Question

我有一张球员名单：

id  points  last_online
 1     320  2011-07-10
 2    1025  2011-07-05
 3     750  2011-04-25
 4    5000  2011-07-10
 5     525  2011-05-01

为了获得玩家等级（基于分数），我有以下mySQL选择：

SELECT Player.*,
    ( SELECT COUNT(*)
      FROM players Player_i
      WHERE (Player_i.points, Player_i.id) >= (Player.points, Player.id)
     ) AS rank
FROM players AS Player
WHERE Player.id = 1

工作正常。因此，玩家ID 1的等级为5。

但我想只考虑过去30天内上次在线的球员（考虑今天是2011-07-10）。这样，玩家ID 1的等级将是3，因为玩家3和5被排除在等级之外。

知道我该怎么做吗？

Answer 1

两种方式。第一种方法，稍微编辑您的查询：

SELECT Player.*,
    ( SELECT COUNT(*)
      FROM players Player_i
      WHERE Player_i.points >= Player.points
      and last_online > subdate(now(), '30 day')
     ) AS rank
FROM players AS Player
WHERE Player.id = 1
and last_online > subdate(now(), '30 day'); -- EDITED to add this line

第二种方式，以及更好的方式，是使用变量（方式更好的性能，更简单的SQL）：

set @rank := 0;
SELECT *, @rank := @rank + 1
FROM Player
WHERE last_online > subdate(now(), 30 'day')
ORDER BY points desc;

Answer 2

在您的查询中添加WHERE last_online > NOW() - '30 days'：

SELECT Player.*,
    ( SELECT COUNT(*)
    FROM players Player_i
    WHERE (Player_i.points, Player_i.id) >= (Player.points, Player.id)
        AND Player.last_online >= SUBDATE(NOW(), '30 day')
    ) AS rank
FROM players AS Player
WHERE Player.id = 1
    AND Player.last_online >= SUBDATE(NOW(), '30 day');

MySQL让玩家在日期之间排名

2 个答案: