Yii 中的语句有什么问题? [英] What is wrong with the statement,in Yii?

问题描述

我的控制器中有以下代码，

I have a folowing code in my controller ,

   public function actionViewJob() {

        $user_id = Yii::app()->session['user_id'];
        /* For User Authentication */
        if (Yii::app()->user->getId() === null)
            $this->redirect(array('site/login'));
        /* For User Authentication */

         $model=ViewJob::model()->findAll(array('user_id'=>Yii::app()->user->id));
        $params = array('model' => $model,
        );

        $this->render('viewjob', $params);

我收到错误/* 未定义属性CDbCriteria.user_id".*/但是当我使用 findByAttrinute 它的工作 f9 时，但我没有得到结果，即它没有过滤数据，请帮忙.

I am getting error /* Property "CDbCriteria.user_id" is not defined. */ but when I am using findByAttrinute its working f9,but I not getting the result,i.e its not filtering data,pls help.

查看部分:

// not posting full code //


   <?php
        $this->widget('zii.widgets.grid.CGridView', array(
            'dataProvider' =>$model->search(),  

'columns' => array(
            array(
                'name' => 'title',
                'type' => 'raw',
                'value' => 'CHtml::encode($data->title)',
                'htmlOptions' => array('style'=>'width:90px;','class'=>'zzz'),
               // 'filter'=>'false' /* for hiding filter boxes */

         ),?>

推荐答案

错误的方法.您应该使用 CActiveRecord::findAllByAttributes() 代替

Wrong method. You should use CActiveRecord::findAllByAttributes() instead

$model=ViewJob::model()->findAllByAttributes(array('user_id'=>Yii::app()->user->id));

或者如果你仍然想使用 findAll 你应该把属性作为条件传入:

Or if you still want to use findAll you should pass the attribute in as a condition:

$model=ViewJob::model()->findAll('user_id=:user_id',array(':user_id'=>Yii::app()->user->id));

现在来看:

dataProvider 需要一个 CDataProvider 的实例.CActiveRecord::search() 返回一个这样的实例:CActiveDataProvider.但是，$model 是一个数组，而不是 CActiveRecord 的实例.您在这里有两个选择:

dataProvider expects an instance of CDataProvider. CActiveRecord::search() returns one such instance: CActiveDataProvider. However, $model is an array not an instance of CActiveRecord. You have two choices here:

a) 您可以编辑控制器以将 $model 用作 ViewJob 实例而不是 ViewJob 实例的数组:

a) You can to edit your controller to use $model as a ViewJob instance and not an array of ViewJob instances:

public function actionViewJob() {

    $user_id = Yii::app()->session['user_id'];
    /* For User Authentication */
    if (Yii::app()->user->getId() === null)
        $this->redirect(array('site/login'));
    /* For User Authentication */

    $model= new ViewJob;
    $model->user_id = $user_id;
    $params = array('model' => $model,
    );
    ...

b) 将视图中的 $model->search() 替换为

b) Replace the $model->search() in the view with

    'dataProvider' => new CArrayDataProvider($model)

选择 (b) 更容易，但它打破了约定，例如 $model 是一个数组而不是一个对象.此外，任何过滤器/搜索功能都必须手动添加，而不是使用 CActiveRecord::search() 提供的默认功能.

Choice (b) is easier but it breaks conventions e.g $model is an array not an object. Also any filters/search functionality has to be added manually as opposed to using the default functionality provided by CActiveRecord::search().

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