jQuery how do i select img src inside href

Question

i have simple html that looks like this :

<div class="img">
      <a target="_blank" href="kl_big.htm" id="turn-of-cloth-01_lg">
      <img src="images/galery/turn-of-cloth-01_lg.jpg"   width="110" height="90" />
      </a>
      <div class="desc">Add a description of the image here</div>
    </div>

i have like 20 sets of this div with different ids , now the href id im gtting with this

$(document).ready(function(){
    $('a').click(function (event){

        event.preventDefault();
        var linkID =$(this).attr("id"); 
    });
});

i dont know what is the img object id , but i need to get its src value . i tried with Jquery like this :

var imgSrcVal = $(this).siblings("img").attr("src");

but with out result , also tryied inside the click event but again with no results :

imgSrcVal = this.nextSibling.src;   

what im doing wrong here ?

Answer 1

The <img> is not a sibling, it's a child of the anchor.

Try this instead:

$('a').click(function (event){
    var linkID = $(this).attr("id"); 
    var imgSrcVal = $('img', this).attr("src");
    //alert(imgSrcVal);
    return false;
})

Simple JSFiddle demo.

Answer 2

The img is nested, so use .find() or .children().

var imgSrcVal = $(this).find("img").attr("src");

• .find() looks looks through all descendants

• .children() looks only at direct descendants

---

Or with native DOM methods:

var imgSrcVal = this.getElementsByTagName('img')[0].src;

The [0] is because getElementsByTagName returns an array-like collection. So [0] gets the first element found.

Answer 3

I think this is what you want. The image is a child of the anchor tag, not a sibling.

  $('a').click(function (event){
        event.preventDefault();
        var linkID =$(this).attr("id"); 
         alert($(this).children('img').attr('src'))
    });

Check out this fiddle

Answer 4

the img tag in that example is not a sibling of the a tag, it's a child. So to get the value of the src attribute, you'd do $(this).children("img").attr("src")