获取最优惠的房间mysql

Question

我在我的数据库中有这个数据，我想得到大多数room_id得到了用户的青睐。

表名：favorit_rooms

id | userid | room_id| favorit 

    1 | 114  | 1  | 1 
    2 | 45  | 1  | 0 
    3 | 45  | 5  | 1 
    4 | 47  | 1  | 1 
    5 | 114  | 3  | 1 
    6 | 120  | 1  | 1 
    7 | 114  | 2  | 1 
    8 | 45  | 2  | 1 
    9 | 45  | 3  | 1 
10 | 45  | 12  | 1 
11 | 131  | 1  | 0 

我通过ROOM_ID和用户ID，但没有工作tryed到组。然后我想以最低的价格订购它们。当然，请注意那个偏好不是0。

SELECT id,room_id,count(userid) as countusers FROM favorit_rooms 
GROUP BY room_id,favorit 
ORDER by room_id,favorit desc LIMIT 5 

我希望结果是：

 room_id countusers 

     1   3   
     2   2   
     3   2   
     5   1   
     12   1   

用户名和ROOM_ID是指数一起，所以没有dublicates。

我该如何做到这一点，谢谢。

Answer 1

看起来你只需要改变分组并添加一个where子句。这应该做你想要什么：

SELECT room_id, count(userid) AS countusers 
FROM favorit_rooms 
WHERE favorit = 1 
GROUP BY room_id 
ORDER by countusers DESC LIMIT 5 

Answer 2

应该是这个

SELECT room_id, count(distinct userid) as countusers 
FROM favorit_rooms 
WHERE favorit = 1 
GROUP BY room_id,favorit 
ORDER by count(distinct userid) desc LIMIT 5