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Editorial Team
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Asked: 2026-06-12T09:22:47+00:00

I have an image slider on a page that reads an XML file called

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I have an image slider on a page that reads an XML file called “slider.xml” that kinda looks like this:

<slider>
<slide>
<link>http://www.example.com/1</link>
<path>http://image.jpg</path>
</slide>
</slider>

There’s multiple “slide” elements but I didn’t include them for space reasons. I have some HTML that looks like this:

<body>
<div class="slide"></div>
<div class="slide"></div>
<div class="slide"></div>
<div class="slide"></div>
<div class="slide"></div>
</body>

I want to read this XML file and write the “link” attribute to the “div” elements on an HTML page as a title attribute. I want it to look like this:

<body>
<div class="slide" title="http://www.example.com/1"></div>
<div class="slide" title="http://www.example.com/2"></div>
<div class="slide" title="http://www.example.com/3"></div>
<div class="slide" title="http://www.example.com/4"></div>
<div class="slide" title="http://www.example.com/5"></div>
</body>

So far I have tried this but haven’t had any luck:

$(document).ready(function () {
    $.ajax({
        type: "GET",
        url: "slider.xml",
        dataType: "xml",
        success: function(xml) {
            $(xml).find('slide').each(function(){
                var url = $(this).find('link').text();
                $('.slide').attr('title', url);
            }); 
        }   
    }); 
});

I don’t have issues reading the XML file, but run into problems after I parse the XML attributes and attach it to the various divs. Should I create a loop and store the xml attributes in an array? Is there a better way to do this?

Also, I cannot edit the XML or HTML.

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1 Answer

  1. Editorial Team

    All your divs will have the same title since you call $('.slide') which select them all.

    This should work:

    $(document).ready(function () {
    var slide = $('.slide');
    $.ajax({
        type: "GET",
        url: "slider.xml",
        dataType: "xml",
        success: function(xml) {
            $(xml).find('slide').each(function(i){
                var url = $(this).find('link').text();
                slide.eq(i).attr('title', url);
            }); 
        }   
    }); 
});
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