Why do I get a 'Supplied argument is not a valid MySQL result resource' error?

There are a number of possible solutions (which are largely connected with the preceeding SQL statement rather than anything on the line the error number is for (mysql_num_rows() is commonly indicated if it is being used):

1. Make sure the tables and fields exist in your SQL statement.

2. If you are passing a variable in the SQL make sure it is probably escaped (usally in single quotes, but don't forget to addslash() if it came from outside (ie. get/post/cookie) and magic_quotes_gpc is off. Look those up in the php manual if you don't know what they mean).
Not doing this will generate the error:

$sql = "SELECT something

         FROM table

        WHERE something = $variable";

and this will fix things:

$sql = "SELECT something

         FROM table

        WHERE something = '$variable'";

Note: you could also use double quotes, but not within a double quoted string...

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Relevant links:

PHP/MySQL error

Minor mySQL troubles (in PHP)

_____________________
Emperor

(Edited by: Tyberius Prime on Mon 12-May-2003)