How to use contents of array as variables without using eval()?

Question

I have a list of variables or variable names stored in an array. I want to use them in a loop, but I don't want to have to use eval(). How do I do this? If I store the values in an array with quotes, I have to use eval() on the right side of any equation to render the value. If I store just the variable name, I thought I'd be storing the actual variable, but it's not working right.

$(data.xml).find('Question').each(function(){
  var answer_1 = $(this).find("Answers_1").text();
  var answer_2 = $(this).find("Answers_2").text();
  var answer_3 = $(this).find("Answers_3").text();
  var answer_4 = $(this).find("Answers_4").text();
  var theID = $(this).find("ID").text();
  var theBody = $(this).find("Body").text();

  var answerArray = new Array();

  answerArray = [answer_1,answer_2,answer_3,answer_4,theID,theBody]

  for(x=0;x<=5;x++) {
    testme = answerArray[x];

    alert("looking for = " + answerArray[x] + ", which is " + testme)
  }
});

I assume you mean # Answers N.

SLaks
– SLaks

2010-11-22 01:33:41 +00:00
Commented Nov 22, 2010 at 1:33 — SLaks
– SLaks, Commented Nov 22, 2010 at 1:33

SLaks · Accepted Answer · 2010-11-22 01:31:46Z

3

You can put the values themselves in an array:

var answers = [
    $(this).find("Answers_1").text(),
    $(this).find("Answers_2").text(),
    $(this).find("Answers_3").text(),
    $(this).find("Answers_4").text()
];

for(x=0;x<=5;x++) {
    alert("looking for = " + x + ", which is " + answers[x])
}

EDIT: Or even

var answers = $(this)
    .find("Answers_1, Answers_2, Answers_3, Answers_4")
    .map(function() { return $(this).text(); })
    .get();

If your answers share a common class, you can change the selector to $(this).find('.AnswerClass').

If you need variable names, you can use an associate array:

var thing = { 
    a: "Hello",
    b: "World"
};

var name = 'a';
alert(thing[name]);

edited Nov 22, 2010 at 1:31

answered Nov 22, 2010 at 1:20

SLaks

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5 Comments

PleaseStand Over a year ago

JavaScript doesn't have real associative arrays though; you would just be using an object as one. This can fail if you have such keys as "__proto" (in Firefox) and other special names.

Šime Vidas Over a year ago

However, those two code-snippets are not equivalent (just as a sidenote)

aaaaaaaaaaaa Over a year ago

@idealmachine, a JavaScript object is an associative array, it has some abilities beyond a basic associative array, but that doesn't make it less a such. I don't see any odd behavior in Firefox resulting from using "__proto" as a key, would you care to elaborate?

PleaseStand Over a year ago

@eBusiness: I meant __proto__; sorry about the typo. developer.mozilla.org/en/JavaScript/Reference/Global_Objects/… Try var a = {}; a['__proto__'] = "foo"; alert(a);.

aaaaaaaaaaaa Over a year ago

Properties can usually be overloaded, it seems strange that this one can't.

John Hartsock · Accepted Answer · 2010-11-22 01:34:51Z

1

This would make it easier to get the array populated.

var answers = new Array();
$("Answers_1, Answers_2, Answers_3, Answers_4", this).text(function(index, currentText) {
  answers[index] = currentText;
});

edited Nov 22, 2010 at 1:34

answered Nov 22, 2010 at 1:29

John Hartsock

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4 Comments

John Hartsock Over a year ago

@SLaks ...wouldn't this be easier to get the array populated.

SLaks Over a year ago

You can be even cleverer. See my answer.

John Hartsock Over a year ago

@SLaks...yeah that one came to me just after I wrote this.

Šime Vidas Over a year ago

I believe you don't have to supply a jQuery object as the context, just supplying this works just fine.

pkaeding · Accepted Answer · 2010-11-22 01:29:09Z

0

As others have mentioned, if you can put the variables in an array or an object, you will be able to access them more cleanly.

You can, however, access the variables through the window object:

var one = 1;
var two = 2;
var three = 3;

var varName = "one";

alert(window[varName]); // -> equivalent to: alert(one);

Of course, you can assign the varName variable any way like, including while looping through an array.

answered Nov 22, 2010 at 1:29

pkaeding

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