Guest 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. 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()); 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...
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