That looks like it should work, just execute the select query and see what is the output Jack 2008/12/29 Keith Spiller <larentium@xxxxxxxxxxxx> > 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 > > -- J.A. van Zanen
