Oracle SQL 新手在这里和第一次海报。
我以为这很简单,直到我意识到我无法弄清楚如何拆分返回分配。
这是我的任务表:ASGN
ID ST_DT END_DT POS LOCN STATUS WAGE_CD
-- ---------- ---------- ----- ---- ------ -------
A 12-31-2006 08-16-2009 CLERK LAX 3 A
A 08-17-2009 10-04-2009 CLERK LAX 0 Z
A 10-05-2009 06-30-2010 OPR NYC 3 A
A 07-01-2010 12-31-2010 OPR NYC 3 B
A 01-01-2011 06-30-2012 OPR NYC 3 C
A 07-01-2012 04-09-2013 OPR NYC 3 D
A 04-10-2013 06-30-2013 CLERK LAX 3 A
A 07-01-2013 08-10-2014 CLERK LAX 3 B
A 07-01-2013 08-10-2014 CLERK LAX 3 C
B 04-10-2013 05-31-2013 SUP LAX 3 A
B 06-01-2013 06-30-2014 SUP LAX 0 Z
B 07-01-2013 08-10-2014 SUP LAX 3 B
B 08-11-2014 08-11-2014 CLERK NYC 3 A
B 08-12-2014 01-11-2015 SUP LAX 3 A
B 01-12-2015 02-10-2016 SUP LAX 3 B
B 02-11-2016 08-12-2016 OPER SFO 3 A
B 02-11-2016 08-12-2016 OPER SFO 3 B
我已经在下面尝试过这个,结果出乎意料。
SELECT *
FROM (
SELECT ID
,MIN(ST_DT) ST_DT
,MAX(END_DT) END_DT
,POS
,LOCN
,STATUS
FROM ASGN
GROUP BY ID, LOCN, POS, STATUS
) SUBQRY
ORDER BY ID, ST_DT
出乎意料的结果,但在情理之中。这是返回分配不与以前的分配相结合的地方。
ID ST_DT END_DT POS LOCN STATUS
-- ---------- ---------- ----- ---- ------
A 12-31-2006 08-10-2014 CLERK LAX 3
A 08-17-2009 10-04-2009 CLERK LAX 0
A 10-05-2009 04-09-2010 OPR NYC 3
B 04-10-2013 02-10-2015 SUP LAX 3
B 06-01-2013 06-30-2014 SUP LAX 0
B 08-11-2014 08-11-2014 CLERK NYC 3
B 02-11-2016 08-12-2016 OPER SFO 3
结果我想查看每个 ID、POSition、LOCatioN 和 STATUS 中相邻日期的组合位置:
ID ST_DT END_DT POS LOCN STATUS
-- ---------- ---------- ----- ---- ------
A 12-31-2006 08-16-2009 CLERK LAX 3
A 08-17-2009 10-04-2009 CLERK LAX 0
A 10-05-2009 04-09-2010 OPR NYC 3
A 04-10-2013 08-10-2014 CLERK LAX 3
B 04-10-2013 05-31-2013 SUP LAX 3
B 06-01-2013 06-30-2014 SUP LAX 0
B 07-01-2013 08-10-2014 SUP LAX 3
B 08-11-2014 08-11-2014 CLERK NYC 3
B 08-12-2014 02-10-2015 SUP LAX 3
B 02-11-2016 08-12-2016 OPER SFO 3
我问了一个更高级的 Oracle SQL 程序员,他说我必须给我们 PLSQL,但我认为必须有一种方法可以通过 SQL 使这个工作。
测试设置脚本:
create table asgn
(id varchar2(10)
,st_dt date
,end_dt date
,pos varchar2(10)
,locn varchar2(10)
,status number
,wage_cd varchar2(10));
insert into asgn values('A',to_date('12-31-2006','mm-dd-yyyy'),to_date('08-16-2009','mm-dd-yyyy'),'CLERK','LAX',3,'A');
insert into asgn values('A',to_date('08-17-2009','mm-dd-yyyy'),to_date('10-04-2009','mm-dd-yyyy'),'CLERK','LAX',0,'Z');
insert into asgn values('A',to_date('10-05-2009','mm-dd-yyyy'),to_date('06-30-2010','mm-dd-yyyy'),'OPR','NYC',3,'A');
insert into asgn values('A',to_date('07-01-2010','mm-dd-yyyy'),to_date('12-31-2010','mm-dd-yyyy'),'OPR','NYC',3,'B');
