
var jsonString = JSON.stringify(aData);
document.write(jsonString);
/* loop arrray */
for (var x in aData)
alert(aData[x].fullname());
And yet another answer withpush
..
– Boghyon Hoffmann
Apr 17 '18 at 12:50
add a comment |
a=;
a.push(['b','c','d','e','f']);
4
Hi, welcome to SO. Please don't just dump code as an answer. Explain your thoughts so users can better understand what's going on. Thanks.
– Cthulhu
Mar 17 '16 at 9:50
2
ok got it. I will take care of this in future. thanks for advice...
– harjinder singh
Mar 17 '16 at 11:13
2
you can still edit and explain.
– Santosh
Apr 26 '17 at 8:34
add a comment |
The way I made object creator with auto list:
var list = ;
function saveToArray(x)
list.push(x);
;
function newObject ()
saveToArray(this);
;
add a comment |
On alternativ answer is this.
if you have and array like this: var contacts = [bob, mary];
and you want to put another array in this array, you can do that in this way:
Declare the function constructor
function add (firstName,lastName,email,phoneNumber)
this.firstName = firstName;
this.lastName = lastName;
this.email = email;
this.phoneNumber = phoneNumber;
make the object from the function:
var add1 = new add("Alba","Fas","Des@gmail.com","[098] 654365364");
and add the object in to the array:
contacts[contacts.length] = add1;
add a comment |
You are running into a scope problem if you use your code as such. You have to declare it outside the functions if you plan to use it between them (or if calling, pass it as a parameter).
var a = new Array();
var b = new Object();
function first()
a.push(b);
// Alternatively, a[a.length] = b
// both methods work fine
function second()
var c = a[0];
// code
first();
// more code
second();
// even more code
add a comment |
You can use this prototype of javascript like this:
Array.prototype.push.apply(array1, array2,array3);
=> array1 contains the rest of them
add a comment |
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13 Answers
13
active
oldest
votes
13 Answers
13
active
oldest
votes
active
oldest
votes
active
oldest
votes
Put anything into an array using Array.push().
var a=, b=;
a.push(b);
// a[0] === b;
Extra information on Arrays
Add more than one item at a time
var x = ['a'];
x.push('b', 'c');
// x = ['a', 'b', 'c']
Add items to the beginning of an array
var x = ['c', 'd'];
x.unshift('a', 'b');
// x = ['a', 'b', 'c', 'd']
Add the contents of one array to another
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
x.push.apply(x, y);
// x = ['a', 'b', 'c', 'd', 'e', 'f']
// y = ['d', 'e', 'f'] (remains unchanged)
Create a new array from the contents of two arrays
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
var z = x.concat(y);
// x = ['a', 'b', 'c'] (remains unchanged)
// y = ['d', 'e', 'f'] (remains unchanged)
// z = ['a', 'b', 'c', 'd', 'e', 'f']
I also have a question: myArray = ; myArray.push('text': 'some text', 'id' : 13); and now myArray is empty. So if we try get the value from myArray[0]['text'] it will be empty, why? take.ms/jSvbn
– fdrv
Mar 16 '16 at 14:55
Something doesn't seem right. Make sure there isn't a scoping problem. Use let/var to declare myArray. Because myArray.push will always mutate myArray.
– John Strickler
Mar 16 '16 at 19:14
add a comment |
Put anything into an array using Array.push().
var a=, b=;
a.push(b);
// a[0] === b;
Extra information on Arrays
Add more than one item at a time
var x = ['a'];
x.push('b', 'c');
// x = ['a', 'b', 'c']
Add items to the beginning of an array
var x = ['c', 'd'];
x.unshift('a', 'b');
// x = ['a', 'b', 'c', 'd']
Add the contents of one array to another
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
x.push.apply(x, y);
// x = ['a', 'b', 'c', 'd', 'e', 'f']
// y = ['d', 'e', 'f'] (remains unchanged)
Create a new array from the contents of two arrays
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
var z = x.concat(y);
// x = ['a', 'b', 'c'] (remains unchanged)
// y = ['d', 'e', 'f'] (remains unchanged)
// z = ['a', 'b', 'c', 'd', 'e', 'f']
I also have a question: myArray = ; myArray.push('text': 'some text', 'id' : 13); and now myArray is empty. So if we try get the value from myArray[0]['text'] it will be empty, why? take.ms/jSvbn
– fdrv
Mar 16 '16 at 14:55
Something doesn't seem right. Make sure there isn't a scoping problem. Use let/var to declare myArray. Because myArray.push will always mutate myArray.
