ActiveForm 和 GridView 上的 Yii2 Pjax [开始工作]

Question

我有一个网格视图

<?php Pjax::begin(['id' => 'members']) ?> 

<?= GridView::widget([
        'dataProvider' => $dataProvider,
        // 'filterModel' => $searchModel,
        'columns' => [
        ['header'=>'Remove Member',
                        'value'=> function($data)
                        { 
                            return  Html::a(Yii::t('app', ' {modelClass}', [
                                'modelClass' =>'',
                            ]), ['members/stl_remove','id'=>$data->id], 
                            ['class' => 'btn btn-link fa fa-times fa-2x pop']
                            );  
                        },
            'format' => 'raw'
            ],
      ['class' => 'yii\grid\ActionColumn','template'=>'{update}'],
        ],
    ]); ?>
    <?php Pjax::end() ?>

当我单击“删除成员”时 link 出现模式弹出窗口 (delete.php)

<?php Pjax::begin(['id' => 'delete_members']) ?>

    <?php $form = ActiveForm::begin(['options' => ['data-pjax' => true ]]); ?> 

    <div class="row">
        <div class="col-md-5">

            <?php echo $form->field($model, 'remarks')->dropDownList(['death' => 'Death',
                'marriage' => 'Marriage', 'house_shift' => 'House Shift'],['prompt'=>'Select'])
                ->label("Reason For Removal"); ?>

        </div>
    </div>
    <div class="form-group">
        <?= Html::submitButton($model->isNewRecord ? Yii::t('app', 'Remove') : Yii::t('app', 'Update'), ['class' => $model->isNewRecord ? 'btn btn-success' : 'btn btn-primary']) ?>
    </div>

    <?php ActiveForm::end(); ?>
    <?php yii\widgets\Pjax::end() ?>

现在这些是我的控制器代码 [代码]

public function actionStl_remove($id,$ajax=0)
    {
        $model              =   $this->findModel($id);
        $model_delete       =   new Members();
        $model->scenario    =   "remove";

        if ($model_delete->load(Yii::$app->request->post())) 
        {
            $model->remarks =   $model_delete->remarks;
            $model->status  =   1;
            $model->save();
            // $model = new Members();
            // return $this->redirect(['index']);
        }
        else
        {
        return $this->renderAjax('delete', [
                'model' =>  $model_delete,
                'id'    =>  $id,
            ]);
        }
    }

我无法将数据加载到 gridview 中。我单击 link 模态弹出窗口，然后提交，然后进入索引页面，然后 页面重新加载

Answer 1

我就是这样工作的。在 delete.php 中，我进行了以下更改。

<?php $form = ActiveForm::begin([
    'options' => ['data-pjax' => true,
                  'id'=> 'dynamic-form111',
                  // 'validationUrl' => 'validation-rul'
                   ]]); ?> 

    <div class="row">
        <div class="col-md-5">

            <?php echo $form->field($model, 'remarks')->dropDownList(['death' => 'Death',
                'marriage' => 'Marriage', 'house_shift' => 'House Shift'],['prompt'=>'Select'])
                ->label("Reason For Removal"); ?>

        </div>
    </div>
    <div class="form-group">
        <?= Html::submitButton($model->isNewRecord ? Yii::t('app', 'Remove') : Yii::t('app', 'Update'), ['class' => $model->isNewRecord ? 'btn btn-success' : 'btn btn-primary']) ?>
    </div>

    <?php ActiveForm::end(); ?>

基本上我给了来自 dynamic-form111 的一个 ID。然后在 index.php 以及 pjax 中的 gridview 我添加了以下代码

<?php
$this->registerJs(
    'jQuery(document).ready(function($){
                $(document).ready(function () {
                    $("body").on("beforeSubmit", "form#dynamic-form111", function () {
                        var form = $(this);
                        // return false if form still have some validation errors
                        if (form.find(".has-error").length) 
                        {
                            return false;
                        }
                        // submit form
                        $.ajax({
                            url    : form.attr("action"),
                            type   : "post",
                            data   : form.serialize(),
                            success: function (response) 
                            {

                                $("#ajaxModal").modal("toggle");
                                $.pjax.reload({container:"#countries"}); //for pjax update

                            },
                            error  : function () 
                            {
                                console.log("internal server error");
                            }
                        });
                        return false;
                     });
                    });

    });'
);
?>

通过 ajax 检查 delete.php 是否已成功提交，如果是，则使用 pjax 重新加载 gridview 并关闭模式。

ActiveForm 和 GridView 上的 Yii2 Pjax [开始工作]

Yii2 Pjax on ActiveForm and GridView [ Got it Working]

pjax

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