关于sql server：从查询结果中删除重复数据

Remove duplicate data from query results

我有以下查询：

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select
C.ClientID,
C.FirstName + ' ' + C.LastName as ClientName,
CAST(V.StartDate as date) as VisitDate,
count(*) as 'Number of Visits'
from
Visit V
Inner Join Client C on
V.ClientID = C.ClientID
group by
C.ClientID,
C.FirstName + ' ' + C.LastName,
CAST(V.StartDate as date)
having
count(*) > 3
order by
C.ClientID,
CAST(V.StartDate as date)

给出了以下结果(如果有人想知道，名字是假的)

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ClientID ClientName VisitDate Number of Visits
75 Kay Taylor 2016-06-07 4
372 Moses Mcgowan 2016-09-03 4
422 Raven Mckay 2016-03-11 4
422 Raven Mckay 2016-06-14 4
679 Ulysses Booker 2016-01-09 4
696 Timon Turner 2016-07-06 4
1063 Quyn Wall 2016-06-25 4
1142 Garth Moran 2016-11-20 4
1142 Garth Moran 2016-11-21 4
1563 Hedley Gutierrez 2016-01-07 4
1563 Hedley Gutierrez 2016-01-17 4
1563 Hedley Gutierrez 2016-01-21 4
1563 Hedley Gutierrez 2016-01-27 4
1563 Hedley Gutierrez 2016-01-28 4
1563 Hedley Gutierrez 2016-01-30 4
1563 Hedley Gutierrez 2016-02-27 4
1563 Hedley Gutierrez 2016-03-26 4
1563 Hedley Gutierrez 2016-04-06 4
1563 Hedley Gutierrez 2016-04-09 4
1563 Hedley Gutierrez 2016-04-22 4
1563 Hedley Gutierrez 2016-05-06 4
1563 Hedley Gutierrez 2016-05-26 4
1563 Hedley Gutierrez 2016-06-02 4
1563 Hedley Gutierrez 2016-07-14 4
1563 Hedley Gutierrez 2016-07-29 4
1563 Hedley Gutierrez 2016-08-09 7
1563 Hedley Gutierrez 2016-09-01 4
1563 Hedley Gutierrez 2016-09-23 4
1563 Hedley Gutierrez 2016-12-07 4
1636 Kiara Lowery 2016-01-12 4
2917 Cynthia Carr 2016-06-21 4
2917 Cynthia Carr 2016-10-21 4
3219 Alan Monroe 2016-01-02 4
3219 Alan Monroe 016-02-27 4
3219 Alan Monroe 2016-09-01 5
4288 Natalie Mitchell 2016-03-19 4

如何让结果只显示ClientID和ClientName一次，结果是这样的？

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ClientID ClientName VisitDate Number of Visits
75 Kay Taylor 2016-06-07 4
372 Moses Mcgowan 2016-09-03 4
422 Raven Mckay 2016-03-11 4
2016-06-14 4
679 Ulysses Booker 2016-01-09 4
696 Timon Turner 2016-07-06 4
1063 Quyn Wall 2016-06-25 4
1142 Garth Moran 2016-11-20 4
2016-11-21 4
1563 Hedley Gutierrez 2016-01-07 4
2016-01-17 4
2016-01-21 4
2016-01-27 4
2016-01-28 4
2016-01-30 4
2016-02-27 4
2016-03-26 4
2016-04-06 4
2016-04-09 4
2016-04-22 4
2016-05-06 4
2016-05-26 4
2016-06-02 4
2016-07-14 4
2016-07-29 4
2016-08-09 7
2016-09-01 4
2016-09-23 4
2016-12-07 4
1636 Kiara Lowery 2016-01-12 4
2917 Cynthia Carr 2016-06-21 4
2016-10-21 4
3219 Alan Monroe 2016-01-02 4
3219 016-02-27 4
2016-09-01 5
4288 Natalie Mitchell 2016-03-19 4

相关讨论

实际上，你想要的不是删除重复项，而是不显示重复项。

为此，您可以使用带有ROW_NUMBER()的CASE语句并在第1行显示值，并在ELSE分支上显示NULL或''(其他行)：

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select
CASE
WHEN ROW_NUMBER() OVER (PARTITION BY C.ClientID ORDER BY CAST(V.StartDate as date) ASC) = 1
THEN C.ClientID
ELSE NULL
END as ClientID,
CASE
WHEN ROW_NUMBER() OVER (PARTITION BY C.ClientID ORDER BY CAST(V.StartDate as date) ASC) = 1
THEN C.FirstName + ' ' + C.LastName
ELSE NULL
END as ClientName,
CAST(V.StartDate as date) as VisitDate,
count(*) as 'Number of Visits'
from
Visit V
Inner Join Client C on
V.ClientID = C.ClientID
group by
C.ClientID,
C.FirstName + ' ' + C.LastName,
CAST(V.StartDate as date)
having
count(*) > 3
order by
C.ClientID,
CAST(V.StartDate as date)

