Question

在多步骤表单的步骤2中，如果选择了单选按钮并且＆＃34;转到步骤3＆＃34;单击或如果没有选择单选按钮，＆＃34;转到步骤3＆＃34;单击它始终显示以下错误：

{message: "", exception: "Symfony\Component\HttpKernel\Exception\HttpException",…}
exception"Symfony\Component\HttpKernel\Exception\HttpException"
file "/Users/johnw/projects/store/vendor/laravel/framework/src/Illuminate/Foundation/Exceptions/Handler.php" line : 203 message: "" trace:[{,…},…]

你知道问题出在哪里吗？

第2步的Html：

<form method="post" id="step2form" action="">
    <h6>Payment method</h6>
    <div class="form-group">
        <div class="form-check">
            <input class="form-check-input" type="radio" name="payment_method" value="option1">
            <label class="form-check-label d-flex align-items-center" for="exampleRadios1">
                <span class="mr-auto">Payment method 1</span>
            </label>
        </div>
        <br>
        <div class="form-check">
            <input class="form-check-input" type="radio" name="payment_method" value="option1">
            <label class="form-check-label d-flex align-items-center" for="exampleRadios1">
                <span class="mr-auto">Payment method 2</span>
            </label>
        </div>
    </div>
    <br>
    <div class="text-right">
    <button type="button" href="#step3" data-toggle="tab" role="tab"
            class="btn btn-outline-primary prev-step">
        Go back to step 2
    </button>
        <input type="submit" href="#step2" id="goToStep3"
         class="btn btn-primary btn float-right next-step"
               value="Go to step 3"/>
    </div>
</form>

用于处理第2步的ajax请求的PaymentController方法：

public function storePaymentMethods(Request $request, $id, $slug = null, Validator $validator){

        $validator = Validator::make($request->all(),[
            'payment_method' => 'required',
        ]);

        if($validator->passes())

        {
            return response()->json([
                'success' => true,
                'message' => 'success'
            ], 200);
        }
        $errors = $validator->errors();
        $errors =  json_decode($errors);

        return response()->json([
            'success' => false,
            'errors' => $errors
        ], 422);
    }

AJAX：

   $('#goToStep3').on('click', function (event) {
              event.preventDefault();

              var custom_form = $("#" + page_form_id);

              $.ajax({
                  method: "POST",
                  url: '{{ route('products.storePaymentMethods', compact('id','slug') ) }}',
                  data: custom_form.serialize(),
                  datatype: 'json',
                  success: function (data, textStatus, jqXHR) {
                  },
                  error: function (data) {

                      var errors = data.responseJSON;
                      var errorsHtml = '';
                      $.each(errors['errors'], function (index, value) {
                          errorsHtml += '<ul class="alert alert-danger mt-3"><li class="text-danger">' + value + '</li></ul>';
                          console.log(errorsHtml);
                      });

                      $('#response').show().html(errorsHtml);
                  }
              });

Answer 1

您尚未在表单中添加任何csrf令牌。有几种方法可以像下面提到的方式那样做 -

在表单中添加{{ csrf_field() }}
或使用您的ajax请求添加令牌，例如_token:"{{ csrf_token() }}"

Symfony \ Component \ HttpKernel \ Exception \ HttpException- Ajax发布请求

1 个答案: