我需要在任何指定日期之前7天找到日期。开始日期可以是任何日期，但是，结束日期不应该是星期日或标记为假日的日期。只能连续2个假期。
我的代码可行，但是，它非常冗长。是否有更简单的解决方案？
此外，如果所需的日期范围在将来的某个时间点发生变化，或者连续假期的可能数量发生变化，则需要在多个地方更新脚本，如果可以减少，那么这将是很好的。
我已经有一张桌子（日历），其中包含大量日期，一周中的某一天以及日期是否被视为假日。像这样：
START_DATE, DAY_OF_WEEK, HOLIDAY
10-DEC-17 , Sun , 0
11-DEC-17 , Mon , 0
12-DEC-17 , Tue , 0
13-DEC-17 , Wed , 0
14-DEC-17 , Thu , 0
15-DEC-17 , Fri , 0
16-DEC-17 , Sat , 0
17-DEC-17 , Sun , 0
18-DEC-17 , Mon , 0
19-DEC-17 , Tue , 0
20-DEC-17 , Wed , 0
21-DEC-17 , Thu , 0
22-DEC-17 , Fri , 0
23-DEC-17 , Sat , 0
24-DEC-17 , Sun , 0
25-DEC-17 , Mon , 1
26-DEC-17 , Tue , 1
27-DEC-17 , Wed , 0
28-DEC-17 , Thu , 0
29-DEC-17 , Fri , 0
30-DEC-17 , Sat , 0
31-DEC-17 , Sun , 0
01-JAN-18 , Mon , 1
etc...
预期的OUTPUT类似于：
START_DATE, END_DATE
10-DEC-17, 18-DEC-17
11-DEC-17, 18-DEC-17
12-DEC-17, 19-DEC-17
13-DEC-17, 20-DEC-17
14-DEC-17, 21-DEC-17
15-DEC-17, 22-DEC-17
16-DEC-17, 23-DEC-17
17-DEC-17, 27-DEC-17
18-DEC-17, 27-DEC-17
19-DEC-17, 27-DEC-17
20-DEC-17, 27-DEC-17
21-DEC-17, 28-DEC-17
22-DEC-17, 29-DEC-17
23-DEC-17, 30-DEC-17
24-DEC-17, 02-JAN-18
25-DEC-17, 02-JAN-18
26-DEC-17, 02-JAN-18
27-DEC-17, 03-JAN-18
28-DEC-17, 04-JAN-18
etc...
以下是我现有的代码。我的方法是，因为只能连续3天被排除的日子（星期日，然后是2个假日），所以我提前4天检查每个日期，然后选择第一个不是排除日期的日期。其中一个应始终是有效的结束日期。
with temp as
(
select
start_date,
case
when lead(day_of_week, 7) over(order by start_date) = 'Sun'
or lead(holiday, 7) over(order by start_date) = 1
then null
else
lead(start_date, 7) over(order by start_date)
end as days_7,
case
when lead(day_of_week, 8) over(order by start_date) = 'Sun'
or lead(holiday, 8) over(order by start_date) = 1
then null
else
lead(start_date, 8) over(order by start_date)
end as days_8,
case
when lead(day_of_week, 9) over(order by start_date) = 'Sun'
or lead(holiday, 9) over(order by start_date) = 1
then null
else
lead(start_date, 9) over(order by start_date)
end as days_9,
case
when lead(day_of_week, 10) over(order by start_date) = 'Sun'
or lead(holiday, 10) over(order by start_date) = 1
then null
else
lead(start_date, 10) over(order by start_date)
end as days_10
from
calendar
)
select
start_date,
COALESCE(days_7, days_8, days_9, days_10) as end_date
from
temp
答案 0 :(得分：2)
Oracle 11g R2架构设置：
CREATE TABLE Calendar( START_DATE, HOLIDAY ) AS
SELECT TO_DATE( '10-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '11-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '12-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '13-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '14-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '15-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '16-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '17-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '18-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '19-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '20-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '21-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '22-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '23-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '24-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '25-DEC-17', 'DD-MON-YY' ), 1 FROM DUAL UNION ALL
SELECT TO_DATE( '26-DEC-17', 'DD-MON-YY' ), 1 FROM DUAL UNION ALL
SELECT TO_DATE( '27-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '28-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '29-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '30-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '31-DEC-17', 'DD-MON-YY' ), 0 FROM DUAL UNION ALL
SELECT TO_DATE( '01-JAN-18', 'DD-MON-YY' ), 1 FROM DUAL;
查询1 ：
SELECT start_date,
LEAD( end_date, 7 ) OVER ( ORDER BY start_date ) AS end_date
FROM (
SELECT start_date,
LAST_VALUE(
CASE
WHEN Holiday = 0
AND start_date - TRUNC( start_date, 'IW' ) < 6
THEN start_date
