POST variable always stays empty









up vote
-1
down vote

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This is the method in which i manage the data send with jquery.ajax. By default is send an empty string and on each change i watch the change on the input and resend it. Through console things are good but in php $this->searchField always has the value of an empty string



function checkInfo()
if(isset($_POST['searchField']))
$this->searchField = json_decode($_POST['searchField']);

if ($this->searchField != "")
$this->query = "SELECT * FROM myguests WHERE firstName like '%".$searchField."%'";

else if ($this->searchField == "")
$this->query = "SELECT * FROM myguests ORDER BY id";

$this->tableDisplay();
exit();

else
return "error";




This is my jquery ajax function :



function searchGuests(users)
var data =
"searchField": users
;
console.log(data);
$.ajax(
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
)
.done(function(tr)
$("#tableBody").empty();
$("#tableBody").append(tr);
)
.fail(function(error)
console.log(error);
);
;


Any idea what I am missing?










share|improve this question





















  • what does the users variable holding?
    – Ali
    Nov 11 at 11:50






  • 1




    You aren't sending any JSON so no need to use json_decode()
    – charlietfl
    Nov 11 at 11:54














up vote
-1
down vote

favorite












This is the method in which i manage the data send with jquery.ajax. By default is send an empty string and on each change i watch the change on the input and resend it. Through console things are good but in php $this->searchField always has the value of an empty string



function checkInfo()
if(isset($_POST['searchField']))
$this->searchField = json_decode($_POST['searchField']);

if ($this->searchField != "")
$this->query = "SELECT * FROM myguests WHERE firstName like '%".$searchField."%'";

else if ($this->searchField == "")
$this->query = "SELECT * FROM myguests ORDER BY id";

$this->tableDisplay();
exit();

else
return "error";




This is my jquery ajax function :



function searchGuests(users)
var data =
"searchField": users
;
console.log(data);
$.ajax(
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
)
.done(function(tr)
$("#tableBody").empty();
$("#tableBody").append(tr);
)
.fail(function(error)
console.log(error);
);
;


Any idea what I am missing?










share|improve this question





















  • what does the users variable holding?
    – Ali
    Nov 11 at 11:50






  • 1




    You aren't sending any JSON so no need to use json_decode()
    – charlietfl
    Nov 11 at 11:54












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











This is the method in which i manage the data send with jquery.ajax. By default is send an empty string and on each change i watch the change on the input and resend it. Through console things are good but in php $this->searchField always has the value of an empty string



function checkInfo()
if(isset($_POST['searchField']))
$this->searchField = json_decode($_POST['searchField']);

if ($this->searchField != "")
$this->query = "SELECT * FROM myguests WHERE firstName like '%".$searchField."%'";

else if ($this->searchField == "")
$this->query = "SELECT * FROM myguests ORDER BY id";

$this->tableDisplay();
exit();

else
return "error";




This is my jquery ajax function :



function searchGuests(users)
var data =
"searchField": users
;
console.log(data);
$.ajax(
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
)
.done(function(tr)
$("#tableBody").empty();
$("#tableBody").append(tr);
)
.fail(function(error)
console.log(error);
);
;


Any idea what I am missing?










share|improve this question













This is the method in which i manage the data send with jquery.ajax. By default is send an empty string and on each change i watch the change on the input and resend it. Through console things are good but in php $this->searchField always has the value of an empty string



function checkInfo()
if(isset($_POST['searchField']))
$this->searchField = json_decode($_POST['searchField']);

if ($this->searchField != "")
$this->query = "SELECT * FROM myguests WHERE firstName like '%".$searchField."%'";

else if ($this->searchField == "")
$this->query = "SELECT * FROM myguests ORDER BY id";

$this->tableDisplay();
exit();

else
return "error";




This is my jquery ajax function :



function searchGuests(users)
var data =
"searchField": users
;
console.log(data);
$.ajax(
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
)
.done(function(tr)
$("#tableBody").empty();
$("#tableBody").append(tr);
)
.fail(function(error)
console.log(error);
);
;


Any idea what I am missing?







javascript php jquery sql ajax






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 11 at 11:31









juxhin bleta

598




598











  • what does the users variable holding?
    – Ali
    Nov 11 at 11:50






  • 1




    You aren't sending any JSON so no need to use json_decode()
    – charlietfl
    Nov 11 at 11:54
















  • what does the users variable holding?
    – Ali
    Nov 11 at 11:50






  • 1




    You aren't sending any JSON so no need to use json_decode()
    – charlietfl
    Nov 11 at 11:54















what does the users variable holding?
– Ali
Nov 11 at 11:50




what does the users variable holding?
– Ali
Nov 11 at 11:50




1




1




You aren't sending any JSON so no need to use json_decode()
– charlietfl
Nov 11 at 11:54




You aren't sending any JSON so no need to use json_decode()
– charlietfl
Nov 11 at 11:54












1 Answer
1






active

oldest

votes

















up vote
-1
down vote













Try encoding the users to a string before sending it over.



function searchGuests(users)
var data =
"searchField": JSON.stringify(users);
;
console.log(data);
$.ajax(
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
)
.done(function(tr)
$("#tableBody").empty();
$("#tableBody").append(tr);
)
.fail(function(error)
console.log(error);
);
;





share|improve this answer






















  • No such method as JSON.encode() and the idea doesn't make sense either. jQuery uses $.param() to serialize objects by default
    – charlietfl
    Nov 11 at 11:52











