mysql使用內部連接獲取用戶等級

Question

我有三個表如下。我想獲得最高級別的用戶。

1）用戶表作爲

id  | user_id | created_at 
1   | 100  | 2014-11-07 02:54:09 
2   | 102  | 2014-11-08 03:52:40 
3   | 103  | 2014-11-10 02:47:26 
4   | 104  | 2014-11-11 02:54:48 
5   | 105  | 2014-11-14 03:11:23 
6   | 105  | 2014-11-15 00:56:34 

2）user_profile表作爲

id  | user_id | rank 
1   | 100  | 100 
2   | 102  | 500 
3   | 103  | 10 
4   | 104  | 0 
5   | 105  | 11 
6   | 105  | 1000 

3）user_followers表作爲

id  | user_id | followers 
1   | 100  | 10 
2   | 102  | 20 
3   | 103  | 30 
4   | 104  | 40 
5   | 105  | 0 
6   | 105  | 50 

現在我的查詢是我想要得到的名單表2中用戶排名最高。如果在桌面3中擁有最高關注者的領帶用戶將獲勝。如果同一個追隨者首先創建的用戶會贏。

而另一個我想找到用戶級別相同的邏輯傳遞用戶ID。

我已經嘗試過類似

SET @i=0; 
SELECT user_id, rank, @i:=@i+1 AS rank FROM user_profile ORDER BY rank DESC 

Answer 1

SET @rank = 0; 
SELECT 
@rank := @rank + 1 AS rank, * 
FROM 
(
SELECT users.user_id, user_profile.rank, user_followers.followers, users.created_at 
FROM users 
LEFT JOIN user_profile ON users.user_id = user_profile.user_id 
LEFT JOIN user_followers ON users.user_id = user_followers.user_id 
ORDER BY user_profile.rank DESC, user_followers.followers DESC, users.created_at ASC 
) 

Answer 2

阿里昂的回答這個樣子......

SELECT 
users.* 
FROM 
users 
JOIN user_profile 
    ON users.user_id = user_profile.user_id 
JOIN user_followers 
    ON user_profile.user_id=user_followers.user_id 
ORDER BY 
user_profile.rank DESC, 
user_followers.followers DESC, 
users.created_at DESC 

...但是這似乎有點接近你以後。 ...

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