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Php Yii:无法在网格视图中列出数据_Php_Yii - Fatal编程技术网

Php Yii:无法在网格视图中列出数据

Php Yii:无法在网格视图中列出数据,php,yii,Php,Yii,我不能使用yii框架在网格中列出数据。我的控制器是Sitecontroller.php,我的视图是list\u jobseker.php。我遇到错误“致命错误:调用非对象上的成员函数getData()”。有人帮我吗 我的控制器代码 public function actionlist_jobseeker() { $session_id=Yii::app()->session['user_id']; if ($session_id == "") {

我不能使用yii框架在网格中列出数据。我的控制器是Sitecontroller.php,我的视图是list\u jobseker.php。我遇到错误“致命错误:调用非对象上的成员函数getData()”。有人帮我吗

我的控制器代码

  public function actionlist_jobseeker()
  {
  $session_id=Yii::app()->session['user_id']; 
  if ($session_id == "")
    {
        $this->redirect( array('/employee/site/login'));
    }
  $user_id  =$session_id;
  $models = Yii::app()->db->createCommand()
     ->select('*')
     ->from('job_seeker_profile s')
     ->join('job_profile j','s.user_id = j.user_id')
     ->order('s.id')
     ->queryAll();
     $this->render('list_jobseeker',array('models' =>$models));
}

查看-list_jobseker.php

<h1>View Jobseeker</h1>

<div class="flash-success">

</div>

 <div class="form">
 <?php
  $this->widget('zii.widgets.grid.CGridView', array(
    'id'=>'rates-phase-grid',
   'htmlOptions' => array('class' => 'table table-striped table-bordered table-hover'),
    'dataProvider'=>new CArrayDataProvider($models),
    'columns' => array(
        array(
            'name' => 'Name',
            'type' => 'raw',
            'value' => 'CHtml::encode($data[*]->name)',
            'htmlOptions' => array('style'=>'width:90px;','class'=>'zzz'),

     ),
        array(
            'name' => 'Email',
            'type' => 'raw',
            'value' => 'CHtml::encode($data[*]->email)',
            'htmlOptions' => array('style'=>'width:250px;','class'=>'zzz')

        ),
        array(
            'name' => 'Password',
            'type' => 'raw',
            'value' => 'CHtml::encode($data[*]->password)',
            'htmlOptions' => array('style'=>'width:90px;','class'=>'zzz')
        ),
        array(
            'name' => 'Contact No',
            'type' => 'raw',
            'value' => 'CHtml::encode($data[*]->contact_no)',
            'htmlOptions' => array('style'=>'width:40px;','class'=>'zzz')

        ),
         array(
            'name' => 'Gender',
            'type' => 'raw',
            'value' => 'CHtml::encode($data[*]->gender)',
            'htmlOptions' => array('style'=>'width:40px;','class'=>'zzz')

        ),


        array(
            'class' =>'CButtonColumn',
            'deleteConfirmation'=>'Are you sure you want to delte this item?',
            'template'=>'{update}{delete}',

            'buttons' =>array('update'=>array(

             'label'=>'edit',
      'url'=>'Yii::app()->controller->createUrl("UpdateJob",array("id"=>$data["id"]))',

                ),
             'delete'=>array('label'=>'delete',
     'url'=>'Yii::app()->controller->createUrl("DeleteJob",array("id"=>$data["id"]))'),


                )
            )
          ),
   ));
  ?>
查看求职者

dataProvider
应该是一个实例。您正在传入一个数组:
$model
。您可以将其包装为:

其他几个问题:

a) gridview希望
$data
是一个对象而不是数组。将所有
$data->*
实例更改为
$data[*]
或使用
CActiveDataProvider
CActiveRecord
实例作为
求职者档案

b) 按照惯例,
$model
通常指单个
CModel
实例。因此,您的数组应以复数形式命名,例如
$items

我使用以下名称:

    $data = ...::model()->findAll();
    $dataProvider=new CArrayDataProvider('Class of your object');
    $dataProvider->setData($data);
    $dataProvider->keyField = "primary key of your model";


    $this->widget('zii.widgets.grid.CGridView', array(
    'dataProvider'=>$dataProvider,
...
array( 'class'=>'CButtonColumn',
                    'template' => '{ver}',
                    'buttons' => array(
                        'ver' => array(
                            'url'=>'"?id=".$data["..."]',
                        'imageUrl'=> Yii::app()->baseUrl.'/images/view.png',  //Image URL of the button.
                    ),
                )
            ),

您好,现在我遇到了错误“尝试获取非对象C:\wamp\www\yii\u new\framework\base\CComponent.php(612):eval()'d code(1)”您使用的网格错误。我添加了一些问题。在C:\wamp\www\yii\u new\framework\base\CComponent.php(612):eval()'d code on line 1)中出现错误“Parse error:syntax error,意外的'*',应为'].”此外,我已经编辑了视图和控制器页面,请查看。您还没有阅读手册。答案在评论中:和可能的重复
    $data = ...::model()->findAll();
    $dataProvider=new CArrayDataProvider('Class of your object');
    $dataProvider->setData($data);
    $dataProvider->keyField = "primary key of your model";


    $this->widget('zii.widgets.grid.CGridView', array(
    'dataProvider'=>$dataProvider,
...
array( 'class'=>'CButtonColumn',
                    'template' => '{ver}',
                    'buttons' => array(
                        'ver' => array(
                            'url'=>'"?id=".$data["..."]',
                        'imageUrl'=> Yii::app()->baseUrl.'/images/view.png',  //Image URL of the button.
                    ),
                )
            ),