Re: PHP and mySQL

In article <OIBBKNAFKLNGCMHCDGIKKEGHCDAA.sullivan@ccc-e.net>, sullivan@ccc-e.net says... > I am getting a parse error on line 75. I am trying to say: > > if there is a booktitle and a quantity chosen, then go to that booktitle and > adjust the quantity in the database. > > Thanks! > Renee > > > <?php > $user = "adminer"; > $pass = "hoosiers"; > $db = "Book Store1"; > $local = "jolinux"; > $link = mysql_connect( "$local", $user, $pass ); > if (! $link ) > die ( "Couldn't open the database" ); > mysql_select_db( $db, $link ) > or die ( "Couldn't open the $db: ".mysql_error() ); > > if ($submit){ > if( $booktitle, "quantity" ){ > $sql = "UPDATE Book2 SET stock ='$stock-quantity' WHERE booktitle=$booktitle > AND quantity=quantity"; > } > // $result = mysql_query($mysql); > }else if(!$submit){ > echo "Your order has not been placed.<p>"; > } > ?> > </BODY> > </HTML> There don't seem to be 75 lines there? But I think you _might_ be missing a closing } I suspect you will then encounter problems with your SQL: you might want to add mysql_error() after your update call, and ensure that the variable you are using as your sql query is the variable you have assigned the sql query to :-) -- David Robley Temporary Kiwi! Quod subigo farinam