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@xxxxxxxxxxxx>
To: "php_db" <php-db@xxxxxxxxxxxxx>
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
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