insert into asgn values('A',to_date('01-01-2011','mm-dd-yyyy'),to_date('06-30-2012','mm-dd-yyyy'),'OPR','NYC',3,'C');
insert into asgn values('A',to_date('07-01-2012','mm-dd-yyyy'),to_date('04-09-2013','mm-dd-yyyy'),'OPR','NYC',3,'D');
insert into asgn values('A',to_date('04-10-2013','mm-dd-yyyy'),to_date('06-30-2013','mm-dd-yyyy'),'CLERK','LAX',3,'A');
insert into asgn values('A',to_date('07-01-2013','mm-dd-yyyy'),to_date('08-10-2014','mm-dd-yyyy'),'CLERK','LAX',3,'B');
insert into asgn values('A',to_date('07-01-2013','mm-dd-yyyy'),to_date('08-10-2014','mm-dd-yyyy'),'CLERK','LAX',3,'C');
insert into asgn values('B',to_date('04-10-2013','mm-dd-yyyy'),to_date('05-31-2013','mm-dd-yyyy'),'SUP','LAX',3,'A');
insert into asgn values('B',to_date('06-01-2013','mm-dd-yyyy'),to_date('06-30-2014','mm-dd-yyyy'),'SUP','LAX',0,'Z');
insert into asgn values('B',to_date('07-01-2013','mm-dd-yyyy'),to_date('08-10-2014','mm-dd-yyyy'),'SUP','LAX',3,'B');
insert into asgn values('B',to_date('08-11-2014','mm-dd-yyyy'),to_date('08-11-2014','mm-dd-yyyy'),'CLERK','NYC',3,'A');
insert into asgn values('B',to_date('08-12-2014','mm-dd-yyyy'),to_date('01-11-2015','mm-dd-yyyy'),'SUP','LAX',3,'A');
insert into asgn values('B',to_date('01-12-2015','mm-dd-yyyy'),to_date('02-10-2016','mm-dd-yyyy'),'SUP','LAX',3,'B');
insert into asgn values('B',to_date('02-11-2016','mm-dd-yyyy'),to_date('08-12-2016','mm-dd-yyyy'),'OPER','SFO',3,'A');
insert into asgn values('B',to_date('02-11-2016','mm-dd-yyyy'),to_date('08-12-2016','mm-dd-yyyy'),'OPER','SFO',3,'B');
最佳答案
员工 A 在 07-01-2013 到 08-10-2014 有两行;我认为这是一个错误,我删除了其中一行。
除此之外,这是“tabibitosan 方法”的应用,用于解决日期范围的“间隙和岛屿”问题。诀窍在于为相邻间隔创建“组”( gp
)。
with
prep ( id, st_dt, end_dt, gp, pos, locn, status ) as (
select id, st_dt, end_dt,
end_dt - sum( end_dt - st_dt + 1 ) over (partition by id, pos, locn, status
order by st_dt),
pos, locn, status
from asgn
)
select id, min(st_dt) as st_dt, max(end_dt) as end_dt, pos, locn, status
from prep
group by id, gp, pos, locn, status
order by id, st_dt
;
ID ST_DT END_DT POS LOCN STATUS
---------- ---------- ---------- ---------- ---------- ----------
A 12-31-2006 08-16-2009 CLERK LAX 3
A 08-17-2009 10-04-2009 CLERK LAX 0
A 10-05-2009 04-09-2013 OPR NYC 3
A 04-10-2013 08-10-2014 CLERK LAX 3
B 04-10-2013 05-31-2013 SUP LAX 3
B 06-01-2013 06-30-2014 SUP LAX 0
B 07-01-2013 08-10-2014 SUP LAX 3
B 08-11-2014 08-11-2014 CLERK NYC 3
B 08-12-2014 02-10-2016 SUP LAX 3
B 02-11-2016 08-12-2016 OPER SFO 3
10 rows selected
关于sql - 合并日期范围,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39114459/