This is the first time I am trying to create a table in MySQL based on user input. Here is the scenario, The user will enter in the name of a stage, e.g. Harley Davidson. I then look for the first space (to indicate the first word), make it into lower case and then do the following: $new_stage_name = "stage_".$var; This will give $new_stage_name the value of "stage_harley". That works fine. Now I want to run the following query: $SQL = "CREATE TABLE ".$new_stage_name." (set values and their properties)"; If I run this query using PHPMyAdmin, minus the var of course, the table is created. When I run it in my code, nothing happens. I do a die($SQL) to see the final outcome of the query and it's correct and is what I am using to run the query in PHPMyAdmin. This is what my code is doing: I have class RecordSet which makes the db connection and its constructor takes the $SQL as an arg. So, $SQL = "what is listed above"; $objRecordSet = new RecordSet($SQL); //I used this all the time with all kinds of other Select, Insert, Update, and Delete queries and it works perfect. For one, I don't know how to test whether the table has been created or not. I know I could probably do some sort of mysql_list_tables and look for it or whatever but I was looking for something similar to If (mysql_affected_rows() == -1) Because that is very easy. And ideas as to why the table is not being created. =================== Jonathan Villa Application Developer IS Design & Development www.isdesigndev.com 414.429.0327 =================== -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
