Question

您好我的请求有问题。如何使这些功能不会返回第二个中存在的数据。恩。 user_id = 1 friend_id = 2来自sendedRequests，user_id = 2，friend_id = 1来自pendingRequests。

以下是代码：

public function showSendedRequests(){
        $request = DB::table('friends_users')
                ->join('users', 'users.id', '=', 'friends_users.friend_id')
                ->where('who_send', $this->id)
                ->where('user_id', $this->id)
                ->where('accepted',false)
                ->get();

        return $request;
    }
    public function showPendingRequests(){
        $request = DB::table('friends_users')
                ->join('users', 'users.id', '=', 'friends_users.user_id')
                ->where([['friend_id', '=',Auth::user()->id],['accepted', '=',false],['who_send','!=',Auth::user()->id]])
                ->get();

        return $request;
    }

Answer 1

您可以使用<>不等于

执行此类操作

 $request = DB::table('friends_users')
                ->join('users', 'users.id', '=', 'friends_users.friend_id')
                ->where('who_send', $this->id)
                ->where('user_id','<>', $this->id)
                ->where('accepted',false)
                ->get();

        return $request;
    }
//for the next function
    $request = DB::table('friends_users')
                    ->join('users', 'users.id', '=', 'friends_users.user_id')
                    ->where([['friend_id', '=',Auth::user()->id],['accepted', '=',false],['who_send','<>',Auth::user()->id]])
                    ->get();

            return $request;
        }

进一步reference

Laravel功能与数据透视表中的组合

1 个答案: