Laravel/Lumen: 验证电子邮件的数据已经存在并且一般验证

Question

我正在写 API。我想知道我怎样才能 return 正确回应。

数据无效（仅当电子邮件地址存在时）
数据无效（一般验证错误）

到目前为止，这是我的代码：

$data = Input::all();

$validator = Validator::make($request->all(), [
    'email'            => 'required|email|unique:users,email',
    'some_other_field' => 'required'
]);

if ($validator->fails()) {
    // NOTE: using a or b, this are my custom methods
    a) return $this->existsResponse($data);
    b) return $this->badRequestResponse($data);
}

我怎么知道电子邮件是否已经存在？

Answer 1

您应该看看验证器 Class。我认为方法 invalid() 和 valid() 应该适合您。此方法返回一组键。

http://laravel.com/api/5.0/Illuminate/Validation/Validator.html#method_invalid

Answer 2

好吧，您正在尝试进行两种不同类型的验证，因此使用两种不同的验证器是有意义的：

$data = $request->all();

// validate email
$validator = Validator::make($data, [
    'email' => 'required|email|unique:users,email',
]);

if ($validator->fails()) {
    return $this->existsResponse($data);
}

// email valid, so validate other data
$validator = Validator::make($data, [
    'some_other_field' => 'required'
]);

if ($validator->fails()) {
    return $this->badRequestResponse($data);
}

// if here, all data is valid

Answer 3

我是这样解决的

public function store(Request $request)
{
    $data = Input::all();

    // validating data

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

    if ($validator->fails()) {
        return $this->badRequestResponse($data);
    }

    // validating if resource exists

    $userExists = User::where('email', $request->email)->count();

    if($userExists) {
        return $this->existsResponse($data);
    }

    // inserting data

    $user = new User();

    $user->fill($data);
    $user->save();

    if($user->save()) {
        return $this->createdResponse($data);
    }
}

Laravel/Lumen: 验证电子邮件的数据已经存在并且一般验证

Laravel/Lumen: validating data for email already exist and general validation

validation

laravel

lumen