有没有办法操纵 php artisan make:request 发出的请求的响应？

Question

我有一个 Laravel wit FormRequest extend 的请求，并带有以下规则和消息：

public function rules()
{
    return [
        'login' => 'required',
        'password' => 'required',
    ];
}

public function messages()
{
    return [
        'login:required' => 'Login is required',
        'password:required' => 'Password is required'
    ];
}

现在我收到这样的回复：

{
    "message": "The given data was invalid.",
    "errors": {
        "login": [
            "The login field is required."
        ],
        "password": [
            "The password field is required."
        ]
    }
}

但我想要这样的东西：

{
    "status": false,
    "validator": true,
    "msg": {
        "e": {
            "login": [
                "The login field is required."
            ],
            "password": [
                "The password field is required."
            ]
        },
        "type": "error"
    }
}

这可能吗？

因为现在我使用这段代码来得到这个：

$item = Validator::make($request->all(), [
    'login' => 'required',
    'password' => 'required',
]);

if ($item->fails())
    return response()->json([
        'status' => false,
        'validator' => true,
        'msg' => [
            'e' => $item->messages(),
            'type' => 'error'
        ],
    ]);

但我想通过 larave 提出的请求来制作这个（php artisan make:request）

Answer 1

是的，你可以做到。将此方法添加到您的表单请求类：

use Illuminate\Contracts\Validation\Validator;
use Illuminate\Http\Exceptions\HttpResponseException;

// ...

protected function failedValidation(Validator $validator)
{
    $data = [
        'status'    => false,
        'validator' => true,
        'msg'       => [
            'e'    => $validator->errors(),
            'type' => 'error',
        ],
    ];

    throw new HttpResponseException(response()->json($data, 422));
}