– John Strickler
Mar 16 '16 at 19:14
add a comment |
Put anything into an array using Array.push().
var a=, b=;
a.push(b);
// a[0] === b;
Extra information on Arrays
Add more than one item at a time
var x = ['a'];
x.push('b', 'c');
// x = ['a', 'b', 'c']
Add items to the beginning of an array
var x = ['c', 'd'];
x.unshift('a', 'b');
// x = ['a', 'b', 'c', 'd']
Add the contents of one array to another
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
x.push.apply(x, y);
// x = ['a', 'b', 'c', 'd', 'e', 'f']
// y = ['d', 'e', 'f'] (remains unchanged)
Create a new array from the contents of two arrays
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
var z = x.concat(y);
// x = ['a', 'b', 'c'] (remains unchanged)
// y = ['d', 'e', 'f'] (remains unchanged)
// z = ['a', 'b', 'c', 'd', 'e', 'f']
Put anything into an array using Array.push().
var a=, b=;
a.push(b);
// a[0] === b;
Extra information on Arrays
Add more than one item at a time
var x = ['a'];
x.push('b', 'c');
// x = ['a', 'b', 'c']
Add items to the beginning of an array
var x = ['c', 'd'];
x.unshift('a', 'b');
// x = ['a', 'b', 'c', 'd']
Add the contents of one array to another
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
x.push.apply(x, y);
// x = ['a', 'b', 'c', 'd', 'e', 'f']
// y = ['d', 'e', 'f'] (remains unchanged)
Create a new array from the contents of two arrays
var x = ['a', 'b', 'c'];
var y = ['d', 'e', 'f'];
var z = x.concat(y);
// x = ['a', 'b', 'c'] (remains unchanged)
// y = ['d', 'e', 'f'] (remains unchanged)
// z = ['a', 'b', 'c', 'd', 'e', 'f']
edited Mar 27 '14 at 15:17
answered Jun 6 '11 at 15:09
John StricklerJohn Strickler
20.6k44364
20.6k44364
I also have a question: myArray = ; myArray.push('text': 'some text', 'id' : 13); and now myArray is empty. So if we try get the value from myArray[0]['text'] it will be empty, why? take.ms/jSvbn
– fdrv
Mar 16 '16 at 14:55
Something doesn't seem right. Make sure there isn't a scoping problem. Use let/var to declare myArray. Because myArray.push will always mutate myArray.
– John Strickler
Mar 16 '16 at 19:14
add a comment |
I also have a question: myArray = ; myArray.push('text': 'some text', 'id' : 13); and now myArray is empty. So if we try get the value from myArray[0]['text'] it will be empty, why? take.ms/jSvbn
– fdrv
Mar 16 '16 at 14:55
Something doesn't seem right. Make sure there isn't a scoping problem. Use let/var to declare myArray. Because myArray.push will always mutate myArray.
– John Strickler
Mar 16 '16 at 19:14
I also have a question: myArray = ; myArray.push('text': 'some text', 'id' : 13); and now myArray is empty. So if we try get the value from myArray[0]['text'] it will be empty, why? take.ms/jSvbn
– fdrv
Mar 16 '16 at 14:55
I also have a question: myArray = ; myArray.push('text': 'some text', 'id' : 13); and now myArray is empty. So if we try get the value from myArray[0]['text'] it will be empty, why? take.ms/jSvbn
– fdrv
Mar 16 '16 at 14:55
Something doesn't seem right. Make sure there isn't a scoping problem. Use let/var to declare myArray. Because myArray.push will always mutate myArray.
– John Strickler
Mar 16 '16 at 19:14
Something doesn't seem right. Make sure there isn't a scoping problem. Use let/var to declare myArray. Because myArray.push will always mutate myArray.
– John Strickler
Mar 16 '16 at 19:14
add a comment |
First of all, there is no object
or array
. There are Object
and Array
. Secondly, you can do that:
a = new Array();
b = new Object();
a[0] = b;
Now a
will be an array with b
as its only element.
3
+1 for the least overly-complicated answer. I've expanded it below to include an answer to OP's question 2.
– Sam
Jun 6 '11 at 15:34
add a comment |
First of all, there is no object
or array
. There are Object
and Array
. Secondly, you can do that:
a = new Array();
b = new Object();
a[0] = b;
Now a
will be an array with b
as its only element.
3
+1 for the least overly-complicated answer. I've expanded it below to include an answer to OP's question 2.
– Sam
Jun 6 '11 at 15:34
add a comment |
First of all, there is no object
or array
. There are Object
and Array
. Secondly, you can do that:
a = new Array();
b = new Object();
a[0] = b;
Now a
will be an array with b
as its only element.
First of all, there is no object
or array
. There are Object
and Array
. Secondly, you can do that:
a = new Array();
b = new Object();
a[0] = b;
Now a
will be an array with b
as its only element.
edited Jun 22 '15 at 17:15
user3734304
answered Jun 6 '11 at 15:08
Gabi PurcaruGabi Purcaru
24k75982
24k75982
3
+1 for the least overly-complicated answer. I've expanded it below to include an answer to OP's question 2.
– Sam
Jun 6 '11 at 15:34
add a comment |
3
+1 for the least overly-complicated answer. I've expanded it below to include an answer to OP's question 2.
– Sam
Jun 6 '11 at 15:34
3
3
+1 for the least overly-complicated answer. I've expanded it below to include an answer to OP's question 2.
– Sam
Jun 6 '11 at 15:34
+1 for the least overly-complicated answer. I've expanded it below to include an answer to OP's question 2.
