At 12:43 29/12/2008, you wrote:
Message-ID: <008B1179CE594EBEA03CB064801823D1@Dragon>
From: "Keith Spiller" <larentium@xxxxxxxxxxxx>
To: "php_db" <php-db@xxxxxxxxxxxxx>
Date: Sun, 28 Dec 2008 17:39:08 -0700
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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
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