Question

我是yii中的新手。我正在尝试从数据库中选择数据并列出它。我有控制器名称sitecontroller.php和查看页面search_result.php。我可以选择并列出没有条件的数据，但我想选择数据有条件

Controller - Sitecontroller.php

<?php
class SiteController extends Controller
   {
 public function actionsearch_result()
  {
    $title=$_GET['title'];
    $experience=$_GET['experience'];
    $criteria=new CDbCriteria();
$Criteria->condition = "title like '%$title%' or  key_skills like '%$title%'";
    $count=Job::model()->count($criteria);
    $pages=new CPagination($count);
    $pages->pageSize=2;
    $pages->applyLimit($criteria);
    $model=Job::model()->findAll($criteria);
    $this->render('search_result',array('model' =>$model,'pages' => $pages));

  }
}

＆GT;

我的观点文件 - search_result.php

<div style="float:right;margin-right:285px;">
 <h1>Search Results</h1>
<ul style="list-style:none; ">
 <?php
foreach($model as $models)  
  { 
 $job_id=$models->id;
 ?>
  <li><p><?php echo $models->title; ?></p>
  <p><?php echo $models->company_name   ; ?></p>
  <p><?php echo $models->description    ; ?></p>
  <p>Keyskill:<?php echo $models->key_skills    ; ?><p>
    </li>
   <?php
   }
   ?>
   </ul>    
   <p><?php $this->widget('CLinkPager', array(
   'pages' => $pages,
  )) ?></p>   
  </div>

有人帮帮我吗？

Answer 1

用我的代码替换您的代码 -

$Criteria->condition = "title like '%$title%' or  key_skills like '%$title%'";

$criteria->condition = "title like '%$title%' or  key_skills like '%$title%'";

取代＆＃34; C＆＃34;使用＆＃34; c＆＃34;。

我无法使用yii框架选择条件数据

1 个答案: