Prevent Random Frame Label From Being Selected More Than Once as3
The code below does just what I need it to do as far as not selecting a random frame label twice in a row. However, I need to take it a step further. Now, I also need to write this code in a way that any random frame that has been selected drops out of the array and is not selected again, period. Any help with achieving this will be really appreciated. Here is my code:
function getRandomLabel(currentLabel: String): String
var labels: Array = ["e1" , "e2", "e3", "e4", "e5", "e6", "e7", "e8" ];
var currentIndex: int = labels.indexOf(currentLabel);
if (currentIndex > -1)
labels.splice(currentIndex, 1);
var index: Number = Math.floor(Math.random() * labels.length);
return labels[index];
actionscript-3 flash
asked Nov 16 '18 at 1:32
WarrenWarren
176
add a comment |
The code below does just what I need it to do as far as not selecting a random frame label twice in a row. However, I need to take it a step further. Now, I also need to write this code in a way that any random frame that has been selected drops out of the array and is not selected again, period. Any help with achieving this will be really appreciated. Here is my code:
function getRandomLabel(currentLabel: String): String
var labels: Array = ["e1" , "e2", "e3", "e4", "e5", "e6", "e7", "e8" ];
var currentIndex: int = labels.indexOf(currentLabel);
if (currentIndex > -1)
labels.splice(currentIndex, 1);
var index: Number = Math.floor(Math.random() * labels.length);
return labels[index];
actionscript-3 flash
asked Nov 16 '18 at 1:32
WarrenWarren
176
add a comment |
The code below does just what I need it to do as far as not selecting a random frame label twice in a row. However, I need to take it a step further. Now, I also need to write this code in a way that any random frame that has been selected drops out of the array and is not selected again, period. Any help with achieving this will be really appreciated. Here is my code:
function getRandomLabel(currentLabel: String): String
var labels: Array = ["e1" , "e2", "e3", "e4", "e5", "e6", "e7", "e8" ];
var currentIndex: int = labels.indexOf(currentLabel);
if (currentIndex > -1)
labels.splice(currentIndex, 1);
var index: Number = Math.floor(Math.random() * labels.length);
return labels[index];
actionscript-3 flash
asked Nov 16 '18 at 1:32
WarrenWarren
176
The code below does just what I need it to do as far as not selecting a random frame label twice in a row. However, I need to take it a step further. Now, I also need to write this code in a way that any random frame that has been selected drops out of the array and is not selected again, period. Any help with achieving this will be really appreciated. Here is my code:
function getRandomLabel(currentLabel: String): String
var labels: Array = ["e1" , "e2", "e3", "e4", "e5", "e6", "e7", "e8" ];
var currentIndex: int = labels.indexOf(currentLabel);
if (currentIndex > -1)
labels.splice(currentIndex, 1);
var index: Number = Math.floor(Math.random() * labels.length);
return labels[index];
actionscript-3 flash
actionscript-3 flash
asked Nov 16 '18 at 1:32
WarrenWarren
176
asked Nov 16 '18 at 1:32
WarrenWarren
176
asked Nov 16 '18 at 1:32
WarrenWarren
176
asked Nov 16 '18 at 1:32
WarrenWarren
176
asked Nov 16 '18 at 1:32
WarrenWarren
176
176
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Pretty easy. What you need is to keep the labels outside the function (in your code it is a function local variable and it is repopulated each call) and to extract each one upon request so it is not on the list next time.
var labels:Array = ["e1", "e2", "e3", "e4", "e5", "e6", "e7", "e8"];
function getRandomLabel():String
var anIndex:int = Math.random() * labels.length;
var result:String = labels[anIndex];
labels.splice(anIndex, 1);
return result;
answered Nov 16 '18 at 6:08
OrganisOrganis
5,1462610
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Pretty easy. What you need is to keep the labels outside the function (in your code it is a function local variable and it is repopulated each call) and to extract each one upon request so it is not on the list next time.
var labels:Array = ["e1", "e2", "e3", "e4", "e5", "e6", "e7", "e8"];
function getRandomLabel():String
var anIndex:int = Math.random() * labels.length;
var result:String = labels[anIndex];
labels.splice(anIndex, 1);
return result;
answered Nov 16 '18 at 6:08
OrganisOrganis
5,1462610
add a comment |
Pretty easy. What you need is to keep the labels outside the function (in your code it is a function local variable and it is repopulated each call) and to extract each one upon request so it is not on the list next time.
var labels:Array = ["e1", "e2", "e3", "e4", "e5", "e6", "e7", "e8"];
function getRandomLabel():String
var anIndex:int = Math.random() * labels.length;
var result:String = labels[anIndex];
labels.splice(anIndex, 1);
return result;
answered Nov 16 '18 at 6:08
OrganisOrganis
5,1462610
add a comment |
Pretty easy. What you need is to keep the labels outside the function (in your code it is a function local variable and it is repopulated each call) and to extract each one upon request so it is not on the list next time.
var labels:Array = ["e1", "e2", "e3", "e4", "e5", "e6", "e7", "e8"];
function getRandomLabel():String
var anIndex:int = Math.random() * labels.length;
var result:String = labels[anIndex];
labels.splice(anIndex, 1);
return result;
answered Nov 16 '18 at 6:08
OrganisOrganis
5,1462610
Pretty easy. What you need is to keep the labels outside the function (in your code it is a function local variable and it is repopulated each call) and to extract each one upon request so it is not on the list next time.
var labels:Array = ["e1", "e2", "e3", "e4", "e5", "e6", "e7", "e8"];
function getRandomLabel():String
var anIndex:int = Math.random() * labels.length;
var result:String = labels[anIndex];
labels.splice(anIndex, 1);
return result;
answered Nov 16 '18 at 6:08
OrganisOrganis
5,1462610
edited Nov 16 '18 at 6:14
answered Nov 16 '18 at 6:08
OrganisOrganis
5,1462610
answered Nov 16 '18 at 6:08
OrganisOrganis
5,1462610
answered Nov 16 '18 at 6:08
OrganisOrganis
5,1462610
5,1462610
add a comment |
add a comment |
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