– Sam
Jun 6 '11 at 15:34
add a comment |
var years = ;
for (i= 2015;i<=2030;i=i+1)
years.push(operator : i)
here array years is having values like
years[0]=operator:2015
years[1]=operator:2016
it continues like this.
add a comment |
var years = ;
for (i= 2015;i<=2030;i=i+1)
years.push(operator : i)
here array years is having values like
years[0]=operator:2015
years[1]=operator:2016
it continues like this.
add a comment |
var years = ;
for (i= 2015;i<=2030;i=i+1)
years.push(operator : i)
here array years is having values like
years[0]=operator:2015
years[1]=operator:2016
it continues like this.
var years = ;
for (i= 2015;i<=2030;i=i+1)
years.push(operator : i)
here array years is having values like
years[0]=operator:2015
years[1]=operator:2016
it continues like this.
edited Oct 11 '16 at 4:43
answered Nov 19 '15 at 19:06
santhoshsanthosh
1,1051628
1,1051628
add a comment |
add a comment |
- JavaScript is case-sensitive. Calling
new array()
andnew object()
will throw aReferenceError
since they don't exist. - It's better to avoid
new Array()
due to its error-prone behavior.
Instead, assign the new array with= [val1, val2, val_n]
. For objects, use=
. - There are many ways when it comes to extending an array (as shown in John's answer) but the safest way would be just to use
concat
instead ofpush
.concat
returns a new array, leaving the original array untouched.push
mutates the calling array which should be avoided, especially if the array is globally defined. - It's also a good practice to freeze the object as well as the new array in order to avoid unintended mutations. A frozen object is neither mutable nor extensible (shallowly).
Applying those points and to answer your two questions, you could define a function like this:
function appendObjTo(thatArray, newObj)
const frozenObj = Object.freeze(newObj);
return Object.freeze(thatArray.concat(frozenObj));
Usage:
// Given
const myArray = ["A", "B"];
// "save it to a variable"
const newArray = appendObjTo(myArray, hello: "world!");
// returns: ["A", "B", hello: "world!"]. myArray did not change.
"Usage" is what made this my personal preferred answer.
– HPWD
Oct 3 '17 at 20:59
@boghyon, I'm curious about the freeze and mutations. Are you talking about changes to the array that could happen directly because of the concat method, or because of changes that might be made to the array from elsewhere in the script while in the process of performing a '.concat()'? If the second, shouldn't this be automatically prevented due to javascript being single-threaded, even in the case of asynchronous code since control should, at least theoretically, not be returned until the method is completed?
– jdmayfield
Feb 1 '18 at 19:30
@jdmayfield: JS's evaluation strategy is call by object-sharing. When objects are passed around, they're not copied. The mutations in those objects, while being passed around, are always immediately visible to the callers regardless whether the mutator was called asynchronously, completed, or not. Since such side effects often led to bugs, we're returning new object, decoupled from the original one. Making the objects immutable (e.g. viafreeze
) is just an additional step to prevent unintended mutations.
– Boghyon Hoffmann
Feb 1 '18 at 22:36
@jdmayfield: But as you might have guessed already, creating new objects every time might become inefficient quickly when the objects get huge. For such cases, there are helpful libs, which make it efficient by leveraging persistent data structures, such as Mori (kinda dead) and Immutable.js. Here is a short introduction to "why immutability": youtu.be/e-5obm1G_FY
– Boghyon Hoffmann
Feb 1 '18 at 22:45
add a comment |
- JavaScript is case-sensitive. Calling
new array()
andnew object()
will throw aReferenceError
since they don't exist. - It's better to avoid
new Array()
due to its error-prone behavior.
Instead, assign the new array with= [val1, val2, val_n]
. For objects, use=
. - There are many ways when it comes to extending an array (as shown in John's answer) but the safest way would be just to use
concat
instead ofpush
.concat
returns a new array, leaving the original array untouched.push
mutates the calling array which should be avoided, especially if the array is globally defined. - It's also a good practice to freeze the object as well as the new array in order to avoid unintended mutations. A frozen object is neither mutable nor extensible (shallowly).
Applying those points and to answer your two questions, you could define a function like this:
function appendObjTo(thatArray, newObj)
const frozenObj = Object.freeze(newObj);
return Object.freeze(thatArray.concat(frozenObj));
Usage:
// Given
const myArray = ["A", "B"];
// "save it to a variable"
const newArray = appendObjTo(myArray, hello: "world!");
// returns: ["A", "B", hello: "world!"]. myArray did not change.
"Usage" is what made this my personal preferred answer.
– HPWD
Oct 3 '17 at 20:59
@boghyon, I'm curious about the freeze and mutations. Are you talking about changes to the array that could happen directly because of the concat method, or because of changes that might be made to the array from elsewhere in the script while in the process of performing a '.concat()'? If the second, shouldn't this be automatically prevented due to javascript being single-threaded, even in the case of asynchronous code since control should, at least theoretically, not be returned until the method is completed?
