Question

在我的模态框中，我想检查唯一的电子邮件，这只能在提交表单后才能完成。当我触发提交按钮时，控制器操作将发送给我在控制器中创建操作，因为它将返回唯一的电子邮件地址错误。我希望在模式框中显示此错误。所以我使用了Ajax提交。以下是我的代码。

我的表单

<?php yii\widgets\Pjax::begin(['class' => 'new_projects']) ?>
<?php $form = ActiveForm::begin([
        'action' => ['create'],
        'options' => ['class' => 'form-horizontal disable-submit-buttons', 'id'=> 'projectsId'],
             'fieldConfig' => [
                    'template' => "<div class=\"row .field-register-email\">
                                        <div class=\"col-xs-6 .help-block\">{label}</div>\n<div class=\"col-xs-6 text-right\">{hint}</div>
                                    \n<div class=\"col-xs-12 \">{input}</div>
                                    </div>",
                              ],
        ]); ?>

 <?= $form->errorSummary($model, $options = ['header'=>'']); ?>

            <?= $form->field($model, 'user_fname')->textInput(['class'=>'form-control']);?>
            <?= $form->field($model, 'user_lname')->textInput(['class'=>'form-control']); ?>
            <?= $form->field($model,'user_email',['enableAjaxValidation'=> 'true'])->textInput(['class'=>'form-control']); ?>
            <?= $form->field($model, 'user_password_hash')->passwordInput([]); ?>
            <?= $form->field($model, 'user_password_hash_repeat')->passwordInput(['class'=>'form-control']); ?>
            <?php  $items = ArrayHelper::map(SimAuthAssignment::find()->all(),'item_name' ,'item_name');?>
            <?= $form->field($mode, 'item_name')->dropDownList($items,[]);?>




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

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

我的控制器创建操作

 public function actionCreate()
    {
       $model = new SimUser(['scenario' => SimUser::SCENARIO_CREATE]);
       $mode = new \app\modules\auth\models\SimAuthAssignment();

        if ($model->load(Yii::$app->request->post()) && $mode->load(Yii::$app->request->post())){ 
            \Yii::$app->response->format = Response::FORMAT_JSON;
            $model->setPassword($model->user_password_hash);
            $model->generateAuthKey(); 
            $model->company_id = Yii::$app->user->identity->company_id;
            if (!$model->save()){
                ($model->load(Yii::$app->request->post()) && $model->validate());
            }
            else{
            $auth = Yii::$app->authManager;
            $authorRole = $auth->getRole($mode->item_name);
            $auth->assign($authorRole, $model->getId());
             echo BaseJson::encode(array(
                "status"=>true,
                "message"=>"Created Project $model->name Successfully",
            ));
            }
        //    return $this->redirect(['index']);
        }  
            return $this->renderAjax('create', [
                'model' => $model,
                'mode' => $mode,
            ]);

    }

Js档案

$(document).on("beforeSubmit", "#projectsId", function () {
    // submit form
    $.ajax({
        url    : $("#projectsId").attr('action'),
        type   : 'post',
        data   : $("#projectsId").serialize(),
        success: function(response)
        {
            alert("success");
            $.pjax.reload({container:"#projectsGrid"});
            $('#modal').modal("hide");
        },
        error  : function (response) 
        { 
            alert("failure");
            console.log('internal server error');
        }
        });
        return false;
        // Cancel form submitting.

}）;

我如何在模态框中出错。即使有错误，也会触发警报（＆＃34;成功＆＃34;）。如果没有唯一的电子邮件，它就完美无缺。我想弄清楚如何让错误到模态框。感谢!!!

Answer 1

在您的控制器中更改代码，如下所示：

 $conn = new mysqli($servername, $username, $password, $dbname);

     if ($conn->connect_error) {
         die("Connection failed: " . $conn->connect_error);
     } 

     $sql = "select * from user where email= '$email'";
     $result = $conn->query($sql);

     if ($result->num_rows > 0) {

          while($row = $result->fetch_assoc()) {
              echo "<br> id: ". $row["id"]. " - Email: ". $row["email"] . "<br>";
          }
     } else {
          echo "0 results";
     }

和你的js文件添加yii表单验证脚本：

use yii\bootstrap\ActiveForm;

public function actionCreate()
    {
       $model = new SimUser(['scenario' => SimUser::SCENARIO_CREATE]);
       $mode = new \app\modules\auth\models\SimAuthAssignment();

        if ($model->load(Yii::$app->request->post()) && $mode->load(Yii::$app->request->post())){ 
            \Yii::$app->response->format = Response::FORMAT_JSON;
            $model->setPassword($model->user_password_hash);
            $model->generateAuthKey(); 
            $model->company_id = Yii::$app->user->identity->company_id;
            if ($model->validate()){
            $model->save();
            $auth = Yii::$app->authManager;
            $authorRole = $auth->getRole($mode->item_name);
            $auth->assign($authorRole, $model->getId());
             echo BaseJson::encode(array(
                "status"=>'true',
                "message"=>"Created Project $model->name Successfully",
            ));
            }
            else{

                $model=ActiveForm::validate($model);
                return['status'=>'false','errors'=>$model];

            }
        //    return $this->redirect(['index']);
        }  
            return $this->renderAjax('create', [
                'model' => $model,
                'mode' => $mode,
            ]);

    }

了解更多参考资料https://github.com/samdark/yii2-cookbook/blob/master/book/forms-activeform-js.md

模态框内的Yii2服务器端Ajax验证

1 个答案: