Re: Supplied argument is not a valid MySQL result resource

Just a fyi, consider using mysql_error() to solve these type of problems. For example: On Thu, 20 Jun 2002 webmaster@tececo.com wrote: > Here is my code: > > <? Using 'or die' like this is okay, while debugging consider putting mysql_error() in your die(). > $link = mysql_pconnect("localhost", "root", "") or > die("Could not connect"); > mysql_select_db("tececo") or die("Could not select database"); > $fn = explode("/", $_SERVER['PHP_SELF']); > $num_of_s = count($fn) - 1; > $fn = "$fn[$num_of_s]"; // Make it '$fn' not $fn in your query // (use quotes if $fn isn't a number) > $query = "SELECT * FROM meta_data WHERE page_name = $fn"; > $result = mysql_query($query); // mysql_query returns false on failure if (!$result) { echo "DB Error ($query) : " . mysql_error(); exit; } > $num_results = mysql_num_rows($result); // maybe some rows should exist before we move on if ($num_results > 0) { > $row = mysql_fetch_array($result); // instead of the below, you could use extract: // http://www.php.net/extract extract($row); > $id = $row['id']; > $title = $row['title']; > $description = $row['description']; > $keywords = $row['keywords']; > ?> > > I get this error: mysql_error() will give you a more useful meaning. > > Warning: Supplied argument is not a valid MySQL result resource in > C:\TecEco(Converting To PHP)\includes\meta.array_create.php on line 9 > > Warning: Supplied argument is not a valid MySQL result resource in > C:\TecEco(Converting To PHP)\includes\meta.array_create.php on line 10 > > Why? > > Thanks > > JJ Harrison > webmaster@tececo.com > www.tececo.com >