New Table Creation with PHP Variables

From: Date: Mon, 29 Dec 2008 00:39:08 +0000
Subject: New Table Creation with PHP Variables
Groups: php.db 
Request: Send a blank email to php-db+get-45838@lists.php.net to get a copy of this message
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

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