– jdmayfield
Feb 1 '18 at 19:30
@jdmayfield: JS's evaluation strategy is call by object-sharing. When objects are passed around, they're not copied. The mutations in those objects, while being passed around, are always immediately visible to the callers regardless whether the mutator was called asynchronously, completed, or not. Since such side effects often led to bugs, we're returning new object, decoupled from the original one. Making the objects immutable (e.g. viafreeze
) is just an additional step to prevent unintended mutations.
– Boghyon Hoffmann
Feb 1 '18 at 22:36
@jdmayfield: But as you might have guessed already, creating new objects every time might become inefficient quickly when the objects get huge. For such cases, there are helpful libs, which make it efficient by leveraging persistent data structures, such as Mori (kinda dead) and Immutable.js. Here is a short introduction to "why immutability": youtu.be/e-5obm1G_FY
– Boghyon Hoffmann
Feb 1 '18 at 22:45
add a comment |
- JavaScript is case-sensitive. Calling
new array()
andnew object()
will throw aReferenceError
since they don't exist. - It's better to avoid
new Array()
due to its error-prone behavior.
Instead, assign the new array with= [val1, val2, val_n]
. For objects, use=
. - There are many ways when it comes to extending an array (as shown in John's answer) but the safest way would be just to use
concat
instead ofpush
.concat
returns a new array, leaving the original array untouched.push
mutates the calling array which should be avoided, especially if the array is globally defined. - It's also a good practice to freeze the object as well as the new array in order to avoid unintended mutations. A frozen object is neither mutable nor extensible (shallowly).
Applying those points and to answer your two questions, you could define a function like this:
function appendObjTo(thatArray, newObj)
const frozenObj = Object.freeze(newObj);
return Object.freeze(thatArray.concat(frozenObj));
Usage:
// Given
const myArray = ["A", "B"];
// "save it to a variable"
const newArray = appendObjTo(myArray, hello: "world!");
// returns: ["A", "B", hello: "world!"]. myArray did not change.
- JavaScript is case-sensitive. Calling
new array()
andnew object()
will throw aReferenceError
since they don't exist. - It's better to avoid
new Array()
due to its error-prone behavior.
Instead, assign the new array with= [val1, val2, val_n]
. For objects, use=
. - There are many ways when it comes to extending an array (as shown in John's answer) but the safest way would be just to use
concat
instead ofpush
.concat
returns a new array, leaving the original array untouched.push
mutates the calling array which should be avoided, especially if the array is globally defined. - It's also a good practice to freeze the object as well as the new array in order to avoid unintended mutations. A frozen object is neither mutable nor extensible (shallowly).
Applying those points and to answer your two questions, you could define a function like this:
function appendObjTo(thatArray, newObj)
const frozenObj = Object.freeze(newObj);
return Object.freeze(thatArray.concat(frozenObj));
Usage:
// Given
const myArray = ["A", "B"];
// "save it to a variable"
const newArray = appendObjTo(myArray, hello: "world!");
// returns: ["A", "B", hello: "world!"]. myArray did not change.
edited Jan 31 at 9:47
answered Feb 23 '17 at 23:23
Boghyon HoffmannBoghyon Hoffmann
5,98252656
5,98252656
"Usage" is what made this my personal preferred answer.
– HPWD
Oct 3 '17 at 20:59
@boghyon, I'm curious about the freeze and mutations. Are you talking about changes to the array that could happen directly because of the concat method, or because of changes that might be made to the array from elsewhere in the script while in the process of performing a '.concat()'? If the second, shouldn't this be automatically prevented due to javascript being single-threaded, even in the case of asynchronous code since control should, at least theoretically, not be returned until the method is completed?
– jdmayfield
Feb 1 '18 at 19:30
@jdmayfield: JS's evaluation strategy is call by object-sharing. When objects are passed around, they're not copied. The mutations in those objects, while being passed around, are always immediately visible to the callers regardless whether the mutator was called asynchronously, completed, or not. Since such side effects often led to bugs, we're returning new object, decoupled from the original one. Making the objects immutable (e.g. viafreeze
) is just an additional step to prevent unintended mutations.
– Boghyon Hoffmann
Feb 1 '18 at 22:36
@jdmayfield: But as you might have guessed already, creating new objects every time might become inefficient quickly when the objects get huge. For such cases, there are helpful libs, which make it efficient by leveraging persistent data structures, such as Mori (kinda dead) and Immutable.js. Here is a short introduction to "why immutability": youtu.be/e-5obm1G_FY
– Boghyon Hoffmann
Feb 1 '18 at 22:45
add a comment |
"Usage" is what made this my personal preferred answer.
– HPWD
Oct 3 '17 at 20:59
@boghyon, I'm curious about the freeze and mutations. Are you talking about changes to the array that could happen directly because of the concat method, or because of changes that might be made to the array from elsewhere in the script while in the process of performing a '.concat()'? If the second, shouldn't this be automatically prevented due to javascript being single-threaded, even in the case of asynchronous code since control should, at least theoretically, not be returned until the method is completed?
– jdmayfield
Feb 1 '18 at 19:30
@jdmayfield: JS's evaluation strategy is call by object-sharing. When objects are passed around, they're not copied. The mutations in those objects, while being passed around, are always immediately visible to the callers regardless whether the mutator was called asynchronously, completed, or not. Since such side effects often led to bugs, we're returning new object, decoupled from the original one. Making the objects immutable (e.g. viafreeze
) is just an additional step to prevent unintended mutations.
– Boghyon Hoffmann
Feb 1 '18 at 22:36
@jdmayfield: But as you might have guessed already, creating new objects every time might become inefficient quickly when the objects get huge. For such cases, there are helpful libs, which make it efficient by leveraging persistent data structures, such as Mori (kinda dead) and Immutable.js. Here is a short introduction to "why immutability": youtu.be/e-5obm1G_FY
– Boghyon Hoffmann
Feb 1 '18 at 22:45
"Usage" is what made this my personal preferred answer.
– HPWD
Oct 3 '17 at 20:59
"Usage" is what made this my personal preferred answer.
– HPWD
Oct 3 '17 at 20:59
@boghyon, I'm curious about the freeze and mutations. Are you talking about changes to the array that could happen directly because of the concat method, or because of changes that might be made to the array from elsewhere in the script while in the process of performing a '.concat()'? If the second, shouldn't this be automatically prevented due to javascript being single-threaded, even in the case of asynchronous code since control should, at least theoretically, not be returned until the method is completed?
– jdmayfield
Feb 1 '18 at 19:30
@boghyon, I'm curious about the freeze and mutations. Are you talking about changes to the array that could happen directly because of the concat method, or because of changes that might be made to the array from elsewhere in the script while in the process of performing a '.concat()'? If the second, shouldn't this be automatically prevented due to javascript being single-threaded, even in the case of asynchronous code since control should, at least theoretically, not be returned until the method is completed?
– jdmayfield
Feb 1 '18 at 19:30
@jdmayfield: JS's evaluation strategy is call by object-sharing. When objects are passed around, they're not copied. The mutations in those objects, while being passed around, are always immediately visible to the callers regardless whether the mutator was called asynchronously, completed, or not. Since such side effects often led to bugs, we're returning new object, decoupled from the original one. Making the objects immutable (e.g. via
freeze
) is just an additional step to prevent unintended mutations.– Boghyon Hoffmann
Feb 1 '18 at 22:36
@jdmayfield: JS's evaluation strategy is call by object-sharing. When objects are passed around, they're not copied. The mutations in those objects, while being passed around, are always immediately visible to the callers regardless whether the mutator was called asynchronously, completed, or not. Since such side effects often led to bugs, we're returning new object, decoupled from the original one. Making the objects immutable (e.g. via
freeze
) is just an additional step to prevent unintended mutations.– Boghyon Hoffmann
Feb 1 '18 at 22:36
@jdmayfield: But as you might have guessed already, creating new objects every time might become inefficient quickly when the objects get huge. For such cases, there are helpful libs, which make it efficient by leveraging persistent data structures, such as Mori (kinda dead) and Immutable.js. Here is a short introduction to "why immutability": youtu.be/e-5obm1G_FY
– Boghyon Hoffmann
Feb 1 '18 at 22:45
@jdmayfield: But as you might have guessed already, creating new objects every time might become inefficient quickly when the objects get huge. For such cases, there are helpful libs, which make it efficient by leveraging persistent data structures, such as Mori (kinda dead) and Immutable.js. Here is a short introduction to "why immutability": youtu.be/e-5obm1G_FY
– Boghyon Hoffmann
Feb 1 '18 at 22:45
add a comment |
Using ES6
notation, you can do something like this:
For appending you can use the spread operator like this:
var arr1 = [1,2,3]
var obj = 4
var newData = [...arr1, obj] // [1,2,3,4]
console.log(newData);
add a comment |
Using ES6
notation, you can do something like this:
For appending you can use the spread operator like this:
var arr1 = [1,2,3]
var obj = 4
var newData = [...arr1, obj] // [1,2,3,4]
console.log(newData);
add a comment |
Using ES6
notation, you can do something like this:
For appending you can use the spread operator like this:
var arr1 = [1,2,3]
var obj = 4
var newData = [...arr1, obj] // [1,2,3,4]
console.log(newData);
Using ES6
notation, you can do something like this:
For appending you can use the spread operator like this:
var arr1 = [1,2,3]
var obj = 4
var newData = [...arr1, obj] // [1,2,3,4]
console.log(newData);
var arr1 = [1,2,3]
var obj = 4
var newData = [...arr1, obj] // [1,2,3,4]
console.log(newData);
var arr1 = [1,2,3]
var obj = 4
var newData = [...arr1, obj] // [1,2,3,4]
console.log(newData);
edited Jan 31 at 7:40
mitesh7172
165110
165110
answered Jun 19 '18 at 17:26
AayushiAayushi
635618
635618
add a comment |
add a comment |
Expanding Gabi Purcaru's answer to include an answer to number 2.
a = new Array();
b = new Object();
a[0] = b;
var c = a[0]; // c is now the object we inserted into a...
add a comment |
Expanding Gabi Purcaru's answer to include an answer to number 2.
a = new Array();
b = new Object();
a[0] = b;
var c = a[0]; // c is now the object we inserted into a...
add a comment |
Expanding Gabi Purcaru's answer to include an answer to number 2.
a = new Array();
b = new Object();
a[0] = b;
var c = a[0]; // c is now the object we inserted into a...
