Update mysql table using POST

Question

This is the query:

if (isset($_POST['editMessage'])) {
    $result = mysql_query("UPDATE messages SET message = '".htmlspecialchars($editedmessage)."' WHERE id = '".$id."'");
    if ($result) {
        die("<strong>Message has been edited!</strong>");
    } else {
        die("<strong>Error ".mysql_error()."</strong>");
    }
}

Using this form:

    <form action="index.php" method="post">
<textarea name='editedmessage' rows='5' cols='70'><?php echo $_POST['editedmessage'];?></textarea>
    <input type='submit' name='editMessage' value='Edit'>

It's not showing an error, it updates the table field, but doesn't enter the edited message into the field, so the field updates and shows no informtion at all.

Where am I going wrong?

Answer 1

htmlspecialchars($editedmessage)

1. You don't seem to be defining $editedmessage anywhere, did you mean $_POST['message1']
2. That should really be mysql_real_escape_string( htmlspecialchars( ... ) )

Answer 2

Try the other way when its correct you get an ressource back:

if(!$result) {
   die('Died: ' . mysql_error());
} else {
   echo "Edited:";
}

Answer 3

您错过了这一行：

$editedmessage = $_POST['editMessage'];

Answer 4

你错了

$result = mysql_query("UPDATE messages SET message = '".htmlspecialchars($_POST['editedmessage'])."' WHERE id = '".$id."'");

Answer 5

You use $editMessage in the query instead of _POST[editMessage] (unless you have register globals on, apparently you don't).

However, do NOT do this without running mysql_real_escape_string() on editMessage first, and DO NOT run htmlspecialchars() on it! Encoded data does not belong in the DB.

Either do $editMessage = $_POST['editMessage']; , or use _POST in the query directly, but wrap it in mysql_real_escape_string() for goodness sake!

However, you DO want to run htmlspecialchars() , htmlentities() , or at the very least string_tags() on $_POST['message1'] when you echo it out. This page is XSS (cross-site script) vulnerable.