Question

如何将链接放在类代码中以继续进行另一种形式

<?php $this->widget('zii.widgets.grid.CGridView', array(
    'id'=>'classlist-grid',
    'dataProvider'=>$model->search(),
    'filter'=>$model,
    'columns'=>array(
        **'classCode'**,
        'classDesc',
        array(
            'class'=>'CButtonColumn',
        ),
    ),
)); ?>

Answer 1

您可以使用碰撞栏

'columns'=>array(
   .....
    array(
      'class'=>'CLinkColumn',
      'label'=>'your_attribute',
      'url'=>'your_controller/your_action&id='.$model->your_attribute,
     'header'=>'your_column_header'
  ),

Answer 2

您也可以使用此

array(
   'name' => 'you_field',
   'header' => 'label',
   'type' => 'raw',
   'value' => 'CHtml::link($data->you_field,Yii::app()->createUrl("controller/action",array("param"=>$data->you_field)),array("target"=>"_blank"))',
),

Answer 3

您好，请使用此代码，

 array(
                            'header' =>Your Header Name,
                            'value' => '$data->subject->subject_name."  -  ".$data->subject->subject_code',
'url' => 'Yii::app()->createUrl("core/subject/deleteelective", array("id"=>$data->electivesubjectid))',
                            'htmlOptions' => array('width' => '45%'),
                        ),

yii如何在cgridview中将链接放到另一个表单

3 个答案: