让我说我在sql server数据库中有这个表，通过增加hist_pp进行排序：
hist_id hist_yr hist_pp hist_empl_id hist_empl_sect_id
90619 2017 5 00018509 61
92295 2017 6 00018509 61
93991 2017 7 00018509 61
95659 2017 8 00018509 99
103993 2017 9 00018509 99
120779 2017 10 00018509 99
我想找到hist_empl_sect_id从一组数字中的任何值变化的行，例如（60,61,62,63）到另一组数字中的任何值，比如说（98,99,100等） 。它必须是每年，所以对于2017年的值.hist_pp将是一年中增加的数字。 hist_id也是一个id自动编号列。
该员工应该返回
95659 2017 8 00018509 99
我尝试了一些我在其他帖子中看到的例子，用CTE等尝试过，我似乎无法让它发挥作用。
这是我尝试过的一个例子，但没有用，当员工应该只有1时，为员工获得了多行：
select a.hist_id, a.hist_yr, format(cast(a.hist_pp as integer), '0#') as hist_pp, a.hist_empl_id, a.hist_empl_sect_id
from temshist a
where a.hist_empl_sect_id <>
(SELECT top 1 b.hist_empl_sect_id
FROM temshist as b
where a.hist_empl_id = b.hist_empl_id
and a.hist_yr = b.hist_yr
and a.hist_pp > b.hist_pp
Order by b.hist_pp desc
)
order by hist_empl_id
答案 0 :(得分：2)
我怀疑 Lag（）非常适合。
示例
;with cte as (
Select *
,PrevValue= Lag(hist_empl_sect_id,1,hist_empl_sect_id) over (Partition by hist_empl_id Order By hist_pp)
From @YourTable
)
Select *
From cte Where PrevValue/98<>hist_empl_sect_id/98
编辑 - VamsiPrabhala指出
您也可以按年份分区
,PrevValue= Lag(hist_empl_sect_id,1,hist_empl_sect_id) over (Partition by hist_yr,hist_empl_id Order By hist_pp)
答案 1 :(得分：0)
这是另一种选择，我用CTE嘲笑它来模拟你的数据，然后重新加入。
with emp_hist as
(
select 90619 as hist_id, 2017 as hist_yr, 5 as hist_pp, '00018509' as hist_empl_id, 61 as hist_empl_sect_id from dual
union all
select 92295 as hist_id, 2017 as hist_yr, 6 as hist_pp, '00018509' as hist_empl_id, 61 as hist_empl_sect_id from dual
union all
select 93991 as hist_id, 2017 as hist_yr, 7 as hist_pp, '00018509' as hist_empl_id, 61 as hist_empl_sect_id from dual
union all
select 95659 as hist_id, 2017 as hist_yr, 8 as hist_pp, '00018509' as hist_empl_id, 99 as hist_empl_sect_id from dual
union all
select 103993 as hist_id, 2017 as hist_yr, 9 as hist_pp, '00018509' as hist_empl_id, 99 as hist_empl_sect_id from dual
union all
select 120779 as hist_id, 2017 as hist_yr, 10 as hist_pp, '00018509' as hist_empl_id, 99 as hist_empl_sect_id from dual
)
select eh2.*
from emp_hist eh1
join emp_hist eh2
on eh1.hist_empl_id = eh2.hist_empl_id
and eh1.hist_pp = (eh2.hist_pp - 1)
and eh1.hist_yr = eh2.hist_yr
where eh2.hist_empl_sect_id in (98, 99, 100)
and eh1.hist_empl_sect_id in (60, 61, 62, 63)
;
答案 2 :(得分：0)
您可以使用案例陈述（以确定组成员资格）和滞后窗口函数（以比较两个连续行）按雇员和年份划分并按hist_pp排序
这假定 （1）员工ID可以跨越多年 （2）Hist_pp对于每个员工ID，年份组合是唯一的 （3）如果员工ID中的hist_empl_sect_id只有一个唯一值，年份组合（该年度该员工的hist_empl_sect_id不会更改），则结果集不应包含该员工ID，年份组合的任何行。
Hist_pp可能有差距。
select hist_id, hist_yr, hist_pp, hist_empl_id, hist_empl_sect_id
from
(
select a.hist_id, a.hist_yr,
format(cast(a.hist_pp as integer), '0#') as hist_pp,
a.hist_empl_id,
-- hist_empl_sect_id of current row
a.hist_empl_sect_id,
-- hist_empl_sect_id of preceding row, when ordered by hist_pp for each employee year combination
