Reputation: 3547
I have this widget:
<?php
class Search extends CWidget
{
public $dataProvider = null;
public function init()
{
$criteria = new CDbCriteria();
if ( isset($_GET['file']) ) {
$criteria->compare('fileName', $_GET['file'], true, 'OR');
$criteria->compare('tags', $_GET['file'], true, 'OR');
}
$this->dataProvider = new CActiveDataProvider("Files", array(
'criteria' => $criteria,
//'countCriteria' => $criteria,
'pagination'=>array(
// results per page
'pageSize'=>1,
),
));
}
public function run(){
$this->render('site/result', array(
'dataProvider' => $this->dataProvider,
'pages' => $this->dataProvider->pagination,
));
}
}
?>
how I can run this widget only when the search button is clicked, and how to view its result in the view reuslt
that is located in the folder site
Upvotes: 0
Views: 654
Reputation: 3559
Maybe I misunderstood you, but I think you have a little confusion about widget. Avoid making a
lengthy answer, I made a model in below
Question 1: How I can run this widget only when the search button is clicked?
The idea is that you put the widget content into hidden div, and then add a jQuery script to show it after the search button is clicked
<div id='search-result' style="display:none"><?php $this->widget(...) ?></div>
<script>
$('#search-button-id').click({
// do something
$('#search-result').show();
})</script>
Question 2: how to view its result in the view reuslt that is located in the folder site?
Imao, I recommend you don't do this. The widget is independant part, you can use it for many specific purposes & many locations of your project, so just put a simple view for widget to render in its own, instead of trying to render a view outside widget's scope. It should look like below
/your-app/protected/extensions/widgets/search/Search.php
/your-app/protected/extensions/widgets/search/views/result.php // view for rendering
Upvotes: 2