Question

我有一个ajax表单，它被提交两次，因此在db中复制了两次相同的记录。

仅在启用客户端验证时才会发生此特定情况。

我也尝试过使用e.stopImmediatePropagation（），但这也会禁用显示错误消息，这在我的情况下不是解决方案。我希望验证和表单提交都能正常工作而不需要复制。

这是我的代码：

控制器：

public function actionIndex() {

    $model = new Transaction();

    if (isset($_POST['ajax']) && $_POST['ajax'] === 'transaction-form') {
        echo CActiveForm::validate($model);

        Yii::app()->end();
    }

    if (Yii::app()->request->isAjaxRequest) {

        $model->attributes = $_POST['Transaction'];

        if($model->save()){
            echo 'Success';
        }

        Yii::app()->end();
    }

    $this->render('index', array('model' => $model));
}

查看文件：

<div class="form">
<?php
$form = $this->beginWidget('CActiveForm', array(
'id' => 'transaction-form',
'action' => Yii::app()->createUrl('transaction/index'),
'enableAjaxValidation'=>true,
'enableClientValidation'=>true,
'clientOptions' => array(
    'validateOnSubmit' => true,
),
    ));
?>


<p class="note">Fields with <span class="required">*</span> are required.</p>

<?php echo $form->errorSummary($model); ?>

<div class="row">
    <?php echo $form->labelEx($model,'BilledAmount'); ?>
    <?php echo $form->textField($model,'BilledAmount',array('size'=>10,'maxlength'=>10)); ?>
    <?php echo $form->error($model,'BilledAmount'); ?>
</div>

<div class="row">
    <?php echo $form->labelEx($model,'ChargedAmount'); ?>
    <?php echo $form->textField($model,'ChargedAmount',array('size'=>10,'maxlength'=>10)); ?>
    <?php echo $form->error($model,'ChargedAmount'); ?>
</div>

<div class="row">
    <?php echo $form->labelEx($model,'CardExpiry'); ?>
    <?php echo $form->textField($model,'CardExpiry',array('size'=>4,'maxlength'=>4)); ?>
    <?php echo $form->error($model,'CardExpiry'); ?>
</div>

<div class="row buttons">
     <input name="submit" type="submit" value="Submit">
</div>

<?php $this->endWidget(); ?>

</div><!-- form -->

<script>
  $(function () {

  $('form#transaction-form').on('submit', function (e) {

  e.preventDefault();

  var action = $("#transaction-form").attr('action');
  var datas = $("#transaction-form").serialize();

  $.ajax({
    type: 'post',
    url: action,
    data: datas,
    success: function (msg) {

      if(msg == 'Success'){
          location.reload();
      }
    }
  });


  return false;

});

Answer 1

您的行动应如下所示：

public function actionIndex() {
    $model = new Transaction();
    if (Yii::app()->getRequest()->getIsAjaxRequest()) {
        echo CActiveForm::validate($model);
        Yii::app()->end();
    }
    if (isset($_POST['Transaction'])) {
        $model->attributes = $_POST['Transaction'];
        if($model->save()){
        Yii::app()->user->setFlash('success', 'Transaction saved.');
                $this->refresh();
        }else{
            Yii::app()->user->setFlash('error', 'Transaction saving error.');
        }
    }
    $this->render('index', array('model' => $model));
}

Ajax表单在yii 1.1中提交两次

1 个答案: