Jquery push elements in an array

Question

I am not sure if there is a mistake in my existing code or the functionality in Jquery Array is like below:

var categories = [];
$(this).children('categories').each(function() {
    categories.push($(this).find('name').text());
});

Now when I have the below XML node :

<categories>
    <name>a</name>
    <name>b</name>
    <name>c</name>
</categories>

I see that in Firebug the categories array has one element - "abc" but actually it should be as index 2 with values as 'a','b' and 'c'

Is there something wrong in my code?

ajimix · Accepted Answer · 2012-03-20 14:44:08Z

7

$(this).children('categories').each(function() {
    $(this).find('name').each(function(){
        categories.push($(this).text());
    });
});

edited Mar 20, 2012 at 14:44

ajimix

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answered Mar 20, 2012 at 0:24

Frenchi In LA

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Comments

James Montagne · Accepted Answer · 2012-03-20 00:23:32Z

3

var categories = [];
$(this).find('categories name').each(function() {
    categories.push($(this).text());
});

You are looping through each categories element (there is only one) and getting the text of all elements within it named name. Only one call to text means only one value. You need to loop over the names not the categories elements.

answered Mar 20, 2012 at 0:23

James Montagne

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4 Comments

Phil Over a year ago

Actually, calling .text() on a jQuery object containing multiple elements will concatenate the values

James Montagne Over a year ago

Right, into one single value. That's why there's only one value in the array.

Programmer Over a year ago

No I cannot use find('categories name') cause I don't want all of it in the tree node (categories node is repetative down in XML tree). So I have to use .children only

James Montagne Over a year ago

So then use children('categories').find('name') don't use two each loops like the answer you picked.

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