Question

So I have yet another problem,

I am trying to get an AJAX script to work, but upon click the page will reload but fail to UPDATE the database field.

The code im using I have working for other similar scripts on the site but for some reason this one, using the same code doesnt work, the code used follows below;

HTML code to send the call to AJAX:

<input onClick="read('<? echo $id; ?>')" id="read" name="read" type="checkbox" value="1" style="position:relative; top:2px; width: auto">

The code to confirm user selection and send onto a form handling file:

function read(ID) {
    if(confirm('Are you sure you have read this carefully, you will not be alerted again "' + ID + '" ?')) {
        $.get('http://<? echo ROOT . ADMIN . INCLUDES; ?>formHandling.php', { read: ID }, function(data) {
            window.location.href = 'http://<? echo $_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'].""; ?>';
        });
    }
    return false;
}

Lastly the code to handle the SQL query:

if (isset($_GET['read'])) {

    // Pass the GET data and associate them to variables
    $read = trim($_GET['read']);

    $query  = "UPDATE cms_motd SET read='$read' WHERE id='1'";
    $result = mysql_query($query)or die("Database query died: " . mysql_error());

    unset($_GET['readConfirm']);
}

Thanks in advance for all that help.

Regards,

Dan.

Answer 1

Don’t you mean:

$query  = "UPDATE cms_motd SET read='1' WHERE id='$read'";

Instead of:

$query  = "UPDATE cms_motd SET read='$read' WHERE id='1'";

Edit:

I don’t know if it is a copy&past error:

$result = mysql_query($query);or die("Database query died: " . mysql_error());

Needs to be:

$result = mysql_query($query) or die("Database query died: " . mysql_error());