如何在 YII2 中将带有单独 API URL 响应的 ajax 返回给控制器?

Question

我正在尝试使用 AJAX 调用 API URL。我需要验证响应并更新数据库，因此我需要将其返回给控制器。 有什么办法可以做到这一点。这是我的观点，JS和控制器代码。

这是我的视图代码，我有一个单独的 URL 用于验证，即 API URL 视图

<?php $form = ActiveForm::begin([
'action' => ['users/renderstep3'],
'validationUrl' => 'API URL',
'options' => [
'class' => 'comment-form'
]
]); ?>
<?= $form->field($paymentmodel, 'customerId')->hiddenInput(['value'=> $userid])->label(false) ?>
<?= $form->field($paymentmodel, 'owner')->textInput(['maxlength' => true]) ?>
<div class="form-group">
<?= Html::submitButton('Submit', ['class' => 'btn btn-success']) ?>
</div>
<?php ActiveForm::end(); ?>
</div>

.JS

jQuery(document).ready(function($) {
$('body').on('submit', '.comment-form', function(event) {
event.preventDefault(); // stopping submitting
var data = $(this).serializeArray();
data.splice(0,1);
var result = {};
for ( i=0 ; i < data.length ; i++)
{
key = data[i].name.replace("UserPaymentDetails[", "").slice(0,-1);
result[key] = data[i].value;
}
var url = $(this).attr('validationUrl');
$.ajax({
url: url,
type: 'post',
dataType: 'json',
data: JSON.stringify(result)
})
.done(function(response) {
return response;
})
.fail(function() {
console.log("error");
});
});
});

控制器操作

public function actionRenderstep3()
{
$model = new Users();
$detailsmodel = new UserDetails();
$paymentmodel = new UserPaymentDetails();
if (Yii::$app->request->isAjax) {
Yii::$app->response->format = yiiwebResponse::FORMAT_JSON;
$data = Yii::$app->request->post();
print_r($data) ; exit;
}
if ($paymentmodel->load(Yii::$app->request->post()) && $paymentmodel->validate()) 
{
$paymentmodel->Status = 0;
$paymentmodel->save();
return $this->redirect(['index']);
}
return $this->render('renderstep3', [
'model' => $model,
'detailsmodel' => $detailsmodel,
'paymentmodel' => $paymentmodel,
]);   }

提前感谢！！

Answer 1

在控制器中，您必须像这样更改操作才能使用 Ajax 进行验证。我已经编辑了我的答案。请注意，您可以删除自定义 js 代码以便像这样使用。

// ... The View file
<?php
$form = ActiveForm::begin([
'action' => ['users/renderstep3'],
'enableAjaxValidation' => true,
'validationUrl' => 'API URL',
'options' => [
'class' => 'comment-form'
]
]);
?>
<?= $form->field($paymentmodel, 'customerId')->hiddenInput(['value'=> $userid])->label(false) ?>
<?= $form->field($paymentmodel, 'owner')->textInput(['maxlength' => true]) ?>
<div class="form-group">
<?= Html::submitButton('Submit', ['class' => 'btn btn-success']) ?>
</div>
<?php ActiveForm::end(); ?>
// ... Controller
public function actionRenderstep3()
{
$model = new Users();
$detailsmodel = new UserDetails();
$paymentmodel = new UserPaymentDetails();
if (Yii::$app->request->isAjax && $paymentmodel->load(Yii::$app->request->post())) {
Yii::$app->response->format = yiiwebResponse::FORMAT_JSON;
return ActiveForm::validate($model);
}
if ($paymentmodel->load(Yii::$app->request->post())) {
$paymentmodel->Status = 0;
$paymentmodel->save(false); // Validate false, because we did the validation before
return $this->redirect(['index']);
}
return $this->render('renderstep3', [
'model' => $model,
'detailsmodel' => $detailsmodel,
'paymentmodel' => $paymentmodel,
]);
}

您可以在此处找到更多信息 https://www.yiiframework.com/doc/guide/2.0/en/input-validation

Answer 2

<?php
$form = ActiveForm::begin([
'action' => ['users/renderstep3'],
'validationUrl' => 'API URL',//ajax validation hit to validationUrl if provide other wise validationUrl is action Url
'options' => [
'class' => 'comment-form'
]
]);
?>

并更改JS中的一些代码 下面的代码调用befor 表单提交

$('body').on('beforeSubmit', '.comment-form', function(event) 

在控制器中

在单个模型验证的情况下

if ($paymentmodel->load(Yii::$app->request->post())) {
if (Yii::$app->request->isAjax) {
Yii::$app->response->format = yiiwebResponse::FORMAT_JSON;
return yiiwidgetsActiveForm::validate($model);
}
$paymentmodel->Status = 0;
if ($paymentmodel->save(false)) {
return $this->redirect(['index']);
}
}

在多个模型验证的情况下

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

if (Yii::$app->request->isAjax) {
Yii::$app->response->format = yiiwebResponse::FORMAT_JSON;
$return = yiiwidgetsActiveForm::validate($model);
$return = yiihelpersArrayHelper::merge(yiiwidgetsActiveForm::validate($detailsmodel), $return);
$return = yiihelpersArrayHelper::merge(yiiwidgetsActiveForm::validate($paymentmodel), $return);
return $return;
}
//here is data saving or logic
}