“操作”列中的Yii2 Gridview下拉按钮

Question

我想将动作列按钮设置为网格视图中的下拉按钮，这是我的代码

// ... GridView configuration ...
['class' => 'yii\grid\ActionColumn',               
            'template' => '{sell} {delete}',
            'buttons' => [
                'sell' => function ($url, $model) {
                    return Html::a('<button type="button" class="btn btn-success"><i class="glyphicon glyphicon-shopping-cart"></i></button>', $url, [
                                'title' => Yii::t('app', 'Sell Tickets'),
                                'data-toggle' => "modal",
                                'data-target' => "#myModal",
                                'data-method' => 'post',
                    ]);
                },
                'delete' => function ($url, $model) {
                    return Html::a('<button type="button" class="btn btn-danger"><i class="glyphicon glyphicon-remove-sign"></i></button>', $url, [
                                'title' => Yii::t('app', 'Delete'),
                                'data-toggle' => "modal",
                                'data-target' => "#myModal",
                                'data-method' => 'post',
                    ]);
                },
            ],
            'urlCreator' => function ($action, $model, $key, $index) {
                if ($action == 'sell') {
                    $url = Url::toRoute(['trip/sell', 'id' => $model->tripScheduleId]);
                    return $url;
                } else {
                    $url = Url::toRoute(['trip/delete', 'id' => $model->tripScheduleId]);
                    return $url;
                }
            },
        ],

这是视图

我追踪了许多此类或任何来源的信息，但对我无济于事。

Answer 1

buttons是一组渲染函数，只需在此处渲染下拉菜单并删除urlCreator部分即可。

urlCreator用于为模板中的默认按钮构建URL，因此通过使用正确的URL渲染下拉列表，您将不需要它。

'template' => '{actions}',
'buttons' => [
    'actions' => function ($url, $model) {
        //return you dropdown here
    }
],

仅需关心$url参数并在函数内创建URL。