Re: New Table Creation with PHP Variables

At 12:43 29/12/2008, you wrote:
>Message-ID: <008B1179CE594EBEA03CB064801823D1@Dragon>
>From: "Keith Spiller" <[email protected]>
>To: "php_db" <[email protected]>
>Date: Sun, 28 Dec 2008 17:39:08 -0700
>MIME-Version: 1.0
>Content-Type: text/plain;
>         format=flowed;
>         charset="iso-8859-1";
>         reply-type=original
>Content-Transfer-Encoding: 7bit
>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.


http://dev.mysql.com/doc/refman/5.1/en/create-table.html


You can create one table from another by adding a 
<http://dev.mysql.com/doc/refman/5.1/en/select.html>SELECT statement 
at the end of the 
<http://dev.mysql.com/doc/refman/5.1/en/create-table.html>CREATE 
TABLE statement:


CREATE TABLE new_tbl SELECT * FROM orig_tbl;


MySQL creates new columns for all elements in the 
<http://dev.mysql.com/doc/refman/5.1/en/select.html>SELECT. For example:


mysql> CREATE TABLE test (a INT NOT NULL AUTO_INCREMENT,
     ->        PRIMARY KEY (a), KEY(b))
     ->        ENGINE=MyISAM SELECT b,c FROM test2;




HTHCheers - Neil