查询问题排名系统MYSQL

Question

我有一个排名系统，我会在每个游戏日保存用户排名和积分。

现在我的问题是我想要获取用户自去年以来攀登的排名位置数。所以在这个例子中，user_id = 1从昨天起已经下降了3个位置。我当前的查询给了我一些我想要的东西，但有一些额外的计算我想删除。所以我的问题是如何计算每个用户（今天和昨天之间）的排名差异？ SQL FIDDLE

Answer 1

SELECT current.user_id,(last.rank -current.rank)
FROM ranking as current
LEFT JOIN ranking as last ON 
last.user_id = current.user_id
WHERE current.rank_date = (SELECT max(rank_date) FROM ranking)
and
last.rank_date = (SELECT max(rank_date) FROM ranking 
 where rank_date <  (SELECT max(rank_date) FROM ranking)               
                 )

Answer 2

我认为最简单的方法是：

SELECT today.user_id, (yest.rank - today.rank) as diff
FROM ranking today JOIN
     ranking yest
     on today.user_id = yest.user_id 
WHERE today.rank_date = CURRENT_DATE AND
      yest.rank_date = date_sub(CURRENT_DATE, interval 1 day);

Answer 3

  

所以我的问题是我如何计算每个人的等级差异   用户（今天和昨天之间）？

关于你的sqlfiddle演示我觉得你想要这个输出：

user_id diff
1       3
1       -3
2       -1
2       -1

所以试试这个：

SELECT current.user_id,(last.rank - current.rank) as diff
FROM ranking as current
LEFT JOIN ranking as last 
ON last.rank_date = current.rank_date+1
   and last.user_id = current.user_id
where last.rank_date is not null
order by current.user_id