Question

<?php $this->widget('zii.widgets.grid.CGridView', array(
'id'=>'message-grid',
    //'selectionChanged'=>'js:function(id){location.href = "'.$this->createUrl('view').'/id/"+$.fn.yiiGridView.getSelection(id);}', 
    'selectionChanged'=>'function(id){location.href = "'.$this->createUrl('view').'&id="+$.fn.yiiGridView.getSelection(id)+"&f=s";}',
//'dataProvider'=>$model->sentitem(),
    'dataProvider'=>$model->search(),    
'filter'=>$model,
    'columns'=>array(
            array(      
                        'class'=>'CCheckBoxColumn',
                        'id'=>'chk',
                        'selectableRows'=>'2',
                        //'htmlOptions'=>array('width'=>'40px'),

            ),

），））;

我已经编写了单击通过选择更改的代码我试图通过jquery dbclick函数访问id，我不知道如何使用dbclick中更改的选择来获取id

Answer 1

试试这个：

$('#message-grid tbody tr').live('dbclick', function() {
    var id = $.fn.yiiGridView.getKey(
        'message-grid',
        $(this).prevAll().length 
    );
});

Answer 2

下面的代码将加载项目的视图页面。（url =＆gt; itemcontroller / view / id / + selectedId）

<?php $this->widget('zii.widgets.grid.CGridView', array(
    'id'=>'id-grid',
    'dataProvider'=>$dataProvider,
    'filter'=>$model,

    'selectableRows'=>1,
    'selectionChanged'=>'function(id){ location.href = "'.$this->createUrl('view').'/id/"+$.fn.yiiGridView.getSelection(id);}',

如何通过双击访问Yii Cgridview中所选行的ID？

2 个答案: