RE: [PHP-DB] Another dynamic sql.

Seems like it will NOT work because you are selecting $car which is already chosen by the user (i.e. which is a known value). I don't know how you laid out your table but shouldn't your query be something like "SELECT price from FROM varetabell where carid=$car" ?? Gurhan -----Original Message----- From: Raymond Lilleodegard [mailto:raymol@online.no] Sent: Friday, January 25, 2002 1:48 PM To: php-db@lists.php.net Subject: [PHP-DB] Another dynamic sql. Hi all! I have this form with some choices: <form> Enter how many cars you want:<input type="text" name="number" > <select size="1" name="car"> <option selected>ford</option> <option>bmw</option> <option>mercedes</option> </select> And then I am trying to get the price out of a table in my database with this code: $sql = mysql_query("SELECT '$car' FROM varetabell where carid='$carid' "); $myrow= mysql_fetch_array($sql); $x = $myrow["$car"]; $price = $x * $number; Shouldn't this work? Or am I missing something here? Best regards Raymond -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-db-unsubscribe@lists.php.net For additional commands, e-mail: php-db-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net