remove button from CGridView with condition - yii

Question

Hi I have CRUD generated CGridView in yii. I need to add a new button to CGridView rows and hide it if appointment_status(one of CGridView column) value equals 0
This is my code of CGridView,
$this->widget('zii.widgets.grid.CGridView', array(
'id' => 'bookings-grid',
'dataProvider' => $model->search(),
'filter' => $model,
'columns' => array(
'id',
'name',
'email',
'telephone',
'time',
'employee',
'appointment_status',
'client_ip',
'link' => array(
'header' => 'Confirmation',
'type' => 'raw',
'value' => 'CHtml::button("$data->appointment_status",array("onclick"=>"document.location.href=\'".Yii::app()->controller->createUrl("controller/action",array("id"=>$data->id))."\'"))',
'visible'=>$data->appointment_status==1,
),
array(
'class' => 'CButtonColumn',
),
),
));
But all I'm getting is error stating,
Undefined variable: data
It would be great help if someone can look into it.

Answer 1

you can use like this:
$this->widget('zii.widgets.grid.CGridView', array(
'id' => 'bookings-grid',
'dataProvider' => $model->search(),
'filter' => $model,
'columns' => array(
'id',
'name',
'email',
'telephone',
'time',
'employee',
'appointment_status',
'client_ip',
'link' => array(
'header' => 'Confirmation',
'type' => 'raw',
'value' => function ($data) {
if ($data->appointment_status == 1) {
return CHtml::button("$data->appointment_status", array("onclick" => "document.location.href=\'" . Yii::app()->controller->createUrl("controller/action", array("id" => $data->id)) . "\'"));
} else {
return;
}
}
),
array(
'class' => 'CButtonColumn',
),
),
));

Answer 2

Your 'visible' handling the column visibility and not the button, you can use custom attribute on model to create and handle the button visibility.
add to your model:
public function getConfirmationButton()
{
if ($data->appointment_status == 1) {
return CHtml::button($this->appointment_status,array("onclick"=>"document.location.href=\'".Yii::app()->controller->createUrl("controller/action",array("id"=>$this->id))."\'"));
} else {
return '';
}
}
and call it in your view:
..........
'link' => array(
'header' => 'Confirmation',
'type' => 'raw',
'value' => '$data->confirmationButton',
),
...........

Answer 3

visible is a boolean or a PHP expression that will be evaluated to give a boolean. During the evaluation $data is assigned to the current item from the dataProvider used. $data doesn't exist outside of the evaluation function evaluateExpression(). As such the implementation should be:
`visible` => '$data->appointment_status == 1',

Answer 4

You need to quote value of visible key in link array. So instead of this:
'visible'=>$data->appointment_status==1
It should be:
'visible'=>'$data->appointment_status==1'
it should work now.

Answer 5

You will get undefined variable because visible not allow any callback.
Try this solution, it's yii2 code and i don't know much of Yii.
'delete' => function ($url, $model) {
return ($model->isVisible($model)) ?
Html::a('<span class="glyphicon glyphicon-trash"></span>',
$url,
['title' => Yii::t('app', 'Delete')]) : '';
public static function isVisible($data)
{
return ($data->appointment_status == 1) ? true : false;
}