Conditions in CakePHP RC6

18 comments baked

• Nate March 02, 2006 at 16:00

  A couple technical details:
  When using ‘=’ as a comparison operator, you actually don’t need to include it, as that is the default, so:

  $this->User->findAll(array(‘User.firstname’ => ‘= Daniel’));
  becomes
  $this->User->findAll(array(‘User.firstname’ => ‘Daniel’));

  Also, you can now use nested arrays for complex conditions, so:
  $this->User->findAll(“User.firstname = ‘Daniel’ OR User.firstname = ‘Hugo’”);
  could be
  $this->User->findAll(array(‘or’ => array(“User.firstname” => ‘Daniel’), array(‘User.firstname = ‘Hugo’”)));

• Sam D March 02, 2006 at 16:33

  Ok now that is more like it. I have been waiting for nesting! Hey Daniel, thanks for this site it is a great little house o’ cake!

  Sam D

• cakebaker March 03, 2006 at 09:11

  @Nate: Thanks for the further explanations.

  @Sam D: Nice to hear you like it. Thanks :)

• Diego January 16, 2008 at 02:27

  Great insight into the conditions array, i was toying arround with 1.2.0.6311 beta, and wondering if this still works.

  $this->User->findAll(array(’or’ => array(”User.firstname” => ‘Daniel’), array(’User.firstname = ‘Hugo’”)));

• cakebaker January 16, 2008 at 18:47

  @Diego: No, this no longer works. Now you can use:

  $this->User->findAll(array('User.firstname' => array('Daniel', 'Hugo')));
  

  or if you want to use different column names you could use:

  $this->User->findAll(array('or' => array('User.id' => '1', 'User.firstname' => 'Daniel')));
  

  HTH

• Sumon March 09, 2008 at 04:45

  How would you put a condition for distinct?

  $this->Model->find(‘list’, ‘distinct’)

• cakebaker March 09, 2008 at 19:45

  @Sumon: I think this is not possible with this method, but I could be wrong. You may want to ask in the Google Group.

• olda July 28, 2008 at 18:04

  what about multiple ‘or’ key in array? for multiple ‘or’ in conditions, like : exp (A=B OR B=C) AND(A=D OR B=D)? how can i write this in cake find conditions?

• cakebaker July 29, 2008 at 16:56

  @olda: You could write it in the following way:
  $this->YourModel->find('all', array('conditions' => '(A=B OR B=C) AND (A=D OR B=D)'));
  But you have to be careful with this approach. It is very easy to introduce an SQL injection vulnerability because the values are not automatically escaped by cake!

  Anyway, hope it helps!

• olda July 29, 2008 at 23:16

  oh je, but i mean somthing more like specify it in model class in $conditions property
  

  array ("or" =>
      array
      (
          "A" => "B"
          "B" => "C"
      )
  )

  
  and some far in code use find function with additional conditions.
  btw, conditions that i use in call find function will join with conditions specify in model?

• cakebaker July 31, 2008 at 17:23

  @olda: Hm, what do you mean with “$conditions property”? The model doesn’t contain such a property. Or do you mean the “conditions” option you can specify when defining an association? In that case you can specify the conditions as a string as shown in my previous comment.

  Hope that helps!

• Ellen September 17, 2008 at 14:45

  I can’t do it like your example, but I can it when I change somethings:

  $this->set(“country”, $this->Country->find(“all”, array(‘conditions’=>array(‘Country.nom_country LIKE’=>’%South%’))));

  Thank you so much!!!

• cakebaker September 17, 2008 at 17:06

  @Ellen: Yes, the syntax was changed recently so I modified the article accordingly. Thanks for the hint!

• Sarah March 08, 2010 at 01:08

  I know this is old but it’s the best Google could find me.

  Basically I want my new model to be a “view” of an existing model.

  I therefore to add to the join conditions so I have created a hasmany condition in the parent model
  

  'Completeorder' => array(
      'className'     => 'Completeorder',
      'conditions'    => "Completeorder.status in ('Pending Pickup','Complete')",
      'order'         => 'Completeorder.pickupdate DESC'
  )

  
  However when I view the query it’s put the conditions in the where statement. This means that my left join isn’t returning null for those parents with no complete orders – its omitting them altogether.

  I’m happy to have a custom query but I’d also like to use Cake where possible. Anyway around this?

• cakebaker March 09, 2010 at 17:40

  @Sarah: Thanks for your comment.

  Right now I don’t see a way around this, though I have to admit I don’t understand your approach completely. Having a relationship YourModel hasMany CompleteOrder seems a bit odd to me ;-)

• Sarah March 09, 2010 at 19:52

  Thanks for the reply. It was part of an attempt to build queries based on pseudo “views” rather than having custom sql. On its own it might seem odd but when you put the building blocks together and are looking for available resources (those with complete/finished orders) it did make sense.

  :)

• fairuz May 14, 2010 at 04:17

  i have some problems about my query in controller… in search.ctp file i am generate my code like this

  //search.ctp
  $selectSearchFilter = array('a'=>'column1','b'=>'column b', 'c'=>'column c ' );
  $form->select('searchFilter',$selectFitlerSearch);
  $form->input('keyword',array('type'=>'text'));
  $form->end('Search');
  
  //in tempModels_controllers
  
  function search()
  {
    if(!empty($this->data()))
       {
          $keyword = $this->data['Pruf']['keyword'];
          $selectFilter = $this->data['Pruf']['searchFitler'];
  
          // i dont know how to generate filter SQL... :-(
         // select * from table where $searchFilter Like %$keyword%
  
  
          $result = $this->find('all',$conditions);
        }

  Questions : How to generate cakephp sql like about statements??

• cakebaker May 14, 2010 at 16:59

  @fairuz: Something like:

  $conditions = array('YourModel.' . $searchFilter . ' LIKE' => '%' . $keyword . '%');
  

  should do the trick (make sure $searchFilter is an allowed value before using it).

  Hope this helps!

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