Jquery / javascript检查图像是否存在并更正它

Question

Given 3 image url below

http://www.vbarter.com/images/content/3/2/32822.JPG (loads) http://www.vbarter.com/images/content/3/2/32784.png (loads)

http://www.bikewalls.com/pictures/abc.jpg (does not load)

I need a javascript/jquery function/code to automatically change the extension that works. Taking the case of above example will be very helpful. Here is my code snippet:

function imgError(image) {

image.onerror = "";
var image_src=image.src.substring(0, image.src.length-3);
image.src = image_src+"png";
$.ajax({
  type: "HEAD",
  url: image.src, //or your url
  success: function(data){
  },
  error: function(data){
      image.src = image_src+"JPG";
            $.ajax({
    type: "HEAD",
    url: image.src, //or your url
    success: function(data){
    },
    error: function(data){
        alert("image fails");

    },
      });
  },
});
return true;
}

HTML part
<img src="above url" onerror="imgError(this);">

Consider the situation where image src is vbarter.com/images/content/3/2/32822.jpg . Inside the inner ajax function, it enters into the error section and alerts "image fails". But vbarter.com/images/content/3/2/32822.JPG exists.

Answer 1

In fact your code is correct, but problem is due to the cross domain request issue using the ajax on jsfiddle site

I have modified your jsfiddle example, change the image source to an image that is under the same domain. You will find that the result is same as what you expected.

<div style="background-color:black;">
    <img src="http://fiddle.jshell.net/img/logo.jpg" onerror="imgError(this);" />
</div>

<script>
function imgError(image) {
    image.onerror = "";
    var image_src=image.src.substring(0, image.src.length-3);
    image.src = image_src+"jpg";
    $.ajax({
       type: "HEAD",
       url: image.src, //or your url
       success: function(data){
           alert("png sucess");
       },
       error: function(data){
           image.src = image_src+"png";
           .ajax({
               type: "HEAD",
               url: image.src, //or your url
               success: function(data){
                  alert("JPG success");
               },
               error: function(data){
                  alert("JPG fails");

               },
           });
      }
    });
    return true;
}
</script>

http://jsfiddle.net/ceN8R/6/