Lamez Posted July 23, 2008 Share Posted July 23, 2008 ok my scripts shows who is online, and there characters they are using. But if no one is online, I get a mysql error, here is the error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Server\htdocs\main\include\cons\head.php on line 237 Total Online: it is talking about this area: <?php while($char = mysql_fetch_array($sql)){ echo $char['login']; echo "(".$name.")"; echo " "; } ?> here is the rest of the code: <?php $sql = mysql_query("SELECT * FROM logon.characters WHERE `online` = '1'")or die(mysql_error()); $char = mysql_fetch_array($sql); $id = $char['acct']; $name = $char['name']; //echo $id; $sql = mysql_query("SELECT * FROM logon.accounts WHERE `acct` = $id"); $on_num = mysql_num_rows($sql); ?> if someone is online it would show something like: account(char_name) Total Online: 1 but I get an error if no one is online. -Thanks Quote Link to comment Share on other sites More sharing options...
Barand Posted July 23, 2008 Share Posted July 23, 2008 Returning 0 rows won't give that error, but an error in your SQL code will. Check for mysql_error() on other query too Quote Link to comment Share on other sites More sharing options...
Lamez Posted July 23, 2008 Author Share Posted July 23, 2008 which query? Quote Link to comment Share on other sites More sharing options...
Lamez Posted July 23, 2008 Author Share Posted July 23, 2008 gosh I forgot the ' ' around $id. I cannot believe that caused it! Quote Link to comment Share on other sites More sharing options...
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