Question

in my YII form i have a dropdown depending on a table to another table which saves the id of the depended table like below

<div class="column">
    <?php echo $form->labelEx($model,'employee'); ?>
<?php echo $form->dropDownList($model,'employee',CHtml::listData(Employee::model()->findAllByAttributes(array('status'=>'Active')),'emp_id','emp_name'),array('empty'=>'select')); ?>
    <?php echo $form->error($model,'employee'); ?>
</div>

from above the form will save emp_id of the emp_name.in grid view also it showing same id only.but i just need to display emp_name of corresponding emp_id and should saved as id in database.anyone have idea post i am new to this thanks in advance

Answer 1

In your grid view

'columns'=>array(
              ....
//    'employee',
       array(
            'name'  => 'employee',
            'value' => 'Employee::model()->findByPk($data->employee)->emp_name',
            ),

And if relation to Employee model is defined in your model then you can use

'value'=>'$data->relation_name->emp_name'