END
) IGNORE NULLS OVER ( ORDER BY start_date DESC ) AS End_date
FROM Calendar
)
<强> Results ：
| START_DATE | END_DATE |
|----------------------|----------------------|
| 2017-12-10T00:00:00Z | 2017-12-18T00:00:00Z |
| 2017-12-11T00:00:00Z | 2017-12-18T00:00:00Z |
| 2017-12-12T00:00:00Z | 2017-12-19T00:00:00Z |
| 2017-12-13T00:00:00Z | 2017-12-20T00:00:00Z |
| 2017-12-14T00:00:00Z | 2017-12-21T00:00:00Z |
| 2017-12-15T00:00:00Z | 2017-12-22T00:00:00Z |
| 2017-12-16T00:00:00Z | 2017-12-23T00:00:00Z |
| 2017-12-17T00:00:00Z | 2017-12-27T00:00:00Z |
| 2017-12-18T00:00:00Z | 2017-12-27T00:00:00Z |
| 2017-12-19T00:00:00Z | 2017-12-27T00:00:00Z |
| 2017-12-20T00:00:00Z | 2017-12-27T00:00:00Z |
| 2017-12-21T00:00:00Z | 2017-12-28T00:00:00Z |
| 2017-12-22T00:00:00Z | 2017-12-29T00:00:00Z |
| 2017-12-23T00:00:00Z | 2017-12-30T00:00:00Z |
| 2017-12-24T00:00:00Z | (null) |
| 2017-12-25T00:00:00Z | (null) |
| 2017-12-26T00:00:00Z | (null) |
| 2017-12-27T00:00:00Z | (null) |
| 2017-12-28T00:00:00Z | (null) |
| 2017-12-29T00:00:00Z | (null) |
| 2017-12-30T00:00:00Z | (null) |
| 2017-12-31T00:00:00Z | (null) |
| 2018-01-01T00:00:00Z | (null) |
答案 1 :(得分：0)
给出假期表：
with holidays as (
select to_date('10-DEC-17','dd-MON-YY') as mydate , 'Sun' as myday, 0 holiday from dual
union all
select to_date('11-DEC-17','dd-MON-YY') as mydate , 'Mon' as myday, 0 holiday from dual
union all
select to_date('12-DEC-17','dd-MON-YY') as mydate , 'Tue' as myday, 0 holiday from dual
union all
select to_date('13-DEC-17','dd-MON-YY') as mydate , 'Wed' as myday, 0 holiday from dual
union all
select to_date('14-DEC-17','dd-MON-YY') as mydate , 'Thu' as myday, 0 holiday from dual
union all
select to_date('15-DEC-17','dd-MON-YY') as mydate , 'Fri' as myday, 0 holiday from dual
union all
select to_date('16-DEC-17','dd-MON-YY') as mydate , 'Sat' as myday, 0 holiday from dual
union all
select to_date('17-DEC-17','dd-MON-YY') as mydate , 'Sun' as myday, 0 holiday from dual
union all
select to_date('18-DEC-17','dd-MON-YY') as mydate , 'Mon' as myday, 0 holiday from dual
union all
select to_date('19-DEC-17','dd-MON-YY') as mydate , 'Tue' as myday, 0 holiday from dual
union all
select to_date('20-DEC-17','dd-MON-YY') as mydate , 'Wed' as myday, 0 holiday from dual
union all
select to_date('21-DEC-17','dd-MON-YY') as mydate , 'Thu' as myday, 0 holiday from dual
union all
select to_date('22-DEC-17','dd-MON-YY') as mydate , 'Fri' as myday, 0 holiday from dual
union all
select to_date('23-DEC-17','dd-MON-YY') as mydate , 'Sat' as myday, 0 holiday from dual
union all
select to_date('24-DEC-17','dd-MON-YY') as mydate , 'Sun' as myday, 0 holiday from dual
union all
select to_date('25-DEC-17','dd-MON-YY') as mydate , 'Mon' as myday, 1 holiday from dual
union all
select to_date('26-DEC-17','dd-MON-YY') as mydate , 'Tue' as myday, 1 holiday from dual
union all
select to_date('27-DEC-17','dd-MON-YY') as mydate , 'Wed' as myday, 0 holiday from dual
union all
select to_date('28-DEC-17','dd-MON-YY') as mydate , 'Thu' as myday, 0 holiday from dual
union all
select to_date('29-DEC-17','dd-MON-YY') as mydate , 'Fri' as myday, 0 holiday from dual
union all
select to_date('30-DEC-17','dd-MON-YY') as mydate , 'Sat' as myday, 0 holiday from dual
union all
select to_date('31-DEC-17','dd-MON-YY') as mydate , 'Sun' as myday, 0 holiday from dual
union all
select to_date('01-JAN-18','dd-MON-YY') as mydate , 'Mon' as myday, 1 holiday from dual),
在这里，我找到工作日，然后使用第一行来获得所需的日期
working as (
select holidays.*, case when myday='Sun' or holiday=1
then 0 else 1 end as working_day from holidays
)
select mydate from (
select mydate from working where
mydate>=to_date('21/12/2017','dd/mm/rrrr')+7
order by (case when working_day=0 then null else mydate end) asc nulls last
) where rownum=1