首先感谢帮助过这个复杂难查询的朋友们.
First of all I would like to thank the friends who helped this complex and difficult query.
我有三张桌子
表一
StaffId FirstName LastName staffType
---------------------------------------
1 Adam Sorme Student
2 Lara Sandra Teacher
3 Jack Jones Student
表 2
GateId GateName
---------------------------------------
1 frontDoor
2 superDoor
表 3
Id transitionDate GateId StaffId
---------------------------------------
1 2018-01-1 08:00:00 1 1
2 2018-01-1 10:00:00 2 1
3 2018-01-1 20:00:00 2 1
4 2018-01-2 07:00:00 1 2
5 2018-01-2 10:00:00 1 3
6 2018-01-9 12:00:00 2 2
我想要学生每天的第一个和最后一个动作.如果在指定日期之间没有可用的移动,则必须将值设置为 null
I want the first and last movements of students for each day. Value must be set to null if no movement is available between the specified dates
transitionDate> '2018-01-1 00:00:00 000'
and transitionDate< '2018-01-03 00:00:00 000'
输出:
Id Date MinTransitionDate MaxTransitionDate FirstGateName LastGateName StaffId StaffType
1 2018-01-01 2018-01-1 08:00:00 2018-01-1 20:00:00 frontDoor superDoor 1 Student
2 2018-01-01 null null null null 3 student
3 2018-01-02 null null null null 1 student
4 2018-01-02 2018-01-2 10:00:00 null frontDoor null 3 student
你可以试试像下面这样的查询
you can try a query like below
查看工作演示
create table staff(StaffId int, FirstName nvarchar(10), LastName nvarchar(10), staffType nvarchar(10))
insert into staff values
(1,'Adam','Sorme','Student')
,(2,'Lara','Sandra','Teacher')
,(3,'Jack','Jones','Student')
go
create table gate(GateId int, GateName nvarchar(10))
insert into gate values
(1,'frontDoor')
,(2,'superDoor')
go
create table logs
(Id int, transitionDate datetime, GateId int, StaffId int)
insert into logs values
(1,'2018-01-1 08:00:00',1,1)
,(2,'2018-01-1 10:00:00',2,1)
,(3,'2018-01-1 20:00:00',2,1)
,(4,'2018-01-2 07:00:00',1,2)
,(5,'2018-01-2 10:00:00',1,3)
,(6,'2018-01-9 12:00:00',2,2)
go
declare @startdate datetime, @enddate datetime
select @startdate='2018-01-1 00:00:00' , @enddate='2018-01-03 00:00:00'
; with tempSet as
(
select
transitionDatetime=l.transitionDate,
gateName=g.gateName,
staffid=l.staffid,
idx=
row_number() over(partition by l.staffid order by l.transitionDate ) -
row_number() over(partition by l.staffid,cast(l.transitionDate as date) order by l.transitionDate ),
transitionDate=cast(l.transitionDate as date)
from
logs l inner join staff s on
l.staffid=s.staffid and staffType='Student'
join gate g on g.gateid=l.gateid
)
, groupedSet as
(
select
t1.*,
FirstGateName=t2.gatename,
lastGateName=t3.gatename
from
(
select
staffid,
mintransitionDate=min(transitionDatetime),
maxtransitionDate= case when count(1)>1 then max(transitionDatetime) else null end,
transitionDate=max(transitionDate),
idx
from
tempSet
group by staffid,idx
) t1
left join
tempSet t2
on t1.idx=t2.idx
and t1.staffid=t2.staffid and t1.mintransitionDate=t2.transitionDatetime
left join
tempSet t3
on t1.idx=t3.idx
and t1.staffid=t3.staffid and t1.maxtransitionDate=t3.transitionDatetime
where t1.transitionDate between @startdate and @enddate
)
select
t.*,
g.mintransitionDate,
g.maxtransitionDate,
g.FirstGateName,
g.LastGateName
from
groupedSet g
right join
(
select
d,
staffid
from
(
select
top (select datediff(d,@startdate, @endDate))
d=dateadd(d,row_number() over(order by (select null))-1, @startDate)
from
sys.objects o1 cross join sys.objects o2
)tally
cross join
staff
where staff.stafftype='Student'
)t
on cast(t.d as date)=cast(g.transitionDate as date) and t.staffid=g.staffid
order by t.d asc, t.staffid asc
这篇关于T-SQL 最大日期和最小日期值单行的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持html5模板网!
修改现有小数位信息Modify Existing decimal places info(修改现有小数位信息)
多次指定相关名称“CONVERT"The correlation name #39;CONVERT#39; is specified multiple times(多次指定相关名称“CONVERT)
T-SQL 左连接不返回空列T-SQL left join not returning null columns(T-SQL 左连接不返回空列)
从逗号或管道运算符字符串中删除重复项remove duplicates from comma or pipeline operator string(从逗号或管道运算符字符串中删除重复项)
将迭代查询更改为基于关系集的查询Change an iterative query to a relational set-based query(将迭代查询更改为基于关系集的查询)
将零连接到 sql server 选择值仍然显示 4 位而不是concatenate a zero onto sql server select value shows 4 digits still and not 5(将零连接到 sql server 选择值仍然显示 4 位而不是 5)