  • I solved it indeed it didn't needed to be json_decode -ed
    – juxhin bleta
    Nov 11 at 11:52










  • confused it with php json_encode. it is JSON.stringify
    – Edwin Dijas Chiwona
    Nov 11 at 12:02










  • please share your answer. Thanks
    – Edwin Dijas Chiwona
    Nov 11 at 12:07










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active

oldest

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up vote
-1
down vote













Try encoding the users to a string before sending it over.



function searchGuests(users)
var data =
"searchField": JSON.stringify(users);
;
console.log(data);
$.ajax(
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
)
.done(function(tr)
$("#tableBody").empty();
$("#tableBody").append(tr);
)
.fail(function(error)
console.log(error);
);
;





share|improve this answer






















  • No such method as JSON.encode() and the idea doesn't make sense either. jQuery uses $.param() to serialize objects by default
    – charlietfl
    Nov 11 at 11:52











  • I solved it indeed it didn't needed to be json_decode -ed
    – juxhin bleta
    Nov 11 at 11:52










  • confused it with php json_encode. it is JSON.stringify
    – Edwin Dijas Chiwona
    Nov 11 at 12:02










  • please share your answer. Thanks
    – Edwin Dijas Chiwona
    Nov 11 at 12:07














up vote
-1
down vote













Try encoding the users to a string before sending it over.



function searchGuests(users)
var data =
"searchField": JSON.stringify(users);
;
console.log(data);
$.ajax(
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
)
.done(function(tr)
$("#tableBody").empty();
$("#tableBody").append(tr);
)
.fail(function(error)
console.log(error);
);
;





share|improve this answer






















  • No such method as JSON.encode() and the idea doesn't make sense either. jQuery uses $.param() to serialize objects by default
    – charlietfl
    Nov 11 at 11:52











  • I solved it indeed it didn't needed to be json_decode -ed
    – juxhin bleta
    Nov 11 at 11:52










  • confused it with php json_encode. it is JSON.stringify
    – Edwin Dijas Chiwona
    Nov 11 at 12:02










  • please share your answer. Thanks
    – Edwin Dijas Chiwona
    Nov 11 at 12:07












up vote
-1
down vote










up vote
-1
down vote









Try encoding the users to a string before sending it over.



function searchGuests(users)
var data =
"searchField": JSON.stringify(users);
;
console.log(data);
$.ajax(
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
)
.done(function(tr)
$("#tableBody").empty();
$("#tableBody").append(tr);
)
.fail(function(error)
console.log(error);
);
;





share|improve this answer














Try encoding the users to a string before sending it over.



function searchGuests(users)
var data =
"searchField": JSON.stringify(users);
;
console.log(data);
$.ajax(
type: "POST",
dataType: "text",
url: "controllers/manageTable.php",
data: data
)
.done(function(tr)
$("#tableBody").empty();
$("#tableBody").append(tr);
)
.fail(function(error)
console.log(error);
);
;






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 11 at 12:06

























answered Nov 11 at 11:46









Edwin Dijas Chiwona

34117




34117











  • No such method as JSON.encode() and the idea doesn't make sense either. jQuery uses $.param() to serialize objects by default
    – charlietfl
    Nov 11 at 11:52











  • I solved it indeed it didn't needed to be json_decode -ed
    – juxhin bleta
    Nov 11 at 11:52










  • confused it with php json_encode. it is JSON.stringify
    – Edwin Dijas Chiwona
    Nov 11 at 12:02










  • please share your answer. Thanks
    – Edwin Dijas Chiwona
    Nov 11 at 12:07
















  • No such method as JSON.encode() and the idea doesn't make sense either. jQuery uses $.param() to serialize objects by default
    – charlietfl
    Nov 11 at 11:52











  • I solved it indeed it didn't needed to be json_decode -ed
    – juxhin bleta
    Nov 11 at 11:52










  • confused it with php json_encode. it is JSON.stringify
    – Edwin Dijas Chiwona
    Nov 11 at 12:02










  • please share your answer. Thanks
    – Edwin Dijas Chiwona
    Nov 11 at 12:07















No such method as JSON.encode() and the idea doesn't make sense either. jQuery uses $.param() to serialize objects by default
– charlietfl
Nov 11 at 11:52





No such method as JSON.encode() and the idea doesn't make sense either. jQuery uses $.param() to serialize objects by default
– charlietfl
Nov 11 at 11:52













I solved it indeed it didn't needed to be json_decode -ed
– juxhin bleta
Nov 11 at 11:52




I solved it indeed it didn't needed to be json_decode -ed
– juxhin bleta
Nov 11 at 11:52












confused it with php json_encode. it is JSON.stringify
– Edwin Dijas Chiwona
Nov 11 at 12:02




confused it with php json_encode. it is JSON.stringify
– Edwin Dijas Chiwona
Nov 11 at 12:02












please share your answer. Thanks
– Edwin Dijas Chiwona
Nov 11 at 12:07




please share your answer. Thanks
– Edwin Dijas Chiwona
Nov 11 at 12:07

















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