我在用户迁移中声明用户必须是唯一的:
$table->string('email', 30)->unique();
但是当我第一次在邮差中创建用户时,我收到以下错误:
{
"message": "SQLSTATE[23000]: Integrity constraint violation: 1062 Duplicate entry 'james@gest.com' for key 'user_email_unique' (SQL: insert into `user` (`firstname`, `lastname`, `displayname`, `email`, `phonenumber`, `birthdate`, `profilepicture`, `suspended`, `role_id`, `company_id`, `updated_at`, `created_at`) values (james, egen, james egen, james@gest.com, 0628383493, , , 1, 1, 1, 2016-05-06 13:03:31, 2016-05-06 13:03:31))",
"code": "23000",
"status_code": 500,
当我查看我的数据库时,用户是这样做的,为什么它会给我这个错误?
控制器:
public function store(Request $request)
{
$result = $this->userRepo->store($request);
if (!$this->userRepo->store($request) != null) {
return response()->json(['message' => 'Gebruiker is succesvol aangemaakt'], 200);
}
return response()->json(['message' => $result], 406);
}
userRepo:
public function store(Request $request)
{
$validator = $this->validateUser($request);
if(!$validator != null)
{
$user = new User([
'firstname' => $request->firstname,
'lastname' => $request->lastname,
'displayname' => $request->displayname,
'email' => $request->email,
'phonenumber' => $request->phonenumber,
'birthdate' => $request->birthdate,
'profilepicture' => $request->profilepicture,
'suspended' => $request->suspended,
'role_id' => $request->role_id,
'company_id' => $request->company_id
]);
$user->save();
return null;
}
return $validator;
}
我正在使用dingo和JWT-Auth。
答案 0 :(得分:1)
查找
`if(!$validator != null)`
你不需要双重否定。你可以给:
`if(is_null($validator))`
这就是你获得双save
行动的方式。