Returning only the first node when parsing XML using jQuery

Question

I would like my jQuery script to only return ONLY the top node it finds instead of iterating through all the nodes. What do I change in my jQuery script to accomplish that? I know it has to do with the each(function) {} but i don't know exactly what should change. Thanks in advance!

jQuery Code:

$.ajax({
url: 'xml/timeline.xml',
dataType: 'xml',
success: function(data) {
    $(data).find('channel item').each(function() {
        var fbFeedWrapper = $('<div/>').addClass('fb-feed-wrapper');
        var fbFeedItem = $('<div />').addClass('fb-feed-item');
        $('.socialmedia-list#simlearn').append(fbFeedWrapper);
        fbFeedWrapper.append(fbFeedItem);

        var url = $(this).find('link').text();
        var title = $(this).find('title').text();           
        fbFeedItem.append($('<h1/>').html("<a href='" + url + "'>" + title + "</a>"));                
        var fbItemIcon = $('<div />').addClass('fb-item-icon');
        fbFeedItem.append(fbItemIcon);
        var description = $(this).find('description').text();
                        if (description.length > 80) {
                            description = description.substr(0, 80) + "...";
                        }
        fbFeedItem.append($('<div/>').addClass('fb-item-details').html(description));                            

    })
}


});

XML File format:

<channel>
<item> // <-- If i Only want the top <item> node returned what should i do?
    <title></title>
    <link></link>
    <description></description>
</item>
<item>
    <title></title>
    <link></link>
    <description></description>
</item>
<item>
    <title></title>
    <link></link>
    <description></description>
</item>

Nunners · Accepted Answer · 2014-01-15 14:45:34Z

Instead of using .each(); simply find the first of the matched elements. Assign a variable with this object and then change of the this references from inside the .each() to be the name of the variable.

Example (In the example I name the variable self) :

Change the line :

$(data).find('channel item').each(function() {

to be :

var self = $(data).find('channel item').first();

Then change all of the occurrences of this to be self. Also remove the ending brace and bracket of the .each function (}))

$.ajax({
    url: 'xml/timeline.xml',
    dataType: 'xml',
    success: function (data) {
        var self = $(data).find('channel item').first();
        var fbFeedWrapper = $('<div/>').addClass('fb-feed-wrapper');
        var fbFeedItem = $('<div />').addClass('fb-feed-item');
        $('.socialmedia-list#simlearn').append(fbFeedWrapper);
        fbFeedWrapper.append(fbFeedItem);

        var url = $(self).find('link').text();
        var title = $(self).find('title').text();
        fbFeedItem.append($('<h1/>').html("<a href='" + url + "'>" + title + "</a>"));
        var fbItemIcon = $('<div />').addClass('fb-item-icon');
        fbFeedItem.append(fbItemIcon);
        var description = $(self).find('description').text();
        if (description.length > 80) {
            description = description.substr(0, 80) + "...";
        }
        fbFeedItem.append($('<div/>').addClass('fb-item-details').html(description));
    }
});

Jai · Accepted Answer · 2014-01-15 14:46:42Z

1

Try this:

$(data).find('channel item[0]')
    //--------------------^^^-----add [0] to this

or with :first:

$(data).find('channel item:first')

or with :eq():

$(data).find('channel item:eq(0)')

Even you can try this:

$(data).find('item:eq(0)')

answered Jan 15, 2014 at 14:46

Jai

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2 Comments

alchuang Over a year ago

When I tried replacing $(data).find('channel item').each(function() { with $(data).find('channel item[1]') the jquery won't execute

BrettC Over a year ago

$(data).find('channel item:eq(0)') worked for me, but $(data).find('channel item[0]') did not work for me

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