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 https://forums.phpfreaks.com/topic/116302-solved-if-results-are-0-i-get-mysql-error/ 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 https://forums.phpfreaks.com/topic/116302-solved-if-results-are-0-i-get-mysql-error/#findComment-598047 Share on other sites More sharing options...
Lamez Posted July 23, 2008 Author Share Posted July 23, 2008 which query? Quote Link to comment https://forums.phpfreaks.com/topic/116302-solved-if-results-are-0-i-get-mysql-error/#findComment-598055 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 https://forums.phpfreaks.com/topic/116302-solved-if-results-are-0-i-get-mysql-error/#findComment-598057 Share on other sites More sharing options...
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