3

我有一个在弹出窗口中打开的表单，所以我想通过 ajax 验证来验证我的表单，但是当我单击提交按钮时，我的页面会刷新，所以我没有收到任何验证错误

查看文件：

 <?php $form = ActiveForm::begin([
        'id' => 'signup-form',
        'enableAjaxValidation' => true,
        //'action' => Url::toRoute('user/ajaxregistration'),
        'validationUrl' => Url::toRoute('user/ajaxregistration')

]); ?>

 <div class="col-md-12">
                    <div class="formbox">
                      <div class="inputbox signup">
                        <div class="input-group"> <span class="input-group-addon"><i class="glyphicon name"></i></span>
                          <?= Html::textInput('userFullName', '', ['placeholder' => "Name",'class'=>'form-control']); ?>
                        </div>

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

控制器文件：

public function actionValidate() {

    $model = new SignupForm;

    if (Yii::$app->request->isAjax && $model->load(Yii::$app->request->post())) {
        Yii::$app->response->format = Response::FORMAT_JSON;
        return ActiveForm::validate($model); \Yii::$app->end();

    }


}

型号代码：

 return [

            ['userFullName', 'trim'],
            ['userFullName', 'required'],


        ];

请建议我应该怎么做才能使我的页面不会被刷新并且我会收到验证错误

4

2 回答 2

7

您正在使用没有任何 ActiveField 的 ActiveForm，这意味着在模型中定义的验证规则甚至没有分配给文本输入。我已经为您的问题编写了一个实现：

模型：

use Yii;
use yii\base\Model;

class ExampleForm extends Model
{
    // this 'virtual attribute' defines a form field
    public $userFullName;

    // validation rules
    public function rules()
    {
        return [
            ['userFullName', 'trim'],
            ['userFullName', 'required'],
        ];
    }

}

控制器：

use models\ExampleForm;
use yii\web\Response;
use yii\widgets\ActiveForm;

public function actionExample()
{
    $model = new ExampleForm;

    // validate any AJAX requests fired off by the form
    if (Yii::$app->request->isAjax && $model->load(Yii::$app->request->post())) {
        Yii::$app->response->format = Response::FORMAT_JSON;
        return ActiveForm::validate($model);
    }

    if ($model->load(Yii::$app->request->post())) {

        // code to process form data goes here.. This will execute on a form submission.

    }

    return $this->render('example-form', [
        'model' => $model,
    ]);
}

看法：

<?php
use yii\widgets\ActiveForm;
use yii\helpers\Html;

$this->title = 'Example';
?>
<div class="exampleForm">

    <h1><?= Html::encode($this->title) ?></h1>

    <p>Please fill out the form.</p>

    <!-- this form will submit via POST to the same action that renders it -->
    <?php $form = ActiveForm::begin() ?>

        <!-- this is an active field. Any validation rules for the 'userFullName' field -->
        <!-- that have been defined within the $model object will be applied to this field -->
        <!-- the validation has also been set to validate via ajax within the $options array -->
        <!-- read more about ActiveFields here: http://www.yiiframework.com/doc-2.0/yii-widgets-activefield.html -->
        <?= $form->field($model, 'userFullName', ['enableAjaxValidation' => true]) ?>

        <div class="form-group">
            <?= Html::submitButton('Submit!', ['class' => 'btn btn-primary']) ?>
        </div>

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

</div>

查看是否正在发送 ajax 请求/是否正在验证表单的最佳方法是检查 chromes 开发人员工具，转到网络选项卡并检查活动。

于 2016-11-10T23:53:17.917 回答
1

使用 renderAjax() 方法：

 if (Yii::$app->request->isAjax && $model->load(Yii::$app->request->post())) {
        Yii::$app->response->format = Response::FORMAT_JSON;
        return ActiveForm::validate($model);

 }else {
            return $this->renderAjax('YourViewFileName', [
                'model' => $model,
            ]);
        }
于 2016-11-08T17:58:16.930 回答