JQuery Variable Declaration

Question

Every thing goes fine but when i declaring Variable answeredQ+i Getting Error

var count = 5;
for(var i=0;i<count;i++){
    var className = '.myclass'+i;
    var answeredQ+i = $(className).map(function(){
        return $(this).text();
    }).get();
}

@Dev Why do you need a variable with the name `answeredQ1`, `answeredQ2`, ... ? — Weedoze, Nov 30 '16 at 07:20
@Dev you need to tell us the reason you are choosing to have dynamic variable, We might be able to help you with better approach. — Rajshekar Reddy, Nov 30 '16 at 07:33

score 1 · Accepted Answer · edited May 23 '17 at 12:13

var count = 5;
var answered = {};
for(var i=0;i<count;i++){
    var className = '.myclass'+i;
    answered['Q' + i] = $(className).map(function(){
        return $(this).text();
    }).get();
}

Now you can use answered.Q1 or answered['Q1'] to access first variable, Q2 for 2nd variable, etc.

What you're trying to achieve is known as 'Variable variable'. Many programming languages supports it, but in Javascript there is no straightforward answer, so you have to use an array or object to contain the new variables.

Read about 'variable variable' for more details: "Variable" variables in Javascript?

score 0 · Answer 2 · answered Nov 30 '16 at 07:17

0

use window['answeredQ'+i] but you are probably better assigning the answer to an array, rather than directly to the global scope.

var count = 5;
for (var i = 0; i < count; i++) {
  var className = '.myclass' + i;
  window['answeredQ' + i] = $(className).map(function() {
    return $(this).text();
  }).get();
}

answered Nov 30 '16 at 07:17

synthet1c

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This `window['answeredQ'+i]` will create the variable in global scope, right? – Aruna Nov 30 '16 at 07:20
Yes, as said it's not ideal. – synthet1c Nov 30 '16 at 07:21
how to use this __window['answeredQ'+i]__ outside for loop? – Dev Nov 30 '16 at 07:24
it's the same as any variable declared in the global scope (outside of a function) `answeredQ0` or `window.answeredQ0` or `window['answeredQ0']` – synthet1c Nov 30 '16 at 07:49

Aruna · Answer 3 · 2016-11-30T07:31:09.573

As this variable declaration is invalid, you can use an object to hold all this dynamic keys and use the same outside the for loop as below.

var count = 5;
var answers = {};

for(var i=0;i<count;i++){
    var className = '.myclass'+i;
    answers['answeredQ' + i] = $(className).map(function(){
        return $(this).text();
    }).get();
}

console.log(answers);

Object.keys(answers).forEach(function(key) {
  console.log(key + ': ');
  console.log(answers[key]);
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

JQuery Variable Declaration

3 Answers3