Yii2 Pjax无法正常工作

时间:2015-03-07 04:41:22

标签: yii2 pjax yii2-advanced-app

我想使用Pjax刷新gridview但不知何故它不起作用。这是代码:

_search.php

    <?php

use yii\helpers\Html;
use yii\widgets\ActiveForm;
use yii\widgets\Pjax;

$this->registerJs("
                  $('#btnAjaxSearch').click(function(){
                        $.ajax({
                                type: 'get',
                                data: $('.bank-search form').serializeArray(),
                                success: function (data) {
                                      $.pjax.reload({container:\"#bank\"});
                                },
                                error: function (XMLHttpRequest, textStatus, errorThrown) {
                                      alert('error');
                                }
                        });
                       return false;
                  });
                ", \yii\web\View::POS_END, 'bank-search');
?>

<div class="bank-search">
    <?php Pjax::begin(['id' => 'bank-form']); ?>
    <?php $form = ActiveForm::begin([
        'action' => ['index'],
        'method' => 'get',
    ]); ?>

    <?= $form->field($model, 'bank_name') ?>

    <?= $form->field($model, 'state') ?>

    <?= $form->field($model, 'district') ?>

    <?= $form->field($model, 'city') ?>

    <div class="form-group">
        <?= Html::Button('Search', ['class' => 'btn btn-primary','id' => 'btnAjaxSearch']) ?>
    </div>

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

</div>

的index.php

    <?php

use yii\helpers\Html;
use yii\grid\GridView;
use yii\widgets\Pjax; 

$this->title = 'Banks';
$this->params['breadcrumbs'][] = $this->title;
?>
<div class="bank-index">

    <h1><?= Html::encode($this->title) ?></h1>
    <?php  echo $this->render('_search', ['model' => $searchModel]); ?>

    <p>
        <?= Html::a('Create Bank', ['create'], ['class' => 'btn btn-success']) ?>
    </p>
    <?php Pjax::begin(['id' => 'bank']); ?>
    <?= GridView::widget([
        'dataProvider' => $dataProvider,
        'filterModel' => $searchModel,
        'columns' => [
            ['class' => 'yii\grid\SerialColumn'],

            'id',
            'bank_name',
            'state',
            'district',
            'city',
            // 'branch',

            ['class' => 'yii\grid\ActionColumn'],
        ],
    ]); ?>
    <?php Pjax::end(); ?>
</div>

控制器

 /**
     * Lists all Bank models.
     * @return mixed
     */
    public function actionIndex()
    {
        $searchModel = new BankSearch();
        $dataProvider = $searchModel->search(Yii::$app->request->queryParams);

        return $this->render('index', [
            'searchModel' => $searchModel,
            'dataProvider' => $dataProvider,
        ]);
    }

简单搜索正在运行,但Pjax不是。我是Yii2的新手,所以任何帮助都将不胜感激。谢谢。

2 个答案:

答案 0 :(得分:4)

谢谢爱丁。它帮助我解决了这个问题。这就是我做的。它可能会帮助面临同样问题的人。

正如Edin提到的,你需要将url和搜索参数一起传递给Pjax以刷新gridview。

以下是我编辑过的代码:

    $js = <<<JS
        // get the form id and set the event
        $('#bank-form-id').on('beforeSubmit', function(e) { 
           var form = $(this);
            if(form.find('.has-error').length) {
                return false;
            }
            $.ajax({
                url: form.attr('action'),
                type: 'post',
                data: form.serialize(),
                success: function(response) { 
                    var csrf = yii.getCsrfToken();
                    var bank_name = $('#banksearch-bank_name').val();
                    var state = $('#banksearch-state').val();
                    var district = $('#banksearch-district').val();
                    var city = $('#banksearch-city').val();
                    var url = form.attr('action')+ '&_csrf='+csrf+'&BankSearch[bank_name]='+bank_name+'&BankSearch[state]='+state+'&BankSearch[district]='+district+'&BankSearch[city]='+city;
                    $.pjax.reload({url: url, container:'#bank'});
                }
            });    
        }).on('submit', function(e){
        e.preventDefault();
    });
JS;
$this->registerJs($js);

答案 1 :(得分:2)

Pjax的工作方式是发送另一个带有特殊标头的请求。当检测到pjax请求时,只从服务器返回更新容器所需的html。线

$.pjax.reload({container:\"#bank\"});

将发送另一个请求,并且actionIndex queryParams内部将为空。

您可以通过将搜索参数存储到会话或通过在查询字符串中使用参数指定pjax url来解决此问题。

请尝试以下操作:

  var url = urlWithFilters(); 
  $.pjax({url: url, container: '#bank'});

在这种情况下,您不需要创建自己的ajax调用,只需使用过滤器创建网址。