Expanding Gabi Purcaru's answer to include an answer to number 2.
a = new Array();
b = new Object();
a[0] = b;
var c = a[0]; // c is now the object we inserted into a...
answered Jun 6 '11 at 15:33
SamSam
2,279103452
2,279103452
add a comment |
add a comment |
obejct
is clearly a typo. But both object
and array
need capital letters.
You can use short hands for new Array
and new Object
these are and
You can push data into the array using .push
. This adds it to the end of the array. or you can set an index to contain the data.
function saveToArray()
var o = ;
o.foo = 42;
var arr = ;
arr.push(o);
return arr;
function other()
var arr = saveToArray();
alert(arr[0]);
other();
Educational post with example, good
– GoldBishop
Mar 29 '17 at 13:27
add a comment |
obejct
is clearly a typo. But both object
and array
need capital letters.
You can use short hands for new Array
and new Object
these are and
You can push data into the array using .push
. This adds it to the end of the array. or you can set an index to contain the data.
function saveToArray()
var o = ;
o.foo = 42;
var arr = ;
arr.push(o);
return arr;
function other()
var arr = saveToArray();
alert(arr[0]);
other();
Educational post with example, good
– GoldBishop
Mar 29 '17 at 13:27
add a comment |
obejct
is clearly a typo. But both object
and array
need capital letters.
You can use short hands for new Array
and new Object
these are and
You can push data into the array using .push
. This adds it to the end of the array. or you can set an index to contain the data.
function saveToArray()
var o = ;
o.foo = 42;
var arr = ;
arr.push(o);
return arr;
function other()
var arr = saveToArray();
alert(arr[0]);
other();
obejct
is clearly a typo. But both object
and array
need capital letters.
You can use short hands for new Array
and new Object
these are and
You can push data into the array using .push
. This adds it to the end of the array. or you can set an index to contain the data.
function saveToArray()
var o = ;
o.foo = 42;
var arr = ;
arr.push(o);
return arr;
function other()
var arr = saveToArray();
alert(arr[0]);
other();
answered Jun 6 '11 at 15:10
RaynosRaynos
133k41314372
133k41314372
Educational post with example, good
– GoldBishop
Mar 29 '17 at 13:27
add a comment |
Educational post with example, good
– GoldBishop
Mar 29 '17 at 13:27
Educational post with example, good
– GoldBishop
Mar 29 '17 at 13:27
Educational post with example, good
– GoldBishop
Mar 29 '17 at 13:27
add a comment |
/* array literal */
var aData = ;
/* object constructur */
function Data(firstname, lastname)
this.firstname = firstname;
this.lastname = lastname;
this.fullname = function()
return (this.firstname + " " + this.lastname);
;
/* store object into array */
aData.push(new Data("Jhon", "Doe"));
aData.push(new Data("Anna", "Smith"));
aData.push(new Data("Black", "Pearl"));
/* convert array of object into string json */
var jsonString = JSON.stringify(aData);
document.write(jsonString);
/* loop arrray */
for (var x in aData)
alert(aData[x].fullname());
And yet another answer withpush
..
– Boghyon Hoffmann
Apr 17 '18 at 12:50
add a comment |
/* array literal */
var aData = ;
/* object constructur */
function Data(firstname, lastname)
this.firstname = firstname;
this.lastname = lastname;
this.fullname = function()
return (this.firstname + " " + this.lastname);
;
/* store object into array */
aData.push(new Data("Jhon", "Doe"));
aData.push(new Data("Anna", "Smith"));
aData.push(new Data("Black", "Pearl"));
/* convert array of object into string json */
var jsonString = JSON.stringify(aData);
document.write(jsonString);
/* loop arrray */
for (var x in aData)
alert(aData[x].fullname());
And yet another answer withpush
..
– Boghyon Hoffmann
Apr 17 '18 at 12:50
add a comment |
/* array literal */
var aData = ;
/* object constructur */
function Data(firstname, lastname)
this.firstname = firstname;
this.lastname = lastname;
this.fullname = function()
return (this.firstname + " " + this.lastname);
;
/* store object into array */
aData.push(new Data("Jhon", "Doe"));
aData.push(new Data("Anna", "Smith"));
aData.push(new Data("Black", "Pearl"));
/* convert array of object into string json */
var jsonString = JSON.stringify(aData);
document.write(jsonString);
/* loop arrray */
for (var x in aData)
alert(aData[x].fullname());
/* array literal */
var aData = ;
/* object constructur */
function Data(firstname, lastname)
this.firstname = firstname;
this.lastname = lastname;
this.fullname = function()
return (this.firstname + " " + this.lastname);
;
/* store object into array */
aData.push(new Data("Jhon", "Doe"));
aData.push(new Data("Anna", "Smith"));
aData.push(new Data("Black", "Pearl"));
/* convert array of object into string json */
var jsonString = JSON.stringify(aData);
document.write(jsonString);
/* loop arrray */
for (var x in aData)
alert(aData[x].fullname());
/* array literal */
var aData = ;
/* object constructur */
function Data(firstname, lastname)
this.firstname = firstname;
this.lastname = lastname;
this.fullname = function()
return (this.firstname + " " + this.lastname);
;
/* store object into array */
aData.push(new Data("Jhon", "Doe"));
aData.push(new Data("Anna", "Smith"));
aData.push(new Data("Black", "Pearl"));
/* convert array of object into string json */
var jsonString = JSON.stringify(aData);
document.write(jsonString);
/* loop arrray */
for (var x in aData)
alert(aData[x].fullname());
/* array literal */
var aData = ;
/* object constructur */
function Data(firstname, lastname)
this.firstname = firstname;
this.lastname = lastname;
this.fullname = function()
return (this.firstname + " " + this.lastname);
;
/* store object into array */
aData.push(new Data("Jhon", "Doe"));
aData.push(new Data("Anna", "Smith"));
aData.push(new Data("Black", "Pearl"));
/* convert array of object into string json */
var jsonString = JSON.stringify(aData);
document.write(jsonString);
/* loop arrray */
for (var x in aData)
alert(aData[x].fullname());
edited Jan 7 '18 at 17:46
answered Jan 7 '18 at 17:38
anteloveantelove
70547
70547
And yet another answer withpush
..
– Boghyon Hoffmann
Apr 17 '18 at 12:50
add a comment |
And yet another answer withpush
..
– Boghyon Hoffmann
Apr 17 '18 at 12:50
And yet another answer with
push
..– Boghyon Hoffmann
Apr 17 '18 at 12:50
And yet another answer with
push
..– Boghyon Hoffmann
Apr 17 '18 at 12:50
add a comment |
a=;
a.push(['b','c','d','e','f']);
4
Hi, welcome to SO. Please don't just dump code as an answer. Explain your thoughts so users can better understand what's going on. Thanks.
– Cthulhu
Mar 17 '16 at 9:50
2
ok got it. I will take care of this in future. thanks for advice...
– harjinder singh
Mar 17 '16 at 11:13
2
you can still edit and explain.
– Santosh
Apr 26 '17 at 8:34
add a comment |
a=;
a.push(['b','c','d','e','f']);
4
Hi, welcome to SO. Please don't just dump code as an answer. Explain your thoughts so users can better understand what's going on. Thanks.
– Cthulhu
Mar 17 '16 at 9:50
2
ok got it. I will take care of this in future. thanks for advice...
– harjinder singh
Mar 17 '16 at 11:13
2
you can still edit and explain.
– Santosh
Apr 26 '17 at 8:34
add a comment |
a=;
a.push(['b','c','d','e','f']);
a=;
a.push(['b','c','d','e','f']);
edited Mar 17 '16 at 9:59
Magicprog.fr
3,60742431
3,60742431
answered Mar 17 '16 at 9:31
harjinder singhharjinder singh
211
211
4
Hi, welcome to SO. Please don't just dump code as an answer. Explain your thoughts so users can better understand what's going on. Thanks.
– Cthulhu
Mar 17 '16 at 9:50
2
ok got it. I will take care of this in future. thanks for advice...
– harjinder singh
Mar 17 '16 at 11:13
2
you can still edit and explain.
– Santosh
Apr 26 '17 at 8:34
add a comment |
4
Hi, welcome to SO. Please don't just dump code as an answer. Explain your thoughts so users can better understand what's going on. Thanks.
– Cthulhu
Mar 17 '16 at 9:50
2
ok got it. I will take care of this in future. thanks for advice...
– harjinder singh
Mar 17 '16 at 11:13
2
you can still edit and explain.
– Santosh
Apr 26 '17 at 8:34
4
4
Hi, welcome to SO. Please don't just dump code as an answer. Explain your thoughts so users can better understand what's going on. Thanks.
– Cthulhu
Mar 17 '16 at 9:50
Hi, welcome to SO. Please don't just dump code as an answer. Explain your thoughts so users can better understand what's going on. Thanks.
– Cthulhu
Mar 17 '16 at 9:50
2
2
ok got it. I will take care of this in future. thanks for advice...
– harjinder singh
Mar 17 '16 at 11:13
ok got it. I will take care of this in future. thanks for advice...
– harjinder singh
Mar 17 '16 at 11:13
2
2
you can still edit and explain.
– Santosh
Apr 26 '17 at 8:34
you can still edit and explain.
