JavaScript Eval() generating 'Unexpected Token ) error'

Question

I'm having a problem with the following bit of code and I'm wondering what I'm doing wrong.

I'm getting an Uncaught SyntaxError: Unexpected token ) error on the line where the eval call is made.

function myfunction(){
    var p1 = '';
    var p2 = '';
    var p3 = '';

    for (i=1; i<=3; i++){
        eval("$('#p"+i+"').closest('.filter').find('.vals div').each(function(){if ($.trim(p"+i+").length > 0) {p"+i+" += ',';} p"+i+" += $(this).attr('class');});");
    }
}

Here is the applicable HTML:

<div class="filter">
    <label>Organizations</label>
    <input id="p1" type="text" value="" />
    <div class="vals">
        <div class="3" title="Click to remove">ABC School District</div>
        <div class="4" title="Click to remove">DEF School District</div>
    </div>
</div>
<div class="filter">
    <label>Groups</label>
    <input id="p2" type="text" value="" />
    <div id="vals"></div>
</div>

If it isn't completely obvious I'm also using jQuery here.

Thanks

"I'm wondering what I'm doing wrong" -- My answer would be using eval() in the first place ... You have a hard to debug statement. I don't see why you can't write this without eval() ... — Martin Tournoij
– Martin Tournoij, Commented Oct 26, 2011 at 13:22
You can skip using eval for statements like these "$('#p"+i+"')... by simply using $('#p['+i) as jquery takes the selecter as string. I really do not see any advantage of eval here. — Birey
– Birey, Commented Oct 26, 2011 at 13:30

Dennis · Accepted Answer · 2011-10-26 13:37:32Z

5

You are better off using an array and just executing the code normally:

function myfunction(){
    var p  = [null, '', '', '']; //Empty zeroth element to keep your 1-indexing

    for (i=1; i<=3; i++){
        $('#p'+i).closest('.filter').find('.vals div').each(function(){
            if ($.trim(p[i]).length > 0) {
                p[i] += ',';
            }
            p[i] += $(this).attr('class');
        });
    }
}

edited Oct 26, 2011 at 13:37

answered Oct 26, 2011 at 13:29

Dennis

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4 Comments

locrizak Over a year ago

dammit! +1 for beating me by a minute!

davin Over a year ago

Huh? Totally incorrect! ES5 10.4.2 eval in most scenarios runs in local scope (and in ES3 I think it's in every case).

Dennis Over a year ago

@davin Yea, my test was poorly constructed/I was thinking of setTimeout. Still recommend the removal of eval though.

Ryan Over a year ago

@Dennis: Perfect... Thank you for actually reading through and taking time to understand the purpose of the code (sorry for it being hard to read) and offering up a better solution.

Bart Vangeneugden · Accepted Answer · 2011-10-26 13:27:02Z

4

Your code, no eval

function myfunction(){
    var p1 = '';
    var p2 = '';
    var p3 = '';

    for (i=1; i<=3; i++){
        $("#p"+i).closest(//..... rest here

answered Oct 26, 2011 at 13:27

Bart Vangeneugden

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2 Comments

Ryan Over a year ago

Nope, I'm afraid that won't work. I'm trying to assign values to differently named variables in my eval();

Bart Vangeneugden Over a year ago

You should use an array instead of using different variables when looping

Steve Wellens · Accepted Answer · 2011-10-26 13:24:13Z

1

You might be missing a "

 eval("$('#p" + i + "'")...

FWI: Instead of nesting functions and chaining, you'll find it much easier to break it into several discrete lines of code and use named functions.

answered Oct 26, 2011 at 13:24

Steve Wellens

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