Question

您好我有关于SQL结构的问题。我有一些想通了，但我不能得到其余的。

USERS TABLE

user    OS_signup
1       Mobile
2       Windows
3       Mac
4       Mobile
5       Windows

DUNGEON TABLE

    user  dungeon     progress  status      OS_completed  deaths
    1     Dungeon 1   1         finished    Windows       0
    1     Dungeon 1   2         incomplete  Windows       0
    2     Dungeon 2   1         finished    Windows       1
    3     Dungeon 2   1         finished    Mac           1
    3     Dungeon 1   2         finished    Mac           3
    3     Dungeon 3   3         incomplete  Mac           0
    4     Dungeon 1   1         finished    Mobile        1
    5     Dungeon 1   1         incomplete  Windows       5

在比较完成第一个地牢的第一级进度的用户与在游戏中拥有帐户的用户的百分比时，哪种操作系统最受欢迎？

我有这个查询

SELECT OS_signup, count(*) AS count_total
FROM users
GROUP BY OS_signup

和这一个

SELECT OS_completed, count(*) AS count_completed
FROM dungeons
WHERE dungeon = "Dungeon 1" AND progress >=1 AND status = "finished"
GROUP BY OS_completed

我显然得到了两个表，但是我想编写一个输出组合表的查询，以便得到一个看起来像这样的表：

OS          percent-done
Mobile      0.5
Windows     0.5
Mac         1

如何编写查询来实现上述表格？

谢谢！

Answer 1

对于MySQL，它可能是这样的：

SELECT OS_signup, count(*) AS count_total, COUNT(tmp.user_id) as count_completed, COUNT(tmp.user_id) / COUNT(*) as percent
FROM users
LEFT JOIN (
    SELECT user_id
    FROM dungeons
      WHERE dungeon = 'Dungeon 1' AND progress >= 1 AND status = 'finished'
    GROUP BY user_id
    ) tmp ON tmp.user_id = users.id
GROUP BY OS_signup

Answer 2

只需将这两个查询的结果加在一起：

WITH total(OS, count_total) AS (
  SELECT OS_signup, count(*)
  FROM users
  GROUP BY OS_signup
),
completed(OS, count_completed) AS (
  SELECT OS_completed, count(*)
  FROM dungeons
  WHERE dungeon = 'Dungeon 1'
    AND progress >= 1
    AND status = 'finished'
  GROUP BY OS_completed
)
SELECT OS,
       count_total,
       count_completed,
       CAST(count_completed AS FLOAT) / count_total AS percent
FROM total
JOIN completed USING (OS)
ORDER BY ...;

（MySQL没有公用表表达式;你必须使用视图或子查询。）

SQL两个表中的百分比总和如何随时间变化

2 个答案: