Spring Posted November 6, 2010 Share Posted November 6, 2010 $sql = 'SELECT * FROM account_info WHERE username ="'.$name.'"'; $result = mysql_query($sql) . mysql_error(); $row = mysql_fetch_row($result); I get Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource I'm trying to output the row "wins" for a certain user. Quote Link to comment https://forums.phpfreaks.com/topic/217957-why-wont-this-work-seems-simple-enough/ Share on other sites More sharing options...
PFMaBiSmAd Posted November 6, 2010 Share Posted November 6, 2010 $result = mysql_query($sql) . mysql_error(); ^^^ By concatenating msyql_error() with the result of the mysql_query() in the above line of code, you are likely corrupting the value being put into $result. That line of code should be as follows to get it to display the mysql_error() value if the query fails - $result = mysql_query($sql) or die(mysql_error()); Quote Link to comment https://forums.phpfreaks.com/topic/217957-why-wont-this-work-seems-simple-enough/#findComment-1131207 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.