– Santosh
Apr 26 '17 at 8:34
add a comment |
The way I made object creator with auto list:
var list = ;
function saveToArray(x)
list.push(x);
;
function newObject ()
saveToArray(this);
;
add a comment |
The way I made object creator with auto list:
var list = ;
function saveToArray(x)
list.push(x);
;
function newObject ()
saveToArray(this);
;
add a comment |
The way I made object creator with auto list:
var list = ;
function saveToArray(x)
list.push(x);
;
function newObject ()
saveToArray(this);
;
The way I made object creator with auto list:
var list = ;
function saveToArray(x)
list.push(x);
;
function newObject ()
saveToArray(this);
;
answered May 13 '16 at 19:14
Tadas StundysTadas Stundys
211
211
add a comment |
add a comment |
On alternativ answer is this.
if you have and array like this: var contacts = [bob, mary];
and you want to put another array in this array, you can do that in this way:
Declare the function constructor
function add (firstName,lastName,email,phoneNumber)
this.firstName = firstName;
this.lastName = lastName;
this.email = email;
this.phoneNumber = phoneNumber;
make the object from the function:
var add1 = new add("Alba","Fas","Des@gmail.com","[098] 654365364");
and add the object in to the array:
contacts[contacts.length] = add1;
add a comment |
On alternativ answer is this.
if you have and array like this: var contacts = [bob, mary];
and you want to put another array in this array, you can do that in this way:
Declare the function constructor
function add (firstName,lastName,email,phoneNumber)
this.firstName = firstName;
this.lastName = lastName;
this.email = email;
this.phoneNumber = phoneNumber;
make the object from the function:
var add1 = new add("Alba","Fas","Des@gmail.com","[098] 654365364");
and add the object in to the array:
contacts[contacts.length] = add1;
add a comment |
On alternativ answer is this.
if you have and array like this: var contacts = [bob, mary];
and you want to put another array in this array, you can do that in this way:
Declare the function constructor
function add (firstName,lastName,email,phoneNumber)
this.firstName = firstName;
this.lastName = lastName;
this.email = email;
this.phoneNumber = phoneNumber;
make the object from the function:
var add1 = new add("Alba","Fas","Des@gmail.com","[098] 654365364");
and add the object in to the array:
contacts[contacts.length] = add1;
On alternativ answer is this.
if you have and array like this: var contacts = [bob, mary];
and you want to put another array in this array, you can do that in this way:
Declare the function constructor
function add (firstName,lastName,email,phoneNumber)
this.firstName = firstName;
this.lastName = lastName;
this.email = email;
this.phoneNumber = phoneNumber;
make the object from the function:
var add1 = new add("Alba","Fas","Des@gmail.com","[098] 654365364");
and add the object in to the array:
contacts[contacts.length] = add1;
answered Nov 21 '16 at 13:32
Po PoPo Po
1615
1615
add a comment |
add a comment |
You are running into a scope problem if you use your code as such. You have to declare it outside the functions if you plan to use it between them (or if calling, pass it as a parameter).
var a = new Array();
var b = new Object();
function first()
a.push(b);
// Alternatively, a[a.length] = b
// both methods work fine
function second()
var c = a[0];
// code
first();
// more code
second();
// even more code
add a comment |
You are running into a scope problem if you use your code as such. You have to declare it outside the functions if you plan to use it between them (or if calling, pass it as a parameter).
var a = new Array();
var b = new Object();
function first()
a.push(b);
// Alternatively, a[a.length] = b
// both methods work fine
function second()
var c = a[0];
// code
first();
// more code
second();
// even more code
add a comment |
You are running into a scope problem if you use your code as such. You have to declare it outside the functions if you plan to use it between them (or if calling, pass it as a parameter).
var a = new Array();
var b = new Object();
function first()
a.push(b);
// Alternatively, a[a.length] = b
// both methods work fine
function second()
var c = a[0];
// code
first();
// more code
second();
// even more code
You are running into a scope problem if you use your code as such. You have to declare it outside the functions if you plan to use it between them (or if calling, pass it as a parameter).
var a = new Array();
var b = new Object();
function first()
a.push(b);
// Alternatively, a[a.length] = b
// both methods work fine
function second()
var c = a[0];
// code
first();
// more code
second();
// even more code
answered Jun 6 '11 at 15:13
josh.trowjosh.trow
4,1701628
4,1701628
add a comment |
add a comment |
You can use this prototype of javascript like this:
Array.prototype.push.apply(array1, array2,array3);
=> array1 contains the rest of them
add a comment |
You can use this prototype of javascript like this:
Array.prototype.push.apply(array1, array2,array3);
=> array1 contains the rest of them
add a comment |
You can use this prototype of javascript like this:
Array.prototype.push.apply(array1, array2,array3);
=> array1 contains the rest of them
You can use this prototype of javascript like this:
Array.prototype.push.apply(array1, array2,array3);
=> array1 contains the rest of them
edited Aug 30 '16 at 9:16
Gytis Tenovimas
8,623112647
8,623112647
answered Aug 30 '16 at 4:37
Dat NgoDat Ngo
211
211
add a comment |
add a comment |
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10
Please don't correct errors in posted code, as edit 5 did. If you correct a simple error in the question, often there is no reason for the answer to exist.
– Paul
Aug 6 '15 at 3:01
Also,
push
has been suggested multiple times here already. Please, do not pollute this thread anymore with anotherpush
suggestion.– Boghyon Hoffmann
Apr 14 '18 at 14:17