Join-Abfrage in cakephp

Question

Hallo, ich bin neu in CakePHP und ich habe eine harte Zeit konvertieren mysql-Abfragen in cakephp.I mit haben diese Abfrage, die ich

in CakePHP Syntax konvertieren möchten

SELECT * 
FROM trip 
JOIN countries ON trip.departure_country_id = countries.id 
WHERE countries.country_name LIKE "eng%" 

Dies ist, was ich habe bisher versucht

class Trip extends AppModel 
{ 
    public $useTable = 'trip'; 
    public $primaryKey = 'trip_id'; 

    public $belongsTo = array(
     'User' => array(
      'className' => 'User', 
      'foreignKey' => 'user_id', 
      'fields' => array('User.user_id','User.first_name','User.last_name','User.profile_image','User.country','User.phone_no') 

     ), 
     'departure_country' => array(
      'className' => 'Country', 
      'foreignKey' => 'departure_country_id', 
     ), 
     'arrival_country' => array(
      'className' => 'Country', 
      'foreignKey' => 'arrival_country_id', 
     ) 
    ); 

    public function getLocationBasedTrips($country){ 
     return $this->find('all', array(
      'conditions' => array(
       'OR' => array('Country.country_name Like' => '%'.$country.'%') 
      ), 
     )); 
    } 
} 

Answer 1

Einige Klammern von der vorherigen Antwort entfernt. Versuchen Sie folgendes:

$this->find('all', array(
      'joins' => array(
       array(
        'table' => 'countries', 
        'alias' => 'Country', 
        'conditions' => array(
         'Trip.departure_country_id = Country.id' 
        ) 
       ) 
      ), 
       'conditions' => array(
        'Country.country_name LIKE' => "%$country%" 
       ) 
      ) 
     ); 

Answer 2

Reise Modell:

$this->find('all', 
    array(
     'joins' => array(
      array(
       'table' => 'countries', 
       'alias' => 'Country', 
       'type' => 'INNER', 
       'conditions' => array(
        'Trip.departure_country_id = Country.id' 
       ) 
     ), 
      'conditions' => array(
       'Country.country_name Like' => "%$country%" 
     ) 
    ) 
);