RE: [PHP-DB] Parse Error

Not enough information given to figure out what's going on, but looking at your query i see: "WHERE $cars = "$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dlr _num\""); do you keep 6 different values separated by commas in the field referenced by $cars??? If this is correct what values you are passing for all those variables? Is the $cars variable set to a valid column name in the "test" table?? What is the data type for the column that $cars references? Gurhan -----Original Message----- From: jas [mailto:jlgerfen@hotmail.com] Sent: Thursday, February 21, 2002 12:26 AM To: php-db@lists.php.net Subject: [PHP-DB] Parse Error I hate to post this again but I have looked in a couple of php and mysql books but cannot seem to figure this one out. I am getting a parse error when trying to use php to delete records from a table. The error I am recieving is as follows.... Parse error: parse error in /path/to/php/done2.php3 on line 22 Here is the file that is giving me the error, any help would be great... I think that my problem is that I left out a variable to hold the $cars data but I am not sure. Here is the code... $db_name = "test"; $table_name = "inventory"; $connection = @mysql_connect("localhost", "root", "password") or die ("Could not connect to database. Please try again later."); $db = @mysql_select_db("$db_name",$connection) or die ("Could not select database table. Please try again later."); $sql = "DELETE FROM $table_name WHERE $cars = \"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dl r_num\""); $result = @mysql_query($sql, $connection) or die ("Could not execute query. Any insight would be great... thanks again. Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php