问题
I am Developing a chat system , I have three tables (users , chats , user_chat) where is the user chat is the pivot table ,
I want to check every time that same users want to communicate to back to there chat room and don't create a new chat room
like
thats means that whenever **user_id:15**
and **user_id:20** wish to communicate return them to **chat_room:1** and not create a new one how to do it
I already tried array_intersect() but i didn't get any result
public function store(Request $requests){
//$requests->user()->id;
$data = $requests->all();
$username1 = User::select('username')->where('id',auth()->guard('api')->user()->id )->first();
$username2 = User::select('username')->where('id',$data['recever_id'] )->first();
$user1Conve[] = Conversation::where('user_id' , '=',auth()->guard('api')->user()->id )->select('chat_room_id')->get();
$user2Conve[] = Conversation::where('user_id' , '=',$data['recever_id'] )->select('chat_room_id')->get();
$result = array_intersect($user2Conve, $user1Conve);
if (count($result) == 0) {
$chatroom = ChatRoom::create([
'room_name' => 'محادثة بين' . $username1->username . ' و ' . $username2->username . '', 'sender_id' => auth()->guard('api')->user()->id
]);
$lastId = (int)$chatroom->id;
Conversation::create(['chat_room_id' => $lastId, 'user_id' => auth()->guard('api')->user()->id]);
Conversation::create(['chat_room_id' => $lastId, 'user_id' => $data['recever_id']]);
$message = Messages::create(
['message' => $data['message'],
'user_room_id' => $lastId,
'user_id' => auth()->guard('api')->user()->id,
]);
$success['chat_room_id'] = $chatroom->id;
$success['arrayIntersct'] = $result;
$pusher = new Pusher("418914066f12eac2d5fd", "6e1b5e98b06d7d3ebd7a", "449820", array('cluster' => 'ap2'));
$pusher->trigger('my-channel', 'my-event', array('message' => $message->message));
}
else {
}
return response()->json(['code'=>'success','success'=>$success], $this->successStatus);
}
回答1:
array_intersect serches for all of the values in array1 whose values exist in array two, So I think you should use it with inversed arrays too:
$result1 = array_intersect($user1Conve,$user2Conve);
$result2 = array_intersect($user2Conve, $user1Conve);
if (count($result1) == 0 && count($result2) == 0) {
......
来源：https://stackoverflow.com/questions/48971202/laravel-chat-with-pivot-table-how-to-return-same-users-to-same-chat