[FIXED] CakePHP 3.0: Get model data from two degree model of separation in pagination

Issue

Now I'm very tired so I can't think well. Sorry for anticipated if this a duplicate question and if my English is weird.

In CakePHP 3.0 I have the model "Asistencias". Is asosiated to "Ejecutivos". so, in asistencias table, I have de FK ejecutivo_id and works OK when i try to access data from the ejecutivos table because I made the correct declaration:

$this->paginate = [
            //'contain' => ['Sucursales', 'Cedentes', 'Ejecutivos'],
            'contain' => ['Ejecutivos'],
            'limit'=>100,
            'order'=>['Asistencias.entrada'=>'DESC', 'Ejecutivos.nombre'=>'DESC']               
    ];

As you can see, I've commented the line 2. I take out the asosiation of Cedentes and Sucursales from Asistencias.

Now, the FKs sucursal_id and cedente_id belongs to ejecutivos table, instead asistencias table. It's help me to make others tasks in CakePHP that doesn't interest to you.

My question is how I can access to Sucursales and Cedentes in an efficient way? I guess I can't call them from the contain in the paginate properties.

I want use them in my view like this:

//index.ctp
foreach($asistencias as $asistencia){
  echo $asistencia->ejecutivos->name;
  echo $asistencia->ejecutivos->sucursal->name;
  echo $asistencia->ingreso;
  echo $asistencia->salida;
}

I just can't remember the correct way. Can you help me?

Solution

You can nest the contains to get your desired result.

$this->paginate = [
        'contain' => [
              'Ejecutivos' => [
                     'Sucursales', 'Cedentes', 
               ],
        ],
        'limit'=>100,
        'order'=>['Asistencias.entrada'=>'DESC', 'Ejecutivos.nombre'=>'DESC']               
];

Answered By - chrisShick