Thanks its now working. :-) On Jun 17, 6:55 pm, scs <[email protected]> wrote: > sorry forgot to add the vars > $conditions = 'Practitioner.name LIKE "%'.$name.'%"'; > $conditions .= ' OR Practitioner.surname LIKE "%'.$surname.'%"'; > $record = $this->Practitioner->find('all', array('conditions'=> > $conditions )); > > On Jun 17, 12:54 pm, scs <[email protected]> wrote: > > > > > > > > > Try this > > > $conditions = 'Practitioner.name LIKE "%'..'%"'; > > $conditions .= ' OR Practitioner.surname LIKE "%'..'%"'; > > $record = $this->Practitioner->find('all', array('conditions'=> > > $conditions )); > > > On Jun 17, 10:35 am, mthabisi mlunjwa <[email protected]> wrote: > > > > I'm not sure if I'm missing something here. I'm trying to come up with > > > this sql statement the Cakephp way: > > > > select * from practitioners where name like 'Vermeulen' or surname > > > like 'Vermeulen'; > > > > This is what I've come up with: > > > > find( 'all', $conditions = array("OR"=>array('Practitioner.surname > > > LIKE' => "%"Vermeulen"%",'Practitioner.name LIKE' => > > > "%"Vermeulen"%"))); > > > > This works but it only outputs the whole database. I just want it to > > > output practitioners with surnames or names like Vermeulen. > > > > Please help, thanks in advance.
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