Code returns 'undefined"

Question

Why is this piece of code returning undefined ?

function getMoods(nb) {
        var index;
        var moods;
        var a = ["MOOD0001", "MOOD0002", "MOOD0003", "MOOD0004", "MOOD0005", "MOOD0006", "MOOD0007", "MOOD0008", "MOOD0009", "MOOD0010", "MOOD0011", "MOOD0012", "MOOD0013", "MOOD0014", "MOOD0015", "MOOD0016"];
        for (index=0; index<=nb; ++index) {
                if(index==0 || index==4 || index==8 || index==12) { moods += '<div class="col-xs-4">'; }
                        moods += 
                                '<div class="checkbox">'
                                    +'<label for="'+a[index]+'">'
                                        +'<input type="checkbox" id="'+a[index]+'" class="moods"> '+a[index]
                                    +'</label>'
                                +'</div>';
                if(index==3 || index==7 || index==11) { moods += '</div> '; }
        }
        $("#moods-area").html(moods);
}
getMoods(12)

possible duplicate of $.each() adds "undefined" value

Omar
– Omar

2014-10-13 10:27:00 +00:00
Commented Oct 13, 2014 at 10:27 — Omar
– Omar, Commented Oct 13, 2014 at 10:27

David Thomas · Accepted Answer · 2014-10-12 22:58:26Z

1

Because the initial value of moods is undefined, as it was never itialized (declared, yes; initialised, no); changing:

var moods;

to:

var moods = '';

There is no 'undefined': JS Fiddle demo.

answered Oct 12, 2014 at 22:58

David Thomas

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orhanhenrik · Accepted Answer · 2014-10-12 22:58:17Z

1

You get the "undefined" in the beginning of your string because you are not setting a value for moods, and the adding a string to it. It will then convert moods to a string, which turns out "undefined", and add some content to it. To fix this, use var moods=""; instead of var moods;

answered Oct 12, 2014 at 22:58

orhanhenrik

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Felix Kling · Accepted Answer · 2014-10-12 22:58:24Z

1

The initial value of moods is undefined. See what happens when I add a string to an uninitialized variable:

> var foo;
> foo + 'bar';
"undefinedbar"

Solution: Initialize the variable with an empty string.

answered Oct 12, 2014 at 22:58

Felix Kling

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