lag(a.hist_empl_sect_id, 1)
OVER (
PARTITION BY a.hist_empl_id,a.hist_yr
ORDER BY format(cast(a.hist_pp as integer), '0#')
) as prev_hist_empl_sect_id
from temshist a
) as outr
where
-- group membership of hist_empl_sect_id of current row
(case when hist_empl_sect_id IN (98, 99, 100) then 1 else 0 end)
<>
-- group membership of hist_empl_sect_id of preceding row, ordered by hist_pp for each year
(case when prev_hist_empl_sect_id IN (98, 99, 100) then 1 else 0 end)
AND
-- Preceding row does not belong to a different employee or year
prev_hist_empl_sect_id IS NOT NULL
答案 3 :(得分：0)
查看结果数据，我假设每个hist_yr和hist_empl_id，您需要hist_pp为min且hist_empl_sect_id为{1}}的记录行max，下面是生成所需输出的查询。
SELECT t3.* from
(SELECT t2.*, min(hist_pp) over(partition BY hist_yr, hist_empl_id) AS hist_pp_minValue
FROM
(SELECT hist_id, hist_yr, hist_pp, hist_empl_id, hist_empl_sect_id, r1, max(r1) over (partition BY hist_yr, hist_empl_id) AS maxRank
FROM
(SELECT hist_id, hist_yr, hist_pp, hist_empl_id, hist_empl_sect_id, dense_rank() over(partition BY hist_yr, hist_empl_id
ORDER BY hist_empl_sect_id) AS r1
FROM table1)t1) t2
WHERE t2.maxRank = t2.r1 )t3
WHERE t3.hist_pp_minValue = t3.hist_pp
我已对相关数据进行了测试，结果如下。
hist_id | hist_yr | hist_pp | hist_empl_id | hist_empl_sect_id
---------------------------------------------------------
95659 2017 8 18509 99
为了重新保证，我增加了一些样本数据，如下所示。
insert into table1 values(90619 ,2018,5 ,00018508,62);
insert into table1 values(92295 ,2018,6 ,00018508,62);
insert into table1 values(93991 ,2018,7 ,00018508,62);
insert into table1 values(95659 ,2018,8 ,00018508,91);
insert into table1 values(103993 ,2018,9 ,00018508,91);
insert into table1 values(120779 ,2018,10 ,00018508,91);
以下是生成的结果。
hist_id | hist_yr | hist_pp | hist_empl_id | hist_empl_sect_id
---------------------------------------------------------
95659 2017 8 18509 99
95659 2018 8 18508 91
您可以查看演示here
希望这会有所帮助。
答案 4 :(得分：0)
我认为您想要使用CTE来解决这个问题。这类似于John Cappelletti所做的，但不需要SQL 2012或更高版本。
declare @temshist table
(
hist_id int,
hist_yr int,
hist_pp int,
hist_empl_id varchar(max),
hist_empl_sect_id int
)
insert into @temshist ( hist_id, hist_yr, hist_pp, hist_empl_id, hist_empl_sect_id )
values
( 90619, 2017, 5, '00018509', 61 ),
( 92295, 2017, 6, '00018509', 61 ),
( 93991, 2017, 7, '00018509', 61 ),
( 95659, 2017, 8, '00018509', 99 ),
( 103993, 2017, 9, '00018509', 99 ),
( 120779, 2017, 10, '00018509', 99 )
;with empl_cte as
(
select
row_number() over (partition by hist_empl_id, hist_yr order by hist_pp) as [rn],
hist_id,
hist_yr,
hist_pp,
hist_empl_id,
hist_empl_sect_id
from @temshist
)
select
nxt.hist_id,
nxt.hist_yr,
nxt.hist_pp,
nxt.hist_empl_id,
nxt.hist_empl_sect_id
from empl_cte prv
left join empl_cte nxt on
prv.hist_empl_id = nxt.hist_empl_id and
prv.rn = nxt.rn - 1
where prv.hist_empl_sect_id in (60, 61, 62, 63/*, ...*/) and nxt.hist_empl_sect_id in (98, 99, 100/*, ...*/)