Question

我正在使用Lumen和Dingo API来构建API。我的注册功能检查指定的电子邮件是否已存在。

如果数据库中已存在电子邮件，则使用Dingo API Helpers返回错误响应。

Dingo\Api\Routing\Helpers

isEmailTaken函数：

private function isEmailtaken($email) {
    $userExists = User::where('email', $email)->count();
    if($userExists) {
        $return['error'] = true;
        $return['message'] = "It appears you already have an account with us.";
        return $return;
    } else {
        $return['error'] = false;
        return $return;
    }
}

注册函数调用该函数并返回错误：

    $validateEmail = $this->isEmailtaken($email);
    if ($validateEmail['error'] == true) {
        return $this->response->errorBadRequest($validateEmail['message']);
    }

它在json响应中输出正确的错误消息并带有正确的错误代码：

"{"message":"It appears you already have an account with us.","status_code":400}"

但是，标题中响应生成的状态为OK，代码为200。

Answer 1

status_code 400 bad request将在请求数据无效时返回

下面的代码有助于返回自定义status_code的响应

use Illuminate\Http\Response; 


return response()->json(['message' => 'It appears you already have an account with us.'])->setStatusCode(400);

另一种从laravel验证验证唯一emailID的简短方法，如下所示

 $rules = array(

            'email'=>'required|email|unique:user',
            );  

    $validator = Validator::make($request->all(), $rules);

    if (!$validator->passes()) {  
        return response()->json([ 'message' => $validator->messages()])->setStatusCode(400);

}

具有Dingo API Helpers的流明返回错误的响应状态代码

1 个答案: