存在实体User
,它被记录在表中Users
如何在Laravel中使用修补程序请求?
默认情况下,此表中的某些字段为空。
我需要更新这些字段并设置非空数据。
为此,我尝试使用PATCH
方法Laravel:
路由:
Route::patch('users/update', '[email protected]');
控制器:
public function update(Request $request, $id)
{
$validator = Validator::make($request->all(), [
"name" => 'required|string|min:3|max:50',
"email_work" => 'email|max:255|unique:users',
"surname" => 'required|string|min:3|max:50',
"tel" => 'required|numeric|size:11',
"country" => 'required|integer',
"region" => 'required|integer',
"city" => 'required|integer'
]);
if ($validator->fails()) {
return response()->json(["message" => $validator->errors()->all()], 400);
}
$user = User::where("user_id", $id)->update([
"name" => $request->name,
"surname" => $request->surname,
"tel" => $request->tel,
"country" => $request->country,
"city" => $request->city,
"region" => $request->region,
"email_work" => $request->email
]);
return response()->json(["user" => $user]);
}
这是否意味着我可以传递任何数据更新? 我应该通过$id
参数路由和控制器相对?
如何在Laravel中为PATCH方法使用正确的处理程序?
你能运行'PHP工匠路线:list'向我们展示它告诉你关于你注册这个特定的路线? –
如果你在update()函数中有$ id参数,那么我想你需要添加'{id}'你的路由,就像这样'Route :: patch('users/{id}/update','UsersController @ update ');' –