Re: [PHP-DB] Re: New Table Creation with PHP Variables

That looks like it should work, just execute the select query and see what
is the output


Jack

2008/12/29 Keith Spiller <[email protected]>

> 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
>>
>
>
>
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>


-- 
J.A. van Zanen