您如何更新我的代码在两个表中的值,如下所示:
public function subscriberprofileupdate($ids, Request $request) {
$profile = Subscriberinfo::find($ids);
$id = \Auth::user()->id;
$user_id = \Auth::user()->user_id;
$subscriber_id = \Auth::user()->subscriber_id;
$inputteacher = $request->all();
$profile->contact_person_firstname = $inputteacher['contact_person_firstname'];
$profile->contact_person_lastname = $inputteacher['contact_person_lastname'];
$profile->email = $inputteacher['email'];
$profile->save();
return Redirect('/subscriber/profile');
}
如果我可以在订阅者表中更新电子邮件,则意味着它也将使用user_id更新用户表。 我怎么能做到。
答案 0 :(得分:0)
更好的方法是在您的订阅者表中添加一个名为 user_id 的新列,如果您在模型中给出了适当的关系,则可以获取数据。
答案 1 :(得分:0)
你可以简单地说:
User::where('id',$user_id)->update(['email' => $inputteacher['email']]);
它会更新Users
表{/ 1}} users.id
$user_id
表格电子邮件字段
答案 2 :(得分:0)
例如:
$inputteacher = $request->all();
$userdetail = User::find(1);
$userdetail->firstname = $inputteacher['contact_person_firstname'];
$userdetail->lastname = $inputteacher['contact_person_lastname'];
$profile = Subscriberinfo::find($ids);
$profile->contact_person_firstname = $inputteacher['contact_person_firstname'];
$profile->contact_person_lastname = $inputteacher['contact_person_lastname'];
$userdetail->profiles()->save($profile);
//This will save not only a model, but also all of its relationships
$userdetail->push();