Re: DB_Dataobject and complex joins

The only way to do that is really to manually set the _join string - probably in a method function buildRadCheckJoin() { $this->_join = "LEFT JOIN radcheck ON radcheck.UserName=radacct.UserName AND radcheck.Attribute = 'Password' "; } Regards Alan Ian P. Christian wrote:

Quoting Alan Knowles <alan@akbkhome.com>:

$radUser =& DB_DataObject::factory('Radacct'); $radCheck =& DB_DataObject::factory('Radcheck');

     $radCheck->Attribute = 'Password';

$radUser->joinAdd($radCheck, 'LEFT', 'radcheck');

'should' Do it...

I tried this :) Perhaps I should have mentioned that in my first email. The result of that is as follows: SELECT count(radacct.RadAcctId) as DATAOBJECT_NUM FROM radacct LEFT JOIN radcheck ON radcheck.UserName=radacct.UserName WHERE radcheck.Attribute = 'Password' This actually gives different results from this. (which is what I'm after) SELECT * FROM radacct LEFT JOIN radcheck ON radcheck.Username = radacct.Username AND radcheck.Attribute = 'Password' Notice the last of the WHERE clause. Thanks for your help again AK! Kind Regards, Ian