sethupathy Posted June 11, 2007 Share Posted June 11, 2007 hello guys this query gets me an error pls help i am in a rush Warning: Wrong parameter count for mysql_query() on line 194 line 194 : $result=mysql_query($query,$queryb,$queryc,$queryd,$querye,$queryf) or die("Error in query:".mysql_error()); //if ($result) //echo mysql_affected_rows()." row inserted into the database effectively."; // CLOSE CONNECTION ---> mysql_close($connection); Link to comment https://forums.phpfreaks.com/topic/55100-solved-php-mysql-query-error/ Share on other sites More sharing options...
trq Posted June 11, 2007 Share Posted June 11, 2007 What exactly does your query look like? Link to comment https://forums.phpfreaks.com/topic/55100-solved-php-mysql-query-error/#findComment-272356 Share on other sites More sharing options...
Psycho Posted June 11, 2007 Share Posted June 11, 2007 You can only run one query in mysql_query(). You need to run each query separately: mysql_query($query); mysql_query($queryb); mysql_query($queryc); . . . etc. mysql_query has 1 required parameter (the query) and one optional parameter (the resource link identifier). You are using 6 parameters. http://us.php.net/manual/en/function.mysql-query.php Link to comment https://forums.phpfreaks.com/topic/55100-solved-php-mysql-query-error/#findComment-272357 Share on other sites More sharing options...
sethupathy Posted June 11, 2007 Author Share Posted June 11, 2007 F%/$ yes THANKS A BUNCH Link to comment https://forums.phpfreaks.com/topic/55100-solved-php-mysql-query-error/#findComment-272360 Share on other sites More sharing options...
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