"Strange" PHP/SQL error

[Vegeth]

At least, for me it is.

if (mysql_num_rows(mysql_query("SELECT email FROM users WHERE email='$email'")) > 0) {
    Header("Location: register.php?Agreed=TRUE&ErrorMsg=The E-Mail you have chosen has already been used.");
    die;
}

I don't see anything wrong with that...:...the error I get is on the first like ( if..{ ) and the error I get is:

Warning: Supplied argument is not a valid MySQL result resource in /home/vegeth/www/new/confirmregistration.php on line 12

Anyone know what could be causing this?

[nnichols]

This is what happens when you develop with inadequate debug code.

You should always have some sort of code like this -

$db = mysql_pconnect('blah','blah','blah')
   or die('Failed to connect to db server.<br>MySQL error: '.mysql_error());
mysql_select_db('blah')
   or die('Failed to select db.<br>MySQL error: '.mysql_error());
$sql = 'blah';
$res = mysql_query($sql, $db)
   or die('Query failed.<br>MySQL error:'.mysql_error().'<br>Query: '.$sql);

It makes for more code but it also makes it much easier to debug.

Ideally this should be done in some form of db abstraction layer with better error handling, this is an absolute minimum.

[Vegeth]

Heh, thanks, I divided it into multiple lines (so it did the query and the results in different lines) to help me find out the error, but it didn't help much :....anyways, I will try making some .inc layer for PHP & see if that helps debugging 🙂

Gah, that damn thing has been giving me a headache, have been trying to solve it for days!

[nnichols]

It appears that it is probably because the query is failing due to a syntax error. If you have split the code as I suggested it should tell you exactly why it is failing.