Re: New Table Creation with PHP Variables

Another option that would work if I can figure out the correct syntax is to just NULL certain values if a given condition exists. If product_type='course' then just use the o.product_id value for field4. If product_type != 'course' then use NULL for field4. CREATE TABLE $table[name] SELECT field1, field2, field3, IF(o.product_type='course', o.product_id, NULL) AS field4, field5, field6, field7 FROM table1 as a, table2 as o; Is this right? Thank you for your help. Keith ----- Original Message ----- From: "Keith Spiller" <larentium@hosthive.com> To: "php_db" <php-db@lists.php.net> Sent: Sunday, December 28, 2008 5:39 PM Subject: New Table Creation with PHP Variables

Hi, I'm trying to join multiple tables to then create a new table from the query. I've figured out that part, but some of the fields need to be evaluated and then compared to a php array to derive their data. In this example I am trying to populate the field4 column (from the $product_name array) after evaluating the product_type value on each row. CREATE TABLE $table[name] SELECT field1, field2, field3, IF(o.product_type='course', $product_name[$product_id], NULL) AS field4, field5, field6, field7 FROM table1 as a, table2 as o; Is this possible? Is there another way to accomplish this task? Thanks for your help. Keith