Re: PHP4.0.1 pl2 and mysql_free_result command

On Thu, 07 Sep 2000, Samantha Savvakis wrote: > Is there a problem with this command in the PHP4.0.1 pl2 version? > > It worked under PHP30.16 and now I'm trying to use this command and I get > the error: > > Warning: Supplied argument is not a valid MySQL result resource in .... on > line 21 > > The code looks like this: > > // This is my connection function > $db = db_connection(); > > $result = mysql_query($sql, $db); > if ($result) > { > $p1ID = mysql_insert_id($db); > } > else > { > $error = ".."; > } > mysql_free_result($result); > > Any ideas? > Sam This is a (very) FAQ; see http://au.net.php:81/FAQ.php#7.12 Essentially you need to check the error message being returned from your query with mysql_error() -- David Robley | WEBMASTER & Mail List Admin RESEARCH CENTRE FOR INJURY STUDIES | http://www.nisu.flinders.edu.au/ AusEinet | http://auseinet.flinders.edu.au/ Flinders University, ADELAIDE, SOUTH AUSTRALIA