Question

我收到错误undefined variable model。我在视图和控制器中有以下代码。我的目的是搜索结果并在同一页面中显示。我尝试过以下的东西，但它不起作用。

同样，我这是传统的yii方法，或者我必须使用模型SearchEmployee（）中的search（）函数。？

public function actionSearch()
{
      $model = new SearchEmployee();

       /*Getting Data From Search Form For Processing */
  if (isset($_POST['SearchEmployee'])) {

      $category   =  $_POST['SearchEmployee']['category_id'];
      $skills     =  $_POST['SearchEmployee']['skills'];
      $experience =  $_POST['SearchEmployee']['experience'];

      $model = SearchEmployee::model()->find(array(
      'select' => array('*'), "condition" => "category_id=$category AND key_skills like '%$skills%' AND experience=$experience",
      ));

      $this->render('search',$model);
  }

  /*Getting Data From Search Form For Processing */
  $this->render('search', array('model' => $model));
}

在视图部分：search.php //显示和搜索所需的结果

<div class="row">
   <?php echo $form->labelEx($model,'Category'); ?>
   <?php echo $form->dropDownList($model,'category_id',$list,array('empty' =>'(Select a Category')); ?>
   <?php echo $form->error($model,'category'); ?>
</div>

  <div class="row">
<?php echo $form->labelEx($model,'Experience'); ?>
    <?php echo $form->textField($model,'experience'); ?>
    <?php echo $form->error($model,'experience'); ?>    
</div>


<div class="row buttons">
  <?php echo CHtml::submitButton('Search'); ?>
</div>



<div class="view">

  <h1>Results </h1>

  <div class="view" id="id">

     <h1> Records Display </h1>

     <h4>Name:        <?php echo $form->labelEx($model,'name'); ?></h4>
     <h4>Skills:      <?php echo $form->labelEx($model,'experience'); ?></h4>
     <h4>Experience:  <?php echo $form->labelEx($model,'skills'); ?></h4>
     <h5>&nbsp;&nbsp ;<?php echo $form->labelEx('VIew Details'); ?></h5>
   </div>

 </div>

在视图部分，我使用相同表单中的搜索和查看结果选项。点击搜索按钮后出现错误。这是正确的方法

Answer 1

$model = SearchEmployee::model()->find(array(
                  'select' => array('*'), "condition" => "category_id=$category AND
         key_skills like '%$skills%' AND experience=$experience",
));

这里有错误。你不能以这种方式使用find。

public function find($condition='',$params=array())

$model = SearchEmployee::model()->find("category_id=$category AND
         key_skills like '%$skills%' AND experience=$experience");

但更好

$model = SearchEmployee::model()->find("category_id=:category AND key_skills like :skill AND experience=:experience", array(
    'category'=>$category,
    'skill'=>'%'.$skills.'%',
    'experience'=>$experience
));

Answer 2

删除此行：

$this->render('search',$model);

首先它无效，因为它应该是：

$this->render('search', array('model' => $model));

但也没必要，因为你的代码已经低了它。

yii中未定义的变量模型

2 个答案: