如何在cakephp代码中使用Sql的“like”功能

Question

Hi I have a code in which I am sending a keyword in URL so when we click on link its going to our website so for that I am checking what keyword value is coming in URL in my controller

$keyword= $this->params['url']['keyword'];

$dUsers = $this->Db_user->find(
    'all',
    array(
        'conditions'=>array('Db_user.area_of_expertise'=>$keyword),
        'order'=>array('Db_user.last_name'=>'asc')
    )
);

So my databse has keywod value like : ab ab,bc,ca ab,cd so what is happening by this its giving only ab keyword in result but I need all 3 result which have my keyword in DB so for it I tried to make it like

'Db_user.area_of_expertise'=>'%keyword%'

also tried:

Db_user.area_of_expertise'=>'%'.keyword.'%'

But after it getting no results.

Answer 1

Just use in your conditions array:

'Db_user.area_of_expertise LIKE' => '%'.keyword.'%'