将Laravel验证错误返回给Ajax调用

Question

我试图弄清楚如何回传给我的ajax调用，如果存在验证错误，则可以防止页面继续运行并将这些错误显示给用户。

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

    $credentials = [
        'first_name' => $request->input('first_name'),
        'last_name' => $request->input('last_name'),
        'email'    => $request->input('email'),
        'password' => Hash::make(str_random(8)),
    ];

    $user = Sentinel::registerAndActivate($credentials);

    $user->role()->attach(5);

    return response()->json([
        'success' => true,
        'errors' => null
    ]);
}

Answer 1

您可以通过以下方式使用“ Validate外观”进行尝试：

$validator = \Validator::make($request->all(), [
    'first_name' => 'required',
    'last_name' => 'required',
    'email' => 'required|confirmed|unique:users,email',
    'email_confirmation' => 'required'
]);

// Validate the input and return correct response
if ($validator->fails())
{
    return response()->json(array(
        'success' => false,
        'errors' => $validator->getMessageBag()->toArray()

    ), 422);
}

这将为您提供如下的JSON响应：

{
    'success': false,
    'errors': {
        'first_name': [
            'The first name field is required.'
        ],
        'last_name': [
            'The last name field is required.'
        ]
    }
}