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I have this small gallery (the code is just example):

<div id="gallery_imgs">
    <img src=".png" />
    <img src=".png" />
    <img src=".png" />
</div>
<p></p>

JS code:

var that = $("#gallery_imgs img:eq(1)");
$("p").html($("#gallery_imgs img").find("[src='"+that.attr("src")+"']").length);

This return 0. Any idea why? Is this a bug, or what am I doing wrong?

I have this small gallery (the code is just example):

<div id="gallery_imgs">
    <img src="http://fc01.deviantart/fs71/f/2014/112/0/9/nobody_believes_in_me_by_idjpanda-d7fkd7m.png" />
    <img src="http://fc09.deviantart/fs71/f/2014/112/5/6/feed_me_d__by_idjpanda-d7fgids.png" />
    <img src="http://fc02.deviantart/fs71/f/2014/104/e/5/blar_auction_by_idjpanda-d7ehmoc.png" />
</div>
<p></p>

JS code:

var that = $("#gallery_imgs img:eq(1)");
$("p").html($("#gallery_imgs img").find("[src='"+that.attr("src")+"']").length);

This return 0. Any idea why? Is this a bug, or what am I doing wrong?

• javascript
• jquery
• jquery-selectors

Share Improve this question Follow asked Apr 23, 2014 at 16:19 kleniumklenium 2,62722 gold badges2727 silver badges5252 bronze badges

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2 Answers 2

Sorted by: Reset to default 5

You need .filter()

$("p").html($("#gallery_imgs img").filter("[src='"+that.attr("src")+"']").length);

---

you can use .find() this way

$("p").html($("#gallery_imgs").find("img [src='"+that.attr("src")+"']").length);

---

Problem

$("#gallery_imgs img").find("[src='"+that.attr("src")+"']") you are trying to find the element with src inside the img element not within in the img element

Try

$("p").html($("#gallery_imgs img[src="+that.attr("src")+"]").length);

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