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" <[email protected]>
To: "php_db" <[email protected]>
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