customization in search

this is my code:


/* @var $this BloggerController */

/* @var $model Blogger */





Yii::app()->clientScript->registerScript('search', "



	return false;


$('.search-form form').submit(function(){

	$('#blogger-grid').yiiGridView('update', {

		data: $(this).serialize()


	return false;





<?php echo CHtml::link('Search','#',array('class'=>'search-button')); ?>

<div class="search-form" style="display:none">

<?php $this->renderPartial('_search',array(


)); ?>

</div><!-- search-form -->

<?php $this->widget('bootstrap.widgets.TbGridView', array(






		'blog_name' ,












)); ?>

I have link to search :

<?php echo CHtml::link('Search','#',array('class'=>'search-button')); ?>

but i want that search option directly on page i dont want link how can i do that.

second one i want link on active record for example if i click on blogger name his page should can i gave the link.

  1. this link is just triggering show() for search form it’s on page already.

  2. instead of just '‘blogger_name’, U have to to something like:


    'name' =&gt; 'blogger_name',
    'type' =&gt; 'raw',
    'value' =&gt; '&lt;a href=&quot;Yii::app()-&gt;createUrl(blogger_url)&quot;&gt;&#036;data-&gt;blogger_name&lt;/a&gt;'


please check this

Thank You

I understand the 2nd.

But in first i dont want that link.

I just want to show that page which is on link.

this is container for search form, just remove "style="display:none;".

Than it will be shown on page, and after U can remove link;

<div class="search-form" style="display:none">

<?php $this->renderPartial(’_search’,array(


)); ?>

</div><!-- search-form -->

very very thank you

But sorry

2nd problem is not solving i want to give link in blogger name for example blogger name is amitesh if i click on amitesh it will open amitesh profile.

I get it its working thank you very much