Firebase .indexOn警告

Question

这是我的查询字符串数据结构。

数据结构

{
	"root_firebase_DB": {
		"users": [{
			"meUID": {
				"name": "ME",
				"rooms": [{
					"room_id_01": true
				}]
			}
		}, {
			"friendUID": {
				"name": "FRIEND",
				"rooms": [{
					"room_id_01": true
				}]
			}
		}],
		"rooms": [{
			"room_id_01": {
				"meUID": true,
				"friendUID": true
			}
		}],
		"messages": [{
			"room_id_01": {
				"message_id_01": {
					"text": "Hello world",
					"user": {
						"_id": "meUID",
						"name": "ME"
					}
				},
				"message_id_02": {
					"text": "Xin Chao",
					"user": {
						"_id": "friendUID",
						"name": "FRIEND"
					}
				}
			}
		}]
	}
}

我有meUID和friendUID，我想找一个包含我们的房间。

下面的查询很好，但我收到.indexOn警告

const meUID = 'abc';
const friendUID = 'abc';
/**
* Find room key
* filter by meUID and friendUID
*
* THIS WAY I GOT WARNING .indexON
*
*/
firebase.database().ref()
    .child('rooms')
    .orderByChild(meUID)
    .equalTo(true)
    .on('child_added', snap => {
      if (snap.hasChild(friendUID)) {
        console.log('Got roomKEY', snap.key);
      }
    });

一般来说，我通过meUID得到一系列房间并循环，每个循环我都会过滤以查找此房间是否有friendUID，这样我没有警告有索引，但似乎浪费时间循环和检查所有房间。

/**
* THIS WAY NOT GET ANY WARNING
* BUT IS THERE OTHER GOOD WAY TO GO?
*/
const ref = firebase.database().ref();
const meUID = 'aaa';
const friendUID = 'bbb';
const messages = [];

ref.child(`users/${meUID}/rooms`).once('value', snap => {
  const roomKey = null;
  // loop all room that belong to me
  snap.forEach(room => {
    // with each room, i must going to check is this room belong to friendUID or not
    ref.child(`users/${friendUID}/rooms/${room.key}`).once('value', friendRoomSnap => {
      if (friendRoomSnap.val()) {
        // found room that belong to friendUID too
        roomKey = room.key;
      }
    });
    if (roomKey != null) {
      //if found that room, break the loop
      return true;
    }
  });
  // goto fetching all messages in this room
  ref.child(`messages/${roomKey}`).on('child_added', messageSnap => {
    messages.push(messageSnap.val());
  });
});

//Too much stuff, is there any way better?

请帮助，我是新手。

我的数据结构是否出现问题？