从查询结果中删除重复数据答案

【问题标题】：Remove duplicate data from query results从查询结果中删除重复数据
【发布时间】：2017-07-02 18:44:38
【问题描述】：

我有以下疑问：

 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)

给出以下结果（名字是假的，以防有人想知道）

 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

我怎样才能让结果只显示一次 ClientID 和 ClientName 所以结果是这样的？

 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

【问题讨论】：

你不应该为此使用 sql
在您的应用程序代码中执行此操作。
我同意 HB，这在表示层处理得更好。报告工具将使用正确的设置自动执行此操作。
这将是理想的，但不幸的是，这是一个需要直接进入 Excel 电子表格的数据集，因此这里没有报告层，结果可能有数千个，因此无法手动完成.
@James 你可以在 excel 中使用数据透视表。

标签： sql-server tsql

【解决方案1】：

实际上，您想要的不是删除重复项，而是不显示它们。

为此，您可以使用带有ROW_NUMBER() 的CASE 语句并在第一行显示值，并在ELSE 分支（其他行）上显示NULL 或''：

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)

【讨论】：

【解决方案2】：

试试这个：

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)

希望这会有所帮助。 :)

您可以直接执行此操作以从示例数据中获取示例结果。 :)

【讨论】：

【解决方案3】：

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

在此处查看响应 1097：Using group by on multiple columns

注意：在您的 ORDER BY 中不必使用 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：

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

【讨论】：

这有什么帮助？对 varchar 列进行排序确实会导致奇怪的排序，这取决于日期的格式。如果要按日期排序，转换为日期会有所不同。
这根本不是问的问题，问题是只显示一条记录的名称和id，其他同名/id的行留空
好的。然后我认为您需要对 VisitDate 和 Number of Visits 列使用字符串连接。之后，您可以将每个值拆分为不同的行。见：[链接]codeproject.com/articles/691102/…-stackoverflow.com/questions/17591490/…

【解决方案4】：

我希望下面的查询能做到这一点.... :)

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：

我的测试表中的测试数据：

【讨论】：

【解决方案5】：

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

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

【讨论】：