Javascript "global" variable issue [duplicate]

Question

    var desc;
    $.ajax({
        url: "data/step2.xml",
        dataType: "xml",
        success: function(data){
            $(data).find('date').each(function(){
                var day = $(this).find('day').text();
                var date = $("#txtDate").datepicker("getDate");
                date = (date.getDate()+"-"+date.getMonth()+1+"-"+date.getFullYear());
                if (day==date){
                    $(this).find('availability').each(function(){
                        var prod = $(this).find('product').text();
                        var time = $(this).find('starttime').text();
                        if (prod==label){
                            desc="!";
                        }
                    });
                }
            });
        }
    });

I'm having some issues accessing the desc variable... I want to manipulate it's value as you can see in desc="!", but it returns undefined. Isn't there any way to make the variable global or something? or some other way to access it? ... Oh, and all this code is inside a function.

$.ajax is async by default. Looks like you are trying to access desc before the ajax call completes. — techfoobar
– techfoobar, Commented Mar 1, 2013 at 18:21
You say "all code is inside a function..." which function? Is it all inside the success function? — Explosion Pills
– Explosion Pills, Commented Mar 1, 2013 at 18:24
@techfoobar: but the desc = "1" is INSIDE the success: handler, so by definition the assignment wouldn't happen until after the ajax results come in anyways. — Marc B
– Marc B, Commented Mar 1, 2013 at 18:24

JaredPar · Accepted Answer · 2013-03-01 18:24:39Z

2

In this particular case the problem is not the scoping of the desc variable. The desc variable you manipulate in the success callback is the same one declared before the ajax call.

The problem here is timing. The success method doesn't execute at the same time the ajax call is made. It executes some time later when the ajax call completes. Only at that point is the value written to desc. Any code which handles the desc value must be called from the point where the desc value is set

if (prod == labe) { 
  desc = "!";
  onDescChanged();
}

answered Mar 1, 2013 at 18:24

JaredPar

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1 Comment

zettca Over a year ago

Oh, got it. Thank you! I'll try something else :)

techfoobar · Accepted Answer · 2013-03-01 18:29:41Z

0

I assume your code is attempting something like:

function foo() {
    var desc;
    $.ajax({
        ....
        success: function() {
            ...
            desc = "!";
        }
    });
    return desc; // this will return undefined, as success() hasn't completed
}

To counter this, use a callback function as in JaredPar's answer.

answered Mar 1, 2013 at 18:29

techfoobar

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