条件分页cakephp

2024-05-15 • 问答

我尝试过

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时有效，但是在传递ID时出现此错误

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

分页器仅接受表名称，表实例（\Cake\Datasource\RepositoryInterface）和查询实例（\Cake\Datasource\QueryInterface）。

请勿在查询对象上调用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。

另请参见

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

好的，我为此解决了

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'));
}

条件分页cakephp

ysw9232 回答：条件分页cakephp

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