Question

我是sql的新手。我需要一些帮助来生成夏日信息

会员表

MonthID  | UserID  | TeamID
-----------------------------
  1      |  1       | 1
  1      |  2       | 1
  1      |  3       | 1
  1      |  4       | 1
  1      |  5       | 2
  1      |  6       | 2
  1      |  7       | 2

AND

ReportTable

ID* |  MonthID  | UserID  | IsSend
-----------------------------------
 1  |    1      |    2     | False
 2  |    1      |    3     | True
 3  |    1      |    5     | True

我想生成一个像以下

的总结

TeamID     |  Total Count  |  Send Count | Not Send Count
-----------------------------------------------------------
 1         |      4        |      1      |     3
 2         |      3        |      1      |     2

总计数：团队中的用户数

发送计数：IsSend = True
的团队中的总用户数
不发送计数：总计数 - 发送计数

什么是有效的方式？

Answer 1

尝试一下：

select mt.teamId, count(*) totalCount,
    count(case when rt.isSend = 'True' then 1 end) sendCount,
    count(case when rt.isSend != 'True' then 1 end) notSendCount
from memberTable mt
join reportTable rt on mt.userId = rt.userId
group by mt.teamId

请注意，您的预期结果不会反映您的数据。基于您的数据的结果应该是：

+--------+------------+-----------+--------------+
| TEAMID | TOTALCOUNT | SENDCOUNT | NOTSENDCOUNT |
+--------+------------+-----------+--------------+
|      1 |          2 |         1 |            1 |
|      2 |          1 |         1 |            0 |
+--------+------------+-----------+--------------+

Answer 2

没有表格试试这个，我无法检查这是否有效，但这应该可以帮到你：

SELECT TeamID, count(userID) as "Total count", Sum(IsSend) as "Send Count" FROM MemberTable JOIN ReportTable ON UserID GROUP BY TeamID;

Answer 3

select MT.TeamID,
       count(distinct MT.UserID) as "Total Count",
       count(distinct case when RT.IsSend = 1 then MT.UserID end) as "Send Count",
       count(distinct MT.UserID) - count(distinct case when RT.IsSend = 1 then MT.UserID end) as "Not Send Count"
from MemberTable as MT
  left outer join ReportTable as RT
    on MT.MonthID = RT.MonthID and
       MT.UserID = RT.UserID
group by MT.TeamID

结果：

TeamID      Total Count Send Count  Not Send Count
----------- ----------- ----------- --------------
1           4           1           3
2           3           1           2

在此尝试：http://data.stackexchange.com/stackoverflow/query/66347

在SQL中生成摘要表的有效方法。请看解释

3 个答案: