undefined variable model in yii - php

Question

I am getting the error undefined variable model. I have the following codes in my view and controller. My aim is to search the results and show it in same page. I have tried the following things but its not working.
Again,I this this the conventional yii method to do so,or do I have to use a search() function from model SearchEmployee().?
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));
}
In View Section: search.php//To display and search the desired result
<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 ;<?php echo $form->labelEx('VIew Details'); ?></h5>
</div>
</div>
In view section I am using the search and view results option in the same form.I am getting the error after clicking the search button.Is this the correct way to do

Answer 1

$model = SearchEmployee::model()->find(array(
'select' => array('*'), "condition" => "category_id=$category AND
key_skills like '%$skills%' AND experience=$experience",
));
There is error here. You can't use find in that way.
public function find($condition='',$params=array())
$model = SearchEmployee::model()->find("category_id=$category AND
key_skills like '%$skills%' AND experience=$experience");
But better
$model = SearchEmployee::model()->find("category_id=:category AND key_skills like :skill AND experience=:experience", array(
'category'=>$category,
'skill'=>'%'.$skills.'%',
'experience'=>$experience
));

Answer 2

Remove this line:
$this->render('search',$model);
First it's invalid because it should be:
$this->render('search', array('model' => $model));
But also it's unnecessary because you already have it lower in your code.