Re: Re: New Table Creation with PHP Variables

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On Mon, Dec 29, 2008 at 10:17 AM, Keith Spiller <larentium@xxxxxxxxxxxx> wrote:
> 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.

The idea is right, the format of the query isn't.

I use the case statement for this but it's up to you - work out the
right format for IF from the mysql manual.

(case when o.product_type='course' then o.product_id else null end
case) as field4

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