Question

我试图通过ajax为我的Laravel应用程序提交我的3页表单中的第一步，并且我已将其设置为当此次提交的验证失败时，它将返回验证错误但是它仍然继续进行表单处理的下一步。我到底做错了什么呢？我知道，当ajax响应中存在错误时，我没有做任何事情，但为什么如果出现错误，它仍然会在成功中做些什么。它先检查错误然后检查成功吗？

/*
* Create User After they complete the first part of the form. 
*
*/
public function createUserFromOrder(Request $request)
{
    $validation = $this->validate($request, [
        'first_name' => 'required',
        'last_name' => 'required',
        'email' => 'required|confirm|unique:users,email',
        'email_confirmation' => 'required'
    ]);

    if ($validation->fails()) {
        return Response()::json([
            'success' => false,
            'errors' => $validation->errors()->toArray()
        ], 200);
    }

    $randomPassword = str_random(7);
    $createdSuccessfully = "<div class='alert alert-success' role='alert'><span class='status-available'> User Profile Created.</span></div>";
    $userData = [
        'email' => $request->email,
        'password' => $randomPassword,
        'first_name' => $request->first_name,
        'last_name' => $request->last_name,
    ];

    $user = Sentinel::registerAndActivate($userData);
    $role = Sentinel::findRoleByName('patient');
    $role->users()->attach($user);

    Sentinel::login($user, true);

    $order = Order::create([
        'program_id' => $request->program_id,
        'program_type_id' => $request->program_type_id,
        'amount' => 433,
        'order_type' => 0,
        'paid' => 0,
        'status' => 0
    ]);
    $user->complete($order);
    $order->addAddon($request->input('addons'));

    return $createdSuccessfully;
}


$.ajax({
        type: "POST",            
        url: '{{action('OrderProcessController@createUserFromOrder') }}',
        data:
            {
                email: email,
                first_name: first_name, 
                last_name: last_name,
                program_id: program_id,
                program_type_id: program_type_id,
                amount: amount,
                addons: addons
            },
        success:function(data){            
            $("#user-created-confirmation").html(data);
        },
        error:function (){}
    }, function(){
        setTimeout(function() {

        })
    });

Answer 1

我相信这是因为你可能误解了.ajax（）的“成功”属性它只是意味着AJAX请求成功，而不是天气你的表单通过验证。您应该将错误处理函数放在成功函数中，如下所示：

success:function(data){  
        if(AreThereErrors() == false) {
            $("#user-created-confirmation").html(data);
        } else {
            HandleErrors();
        }
    },

Answer 2

如果您的状态代码不是200，则会调用error ajax调用块。尝试在那里执行你的代码。

例如：

error: function(xhr, status, error) {
  var err = JSON.parse(xhr.responseText);
  console.log(err);
}

通过Ajax提交表单并显示错误

2 个答案: