you could use:
for ($i=0; $i< max; $i++)
{
mysql_query("update Table1, Table2 set Table1.field1 = Table2.field2
where Table1.no = Table2.no");
}
OR if you use mysqlt daemon (support for transactions, in this case is best)
mysql_query('BEGIN');
for ($i=0; $i< max; $i++)
{
if (! mysql_query("update Table1, Table2 set Table1.field1 =
Table2.field2 where Table1.no = Table2.no"))
{
mysql_query('ROLLBACK');
die('ERROR - in query n�: '.$i);
}
}
mysql_query(COMMIT');
Execute all or not execute none. I personally use adodb to connect to
databases and control transactions.
If what you really want is to execute then all in just one function call to
the database, i don't know how can you do it with your mysql version.
HTH
----- Original Message -----
From: "Ng Hwee Hwee" <[EMAIL PROTECTED]>
To: "DBList" <[EMAIL PROTECTED]>
Sent: Wednesday, February 18, 2004 2:14 AM
Subject: [PHP-DB] Update Statement
hi all,
I would like to do the following:
update Table1, Table2 set Table1.field1 = Table2.field2 where Table1.no =
Table2.no
however, i found that multiple updates can only work for MySQL version 4.0.4
but mine is 3.23.54... can anyone enlighten me on how I can achieve the same
results with this older MySQL version? or can PHP 4.2.2 help me in any way?
Thank you so much!
Hwee
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