remove full path, keep filename only

Question

Trying to remove the full url that is being returned to imgurl: Usually returns something like http://localhost/wordpress/wp-content/uploads/filename.jpg or http://localhost/wordpress/wp-content/uploads/images/filename.jpg

I'd like to strip off everything except filename.jpg and return it to ahng_photos_upload_image. Strip off everything to the last forward-slash. How can I do that with Jquery?

window.send_to_editor = function(html) {
imgurl = jQuery('img',html).attr('src');
jQuery('#ahng_photos_upload_image').val(imgurl);
tb_remove();
}

Solution 1

You don't need jQuery for that, just plain old JavaScript will do :)

alert('http://localhost/wordpress/wp-content/uploads/filename.jpg'.split('/').pop());​​

In your case:

var filename = imgurl.split('/').pop();

Solution 2

you can use a regular expression in order to achieve this..

var file = imgUrl.replace(/^.*[\\\/]/, '');

Now the file would consist of only the file name ..

Solution 3

One further option:

var filename = imgurl.substring(imgurl.lastIndexOf('/') + 1);

JS Fiddle demo.

Solution 4

If you're pretty confident that the URLs don't have funny stuff like hashes or parameters, a regex like this would do it:

var filename = imgurl.replace(/^.*\/([^/]*)$/, "$1");

Also: don't forget to declare "imgurl" with var, and you should probably use .prop() instead of .attr() if your version of jQuery is 1.6 or newer:

var imgurl = jQuery('img', html).prop('src');

Also jQuery internally turns the two-argument form of the function into this:

var imgurl = jQuery(html).find('img').prop('src');

so you might as well code it that way.

View more solutions