从查询中选择等于最大值的行

Question

我想知道谁在我拥有的应用程序中拥有最多的朋友（交易），这意味着他可以得到报酬，也可以自己支付给许多其他用户。

我无法进行查询以显示仅拥有最大好友数的人（可以是 1 个或多个，并且可以更改，所以我可以' t 使用限制）。

;with relationships as 
(
    select
      paid as 'auser',
      Member_No as 'afriend'
    from Payments$
    union all
    select
      member_no as 'auser',
      paid as 'afriend'
    from Payments$
),
DistinctRelationships AS (
    SELECT DISTINCT *
    FROM relationships
)
select
  afriend,
  count(*) cnt
from DistinctRelationShips
GROUP BY
  afriend
order by
  count(*) desc

我就是想不通，我试过 count, max(count), where = max, 什么都没用。

这是一个两列table - “Member_No”和“付费” - 会员支付了钱，付费的就是得到钱的人。

Member_No	Paid
14	18
17	1
12	20
12	11
20	8
6	3
2	4
9	20
8	10
5	20
14	16
5	2
12	1
14	10

它来自 Excel，但我将其加载到 sql-server。
这只是一个示例，还有1000行

Answer 1

看来你很over-complicating这个。不需要 self-joining.

只需逆轴旋转每一行，这样您就拥有了关系的两边，然后将它按一边分组并计算另一边的不同

SELECT
-- for just the first then SELECT TOP (1)
-- for all that tie for the top place use SELECT TOP (1) WITH TIES
  v.Id,
  Relationships = COUNT(DISTINCT v.Other),
  TotalTransactions = COUNT(*)
FROM Payments$ p
CROSS APPLY (VALUES
    (p.Member_No, p.Paid),
    (p.Paid, p.Member_No)
) v(Id, Other)
GROUP BY
  v.Id
ORDER BY
  COUNT(DISTINCT v.Other) DESC;

db<>fiddle

从查询中选择等于最大值的行

Choose row that equal to the max value from a query

sql-server

count

subquery

max

with-statement