Sql server 如何划分复杂表

我在SQL Server中有一个名为

CHECKINOUT

的表，该表包含以下列：

PASSID  CHECKTIME               CHECKTYPE   UID
--------------------------------------------------
PS3 2015-08-05 01:12:02.100     0       CAA0322
PS3 2015-08-06 02:17:02.310     1       CAA0322
PS4 2015-08-03 01:02:03.200     0       CAA0322
PS4 2015-08-03 11:11:01.233     1       CAA0322
PS3 2015-08-02 11:11:01.210     0       CAA0322
PS3 2015-08-02 12:02:04.147     1       CAA0322
PS1 2015-09-05 11:11:01.210     0       CAA0322
PS1 2015-09-05 01:12:09.010     1       CAA0322

PASSID

是每次有人访问该场所时，在允许他们进入时，他们的签入时间被注册，并且签入类型变为0时给出的通行证

   CHECKTYPE=0

当他们结帐时，他们结帐的时间也会再次登记在checktime中，但这一次 checktype标志变为1

   CHECKTYPE=0

现在，在一天结束时，我希望看到每个过程的报告，不管它有多少次输入和输出，如

PASSID  CheckInTime             CheckOutTime                     UID
PS1 2015-08-05 01:12:02.100     2015-08-06 02:17:02.310     CAA0322
PS3 2015-08-05 01:12:02.100     2015-08-06 02:17:02.310     CAA0322
PS3 2015-08-02 11:11:01.210     2015-08-02 12:02:04.147     CAA0322
PS4 2015-08-03 01:02:03.200     22015-08-03 11:11:01.233    CAA0322

上述结果显示假定以下情况

• CheckInTime是描述中该ID的checktype=0的第一个checktime
• CheckOutTime是描述中该ID的checktype=1的第一个checktime

我尝试使用分区，但失败了，我只能返回一条记录，但我希望看到我的代码下面的所有记录，但它只返回一条记录，我希望在不指定任何passid的情况下返回所有记录

declare @PassID varchar(30)='PS3';

with LastEntryData(PASSID, CHECKTIME, CHECKTYPE, GateID, UID)
as 
(
    select top 2 
        c.PASSID, c.CHECKTIME, c.CHECKTYPE, c.GateID, c.UID     
    from 
        CHECKINOUT c
    where 
        c.PASSID = @PassID
    order by 
        c.CHECKTIME desc
    ),
    CheckInTime(CHECKTIME) as
    (
        select 
            i.CHECKTIME 
        from 
            LastEntryData i
        where 
            i.CHECKTYPE = 0),
    CheckOutTime(CHECKTIME) as
    (
        select 
            i.CHECKTIME 
        from 
            LastEntryData i
        where 
            i.CHECKTYPE = 1)
select 
    l.PASSID, CheckInTime = i.CHECKTIME,
    CheckOutTime = o.CHECKTIME, l.UID 
from 
    LastEntryData l, CheckInTime i, CheckOutTime o
group by 
    l.PASSID, l.GateID, l.UID, i.CHECKTIME, o.CHECKTIME

让我们做一张桌子。（如果这样做，您将获得更多答案。）

我将做出一些有根据的猜测，因为您的示例数据与预期输出不匹配

此查询（在PostgreSQL中）返回所有签入时间

select uid, passid, checktime as time_in
from checkinout
where checktype = 0
order by uid, passid, checktime;

---

-在ANSI-92 SQL标准（20多年前）中，旧样式的逗号分隔表列表样式被正确的ANSI

JOIN

语法所取代，它的使用是不鼓励的分区与查询无关，它们处理物理存储。我可以通过任何方式获得上述输出，请解释GateID…它不在示例数据中，但在SQL和结果中？数据与预期输出不匹配。在数据中，PS1在2015-08-05根本没有报到。迈克·谢里尔你刚刚救了我一天。子查询是我一直在寻找的，它可以返回一个带有签入的通行证，即使那个人还没有签出

select uid, passid, checktime as time_in
from checkinout
where checktype = 0
order by uid, passid, checktime;

uid passid time_in -- CAA0322 PS1 2015-09-05 11:11:01.21 CAA0322 PS3 2015-08-02 11:11:01.21 CAA0322 PS3 2015-08-05 01:12:02.1 CAA0322 PS4 2015-08-03 01:02:03.2

select 
  t1.uid, t1.passid, t1.checktime,
  (select min(checktime) 
   from checkinout 
   where uid = t1.uid 
     and passid = t1.passid 
     and checktime >= t1.checktime
     and checktype = 1) as time_out
from checkinout t1
where checktype = 0
order by t1.uid, t1.passid, t1.checktime;

uid passid time_in time_out -- CAA0322 PS1 2015-09-05 11:11:01.21 CAA0322 PS3 2015-08-02 11:11:01.21 2015-08-02 12:02:04.147 CAA0322 PS3 2015-08-05 01:12:02.1 2015-08-06 02:17:02.31 CAA0322 PS4 2015-08-03 01:02:03.2 2015-08-03 11:11:01.233

with checkinout_extended as (
  select uid, passid, checktime as time_in, checktype,
  lead(checktime) over (partition by uid, passid order by checktime, checktype) as time_out
  from checkinout
)
select uid, passid, time_in, time_out
from checkinout_extended
where checktype = 0
order by uid, passid, time_in;