SQL 服务器：汇总为单个结果

Question

我有这个问题

SELECT      Client.ClientNo,
            Client.ContactName,
            Deal.Currency,
            MAX(Deal.DealDate)
FROM        Deal
JOIN        Client ON Deal.ClientNo = Client.ClientNo
GROUP BY    Client.ClientNo, Client.ContactName, Deal.Currency;

这给了我一个结果

1   John Smith  EUR 2014-10-07
1   John Smith  GBP 2014-11-12
2   Jane Doe    GBP 2014-09-17
2   Jane Doe    USD 2014-12-23
1   John Smith  USD 2013-11-13
2   Jane Doe    EUR 2012-09-06

问题是，我需要每个客户的最新日期的汇总结果，如下所示：

1   John Smith  GBP 2014-11-12
2   Jane Doe    USD 2014-12-23

如何更改我的查询以实现此目的？

UPDATE 感谢 jarlh 的回答，但是我遗漏了一些东西 - 如果有重复的行 - 它会保留在结果中，如下所示：

1   John Smith  GBP 2014-11-12
1   John Smith  GBP 2014-11-12
2   Jane Doe    USD 2014-12-23

有什么方法可以实现吗？

Answer 1

未经测试，但应该可以。如果客户在同一天有两笔（或更多笔）交易，那么将 return 为该客户记录几行。

SELECT      Client.ClientNo,
            Client.ContactName,
            Deal.Currency,
            Deal.DealDate
FROM        Deal
JOIN        Client ON Deal.ClientNo = Client.ClientNo
WHERE       Deal.DealDate = (select max(DealDate) from Deal
                             where ClientNo = Client.ClientNo)

Answer 2

你可以这样做：

测试数据：

DECLARE @Deal TABLE(ClientNo INT,Currency VARCHAR(10),DealDate DATETIME)
DECLARE @Client TABLE(ClientNo INT,ContactName VARCHAR(100))

INSERT INTO @Deal
VALUES (1,'EUR','2014-10-07'),(1,'GBP','2014-11-12'),(2,'GBP','2014-09-17'),
(2,'USD','2014-12-23'),(1,'USD','2013-11-13'),(2,'EUR','2012-09-06')

INSERT INTO @Client
VALUES (1,'John Smith'),(2,'Jane Doe')

查询：

;WITH latestDeals
AS
(
    SELECT
        ROW_NUMBER() OVER(PARTITION BY ClientNo ORDER BY DealDate DESC) AS RowNbr,
        Deal.*
    FROM
        @Deal AS Deal
)
SELECT
    client.ClientNo,
    client.ContactName,
    latestDeals.Currency,
    latestDeals.DealDate
FROM
    @Client AS client
    JOIN latestDeals
        ON client.ClientNo=latestDeals.ClientNo
        AND latestDeals.RowNbr=1

更新：

如果要使用常规查询。你可以这样做：

SELECT
    client.ClientNo,
    client.ContactName,
    Latestdeal.maxDealDate as DealDate,
    deal.Currency
FROM
    @Client AS client
    JOIN 
    (
        SELECT
            MAX(Deal.DealDate) AS maxDealDate,
            Deal.ClientNo
        FROM
            @Deal AS Deal
        GROUP BY
            Deal.ClientNo
    ) AS Latestdeal
    ON client.ClientNo=Latestdeal.ClientNo
    JOIN @Deal as deal
        ON client.ClientNo=deal.ClientNo
        AND deal.DealDate=Latestdeal.maxDealDate

这将导致相同的输出

结果：

1   John Smith  GBP 2014-11-12 00:00:00.000
2   Jane Doe    USD 2014-12-23 00:00:00.000

Answer 3

试试这个，

测试数据：

CREATE TABLE #YourTable
(
    CLIENT_NO INT,
    CONTACT_NAME VARCHAR(20),
    CURRENCY VARCHAR(10),
    [DEAL_DATE] DATE
)

INSERT INTO #YourTable VALUES
(1,'John Smith','EUR','2014-10-07'),
(1,'John Smith','GBP','2014-11-12'),
(2,'Jane Doe','GBP','2014-09-17'),
(2,'Jane Doe','USD','2014-12-23'),
(1,'John Smith','USD','2013-11-13'),
(2,'Jane Doe','EUR','2012-09-06')

查询：

SELECT CLIENT_NO,CONTACT_NAME,CURRENCY,[DEAL_DATE]
FROM   (SELECT *,
               Row_Number()
                 OVER (
                   PARTITION BY CLIENT_NO
                   ORDER BY [DEAL_DATE] DESC) AS RN
        FROM   #YourTable)A
WHERE  RN = 1

SQL 服务器：汇总为单个结果

SQL Server: aggregate to single result

sql

sql-server

greatest-n-per-group