如何使用javascript捕获损坏的图像URL然后将URL传递给php？

Question

I have a site that displays a lot of external images and thumbnails etc, easily up to 100 on a single page. I crawl and index the urls to the images and save them in mysql and display them with this code inside simple loops from queries.

<img src="<?php echo $row['img_url']; ?>" onerror="this.onerror=null;this.src='http://example.com/image.jpg';" width="150" height="150">

I use that particular code to replace any broken image urls with a default image.

My question is, is it possible to use javascripts onerror or something else to capture the image url that is broken when a broken url is found so that I can pass the url back to php and be able to automatically delete the urls from my database? I am not very good with javascript and after searching I can't seem to find anything similar to what I am looking for, I mostly just find lots of posts on how to replace the broken image.
I am open to any ideas really, the original image urls come from $row['img_url'] as you can see in the code but I know I need javascript or something to catch the errors and then somehow get the urls passed back to php so that I can automate the deletion process instead of just replacing images with default images like my currnt code.

Answer 1

You are almost there. Simply replace this:

this.src='http://example.com/image.jpg'

with:

this.src='http://example.com/image-missing.php?url={$row['primaryKey']}&w=150&h=150'

Now, "image-missing.php" receives the URL of the missing file and the intended size of the missing image. What it needs to do is clean the database (after checking that the call is legitimate (1) and that the referred row exists(2)), and output a replacement image in the proper size.

---

(1) otherwise you have just handed the possibility of deleting your whole image database to the first guy with time on his hands.

(2) someone else might have loaded the same page, and called the same script one millisecond ago.