条件分页 cakephp

Question

我试过了

public function index($id = null)
{
    $this->paginate = [
        'contain' => ['Photos'],
    ];
   // $tareas = $this->paginate($this->Tareas);

    $this->loadModel('Categorias');

    if(isset($id)){
        $tareas2 = $this->Tareas->find()->where(['categoria_id'=>$id])->all();
        $tareas = $this->paginate($tareas2);
    }
    else
    {
        $tareas = $this->paginate($this->Tareas);
    }

        $categorias = $this->Categorias->find()->all();



    $this->set(compact('tareas','categorias'));
}

在不传递 id 时工作，但当我传递 id 时出现此错误

调用未定义的方法 Cake\ORM\ResultSet::getAlias()

Answer 1

分页器只接受表名、表实例 (\Cake\Datasource\RepositoryInterface) 和查询实例 (\Cake\Datasource\QueryInterface)。

不要在查询对象上调用all()，这将执行查询并返回结果集，即\Cake\ORM\ResultSet 的实例，而是传递查询对象。

另见

• Cookbook > Controllers > Components > Pagination > Using Controller::paginate()

Answer 2

好的，我用这个解决了它

public function index($id = null)
{
    if(isset($id)){
        $this->paginate = [
            'contain' => ['Photos'],
            'conditions' => ['categoria_id'=>$id]
        ];
    }
    else{
        $this->paginate = [
            'contain' => ['Photos'],
        ];
    }
    $tareas = $this->paginate($this->Tareas);

    $this->loadModel('Categorias');

    $categorias = $this->Categorias->find()->all();


    $this->set(compact('tareas','categorias'));
}