【发布时间】:2014-04-16 12:01:45
【问题描述】:
我正在工作门户网站,并有一个搜索模块供雇主搜索应用程序。我想在搜索后使用ajax列出应用程序,将结果列出在同一页面中。我应该怎么做。?,下面是模块的截图。尝试了以下的东西。
//控制器代码//
public function actionSearch()
{
$model = new SearchEmployee();
/*Getting Data From Search Form For Processing */
if (isset($_POST['SearchEmployee'])) {
$model->attributes = $_POST['SearchEmployee'];
$category = $_POST['SearchEmployee']['category'];
$skills = $_POST['SearchEmployee']['skills'];
$experience = $_POST['SearchEmployee']['experience'];
$ajaxmodel = SearchEmployee::model()->find(array(
'select' => array('*'), "condition" => "category_id=$category AND key_skills like'%$skills%'AND experience=$experience",
));
if($model==null)
{
Yii::app()->user->setFlash('success', "No Results");
$this->renderPartial('search');
}
else
{
$this->renderPartial('search', array('model' => $ajaxmodel));
Yii::app()->end();
}
}
// 在视图中,不贴完整代码只是显示ajax结果的代码,//
<div class="view">
<h1>Results </h1>
<div class="view" id="id">
<h1> Records Display </h1>
<h4>Name: <?php echo $form->labelEx($model,'skills Required'); ?></h4>
<h4>Skills: <?php echo $form->labelEx($model,'Skills Required'); ?></h4>
<h4>Experience: <?php echo $form->labelEx($model,'Skills Required'); ?></h4>
<h5> <?php echo CHtml::submitButton('VIew Details'); ?></h5>
</div>
</div>
这样可以吗...
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