试试这个：

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DECLARE @Table TABLE (ClientId NVARCHAR(5), ClientName NVARCHAR(6), VisitDate DATE, NumOfVisits INT)

INSERT INTO @Table VALUES ('75' , 'A_NAME' , '2016-06-07' , '4' ),('372' , 'B_NAME' , '2016-09-03' , '4' ),
('422' , 'C_NAME' , '2016-03-11' , '4' ),('500' , 'D_NAME' , '2016-03-15' , '4'),
('500' , 'D_NAME' , '2016-03-19' , '4' ),('500' , 'D_NAME' , '2016-03-20' , '4'),
('500' , 'D_NAME' , '2016-07-15' , '4' ),('500' , 'D_NAME' , '2016-09-13' , '4'),
('600' , 'E_NAME' , '2016-03-19' , '4' ),('600' , 'E_NAME' , '2016-03-20' , '4'),
('600' , 'E_NAME' , '2016-07-15' , '4' ),('600' , 'E_NAME' , '2016-09-13' , '4')

;WITH A AS (
SELECT ROW_NUMBER() OVER(PARTITION BY ClientID ORDER BY ClientID) row_id,* FROM (

-----------------------------------------
SELECT * FROM @Table --- replace this line with your query----
-----------------------------------------

) Main_Result ) SELECT ISNULL(BB.ClientID,'')ClientID,ISNULL(BB.ClientName,'')ClientName,AA.VisitDate,AA.NumOfVisits
FROM A AA LEFT JOIN (SELECT * FROM A BB WHERE BB.row_id=1) BB ON AA.ClientID = BB.ClientID AND AA.row_id =BB.row_id
ORDER BY CONVERT(INT,AA.ClientID)

希望这可以帮助。 :)

您可以直接执行此操作以从样本数据中获取样本结果。 :)

我会将您的初始查询用作CTE或子查询来替换#TMP_DATA。以下是我将如何做到这一点。使用带有LEAD功能的CASE来确定是否应显示ClientID和ClientName中的数据：

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SELECT
CASE WHEN CAST(LAG(T.ClientID,1,'') OVER (PARTITION BY T.ClientID ORDER BY T.ClientID,T.VisitDate) AS VARCHAR) = T.ClientID THEN '' ELSE CAST(T.ClientID AS VARCHAR) END AS ClientID,
CASE WHEN LAG(T.ClientName,1,'') OVER (PARTITION BY T.ClientID ORDER BY T.ClientID,T.VisitDate) = T.ClientName THEN '' ELSE T.ClientName END ClientName,
T.VisitDate,
T.[Number of Visits]
FROM #TMP_DATA AS T

结果集是：
！HTTPS：//i.stack.imgur.com/MBfEn.png

我希望下面的查询能做到这一点.... :)

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WITH CTE AS
(
select top 100 percent
cast(C.ClientID as nvarchar(255)) as ClientID,
C.FirstName + ' ' + C.LastName as ClientName,
CAST(V.StartDate as date) as VisitDate,
count(*) as 'Number of Visits',
row_number() over (partition by C.ClientID,C.FirstName + ' ' + C.LastName ORDER BY CAST(V.StartDate as date) ) as rw_num
from
Visit V
Inner Join Client C on
V.ClientID = C.ClientID
group by
C.ClientID,
C.FirstName + ' ' + C.LastName,
CAST(V.StartDate as date)
having
count(*) > 3
order by
min(C.ClientID),
min(CAST(V.StartDate as date))

)

select case
when rw_num<>1 then '' else ClientID end as ClientID,
case
when rw_num<>1 then '' else ClientName end as ClientName,
VisitDate, [Number of Visits]
from CTE

结果：

enter image description here

我的测试表中的测试数据：

enter image description here

您可以使用GROUP BY解决此问题，按(ClientID，VisitDate)分组。

请参阅此处的响应1097：在多列上使用group by

注意：在ORDER BY中没有必要使用CAST(V.StartDate as date)，您可以使用VisitDate，因为它存在于您的SELECT中：... CAST(V.StartDate as date) as VisitDate,

编辑：
试试这个：

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SELECT
C.ClientID,
C.FirstName + ' ' + C.LastName as ClientName,
CAST(V.StartDate as date) as VisitDate,
count(*) as 'Number of Visits'
from
Visit V
Inner Join Client C on
V.ClientID = C.ClientID
group by
(C.ClientID, VisitDate)
having
count(*) > 3
order by
C.ClientID,
VisitDate