Multiple one-time PHP database updates using jquery's .error()

Question

I have a webpage with some broken images, these images are being show by a database. I am using the following jQuery to hide the images that are broken.

//images are wrapped in an anchor
$("img").error(function() {
    $(this).parent().hide();
});

I would like to utilize the "status" column in the database to set all of the broken images to "hidden". Every anchor that wraps an image on the page has a "id" attribute that matches the primary database key "id".

$("img").error(function() {
    var error = $(this).parent().attr('id');
    $.ajax({
        type: "POST",
        url: "changestatus.php",
        data: "status=hidden&id=".error.""
    });
});

// changestatus.php

<?php 
mysql_connect("localhost", "stackoverflowexampleuser", "stackoverflowexamplepass") or die(mysql_error());
mysql_select_db("stackoverflowexampledatabase") or die(mysql_error());
$id = $_POST['id'];
$status = $_POST['status'];
$query="UPDATE stackoverflowexampletable SET status = '".$status."' WHERE id ='".$id."'";
mysql_query($query) or die ('error');
mysql_close();
header( 'Location: MYSOURCE' ) ;
?>

This is my first stab at ajax, and I know I got some stuff seriously wrong. I saw a couple of examples using KEY VALUE pairs but I don't know what the $_POST['var'] should be.

Can you even request something like this "when the page loads"? I tried wrapping it in an arbitrary button and it didn't work.

Since this just needs to be used once I'm not really focused on using AJAX.

No, although I'm new to debugging ajax with firebug. Nothing is really happening at all. — ThomasReggi
– ThomasReggi, Commented Feb 22, 2011 at 18:14

Chuck Callebs · Accepted Answer · 2011-02-22 18:19:09Z

1

Try this:

$("img").error(function() {
    var error = $(this).parent().attr('id');
    $.ajax({
        type: "POST",
        url: "changestatus.php",
        data: {
             status: "hidden",
             id: error
        }
    });
});

If you're wanting to initiate it on page load, be sure to include it in the $(document).ready() function.

answered Feb 22, 2011 at 18:19

Chuck Callebs

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