Sql 按日期和客户选择最大收入

以下是我的表格中的示例数据：

Rev       Date          Client
1890      2015-11-20    xyz
1536.28   2017-10-27    AAA
934.84    2017-10-27    AAA
919.592   2017-03-22    AAA 
760.985   2014-11-25    xyz

我需要为每个客户选择收入最高的一天

我有以下查询，但它选择了客户每天的最大收入，而不仅仅是每个客户的最高收入

SELECT TOP 1 max(rev)/1000 AS Rev, date, client FROM table1 GROUP BY date, client

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常见问题。我想Rev已经是每天的总和了，我们只是选择了最高的一行。如果有联系，您可以使用

densite\u rank（）

代替

select * from (
    select *, row_number() over (partition by Client order by Rev desc) as rn
    from table1
) t
where rn = 1;

---

一种选择是将WITH TIES子句与行编号（）配合使用

示例

Select top 1 with ties *
 From  YourTable
 Order By Row_Number() over (Partition By Client Order By Rev Desc)

设置：

查询：

SELECT Client
    , t.[Date]
FROM
(
    SELECT Client
        , [Date]
        , SUM(Rev) AS [TotalRev]
        , ROW_NUMBER() OVER (PARTITION BY Client ORDER BY SUM(Rev) DESC) AS RN
    FROM @data
    GROUP BY Client, [Date]
) AS t
WHERE t.RN = 1

结果:

Client    Date
====================
AAA       2017-10-27
xyz       2015-11-20

我会用这个

SELECT DISTINCT ca.*
FROM table1 t
CROSS APPLY (SELECT TOP 1 *
             FROM table1
             WHERE Client = t.Client
             ORDER BY Rev DESC) ca

如果您已经有一个具有唯一ID的

客户机

表，那么它会更干净

SELECT ca.Rev, ca.Date, c.Client
FROM Client c
CROSS APPLY (SELECT TOP 1 Rev, Date
             FROM table1
             WHERE Client = c.Client
             ORDER BY Rev DESC) ca

---

据我所知，michaelg的问题是每天的交易收入最高，因此：

 select * from (
    select *, row_number() over (partition by [date] order by Rev desc) as rn
    from ##table1
) t
where rn = 1;

若您想要您的客户机的总金额，那个么您已经将所有rev分组到子查询中

---

请在我的T-SQL代码中尝试您的脚本

如果您不希望它按客户端分组，请将其从分组中删除。如果您需要客户信息，您必须使用内部查询来获取最大值。我假设TD是Date。如果您

按TD、client分组，则这些就是您将获得的组。
 select * from (
    select *, row_number() over (partition by [date] order by Rev desc) as rn
    from ##table1
) t
where rn = 1;