That looks like it should work, just execute the select query and see what
is the output
2008/12/29 Keith Spiller <larent...@hosthive.com>
> 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.
> ----- Original Message ----- From: "Keith Spiller" <larent...@hosthive.com
> To: "php_db" <email@example.com>
> Sent: Sunday, December 28, 2008 5:39 PM
> Subject: New Table Creation with PHP Variables
>> 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.
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
J.A. van Zanen