Question

是否知道如何使用列显示用户休假的最小到最大日期的输出？

LeaveMain

User  | RequestNo
user1 | 123456

LeaveDetail

RequestNo | Date
123456    | 01/01/2017
123456    | 01/02/2017
123456    | 01/03/2017

期望输出

RequestNo | User  | Leave Date
123456    | user1 | 01/01/2017 - 01/03/2017

Answer 1

假设找到的最高日期被视为date to，最低被视为date from

SELECT b.RequestNo, a.User, CAST(MIN(Date) AS NVARCHAR(10)) + ' - ' + CAST(MAX(Date) AS NVARCHAR(10)) as Leave_Date
FROM
LeaveMain a inner join LeaveDetail b on a.RequestNo = b.requestNo
GROUP BY b.RequestNo, a.user

Answer 2

使用MIN和MAX聚合函数：

SELECT  RequestNo , User , CONVERT(VARCHAR, Mindate, 101) + ' - ' 
    CONVERT(VARCHAR, Maxdate, 101) 
FROM LeaveMain M
JOIN
(
  SELECT RequestNo , MIN(Date) Mindate , MAX(date) Maxdate
  FROM LeaveDetail
  GROUP BY RequestNo
) A ON A.RequestNo = M.RequestNo

Answer 3

只需将结果分组到RequestNo并获取Min和Max日期：

SELECT RequestNo, 
  User, 
  CONVERT(varchar, MIN(Date), 101) 
  + ' - ' CONVERT(varchar, MAX(Date), 101) 
FROM LeaveMain LM 
INNER JOIN LeaveDetail LD ON LM.RequestNo = LD.RequestNo
GROUP BY LM.RequestNo, LM.User

Answer 4

只需使用Join Group by以及Max和Min函数作为下一个演示： -

<强>演示： -

Create table #LeaveMain ([user] nvarchar(10), RequestNo int)
insert into #LeaveMain values ('user1', 123456)
Create table #LeaveDetail ( RequestNo int, [Date] date)
insert into #LeaveDetail values (123456, '01/01/2017')
insert into #LeaveDetail values (123456, '01/02/2017')
insert into #LeaveDetail values (123456, '01/03/2017')

select 
    Max(D.RequestNo) RequestNo , 
    Max([User]) [User] , 
    Convert(nvarchar(10),Min([Date])) + ' - '  
    + Convert(nvarchar(10),Max([Date])) [Leave Date]
from #LeaveMain M inner join #LeaveDetail D
on M.RequestNo = D.RequestNo
group by D.RequestNo

<强>结果： -

RequestNo | User  | Leave Date
123456    | user1 | 01/01/2017 - 01/03/2017

Answer 5

;WITH LeaveMain([User],RequestNo)
AS
(
SELECT 'user1' , 123456
)
,LeaveDetail(RequestNo , Date)
AS
(
SELECT 123456, '01/01/2017' union all
SELECT 123456, '01/02/2017' union all
SELECT 123456, '01/03/2017' 
)
SELECT   RequestNo,[User],CONCAT(MinDate,' - ',MaxDate)AS LeaveDate From
(
SELECT m.[User], m.RequestNo ,MAX(d.[Date])MaxDate,MIn(d.[Date])MinDate from LeaveMain m
INNER JOIN LeaveDetail d
on m.RequestNo=d.RequestNo
group by  m.[User], m.RequestNo
)Dt

Answer 6

尝试使用大量数据，

declare @LeaveDetail table(RequestNo varchar(20),Dates date)
insert into @LeaveDetail VALUES
 ('123456','2017/01/01')
,('123456','2017/01/02')
,('123456','2017/01/03')

declare @LeaveMain table(Users varchar(20),RequestNo varchar(20))
insert into @LeaveMain VALUES
('user1','123456')


select lm.Users,RequestNo
,(select convert(varchar(10),min(Dates),101)+' - '+convert(varchar(10),max(dates),101) 
from @LeaveDetail ld where ld.RequestNo=lm.RequestNo )[Leave Date]
from @LeaveMain lm

在SQL Server中查询以在单行中显示日期和从中显示日期

6 个答案: