我有一个要求
START_DATE：03/01/2018
END_DATE：31/01/2018
我需要一个查询，列出这两个日期之间的所有周开始日期和结束日期，如下所示
StartWeek EndWeek
03/01/2018 04/01/2018
07/01/2018 11/01/2018
14/01/2018 18/01/2018
21/01/2018 25/01/2018
28/01/2018 31/01/2018
这里的工作日是从周日到周四。周从星期日开始
答案 0 :(得分：2)
试试这个，
CREATE OR REPLACE PACKAGE week_pkg
AS
TYPE week_rec IS RECORD (start_week DATE, end_week DATE);
TYPE week_tab IS TABLE OF week_rec;
FUNCTION get_weeks (p_start_date DATE, p_end_date DATE) RETURN week_tab PIPELINED;
END;
/
CREATE OR REPLACE PACKAGE BODY week_pkg
AS
FUNCTION get_weeks (p_start_date DATE, p_end_date DATE) RETURN week_tab PIPELINED
IS
v_date DATE;
v_week_rec week_rec;
BEGIN
v_date := p_start_date - 7;
LOOP
v_week_rec.start_week := NEXT_DAY(v_date, 'SUNDAY');
IF v_week_rec.start_week < p_start_date THEN
v_week_rec.start_week:= p_start_date;
END IF;
v_week_rec.end_week := NEXT_DAY(v_date, 'THURSDAY');
IF v_week_rec.end_week >= p_end_date THEN
v_week_rec.end_week := p_end_date;
PIPE ROW (v_week_rec);
EXIT;
ELSIF v_week_rec.end_week <= p_start_date THEN
v_week_rec.end_week := NEXT_DAY(v_week_rec.start_week, 'THURSDAY');
END IF;
v_date := v_week_rec.end_week;
PIPE ROW (v_week_rec);
END LOOP;
END;
END;
/
SELECT *
FROM table(week_pkg.get_weeks(to_date('03-JAN-2018', 'DD-MON-YYYY'), to_date('31-JAN-2018', 'DD-MON-YYYY')));
输出：
START_WEEK END_WEEK
---------- ---------
03-JAN-18 04-JAN-18
07-JAN-18 11-JAN-18
14-JAN-18 18-JAN-18
21-JAN-18 25-JAN-18
28-JAN-18 31-JAN-18
答案 1 :(得分：0)
您可以使用此SQL。
WITH t (START_DATE, END_DATE)
AS (SELECT TO_DATE ('03/01/2018', 'dd/mm/yyyy'),
TO_DATE ('31/01/2018', 'dd/mm/yyyy')
FROM DUAL)
SELECT DISTINCT
CASE
WHEN LEVEL = 1
THEN
START_DATE
ELSE
CASE
WHEN TRUNC (START_DATE + LEVEL + 7, 'DAY') > END_DATE
THEN
TRUNC (START_DATE + LEVEL, 'DAY')
ELSE
TRUNC (START_DATE + LEVEL + 7, 'DAY')
END
END
START_DATE,
CASE
WHEN NEXT_DAY (
CASE
WHEN LEVEL = 1 THEN START_DATE
ELSE TRUNC (START_DATE + LEVEL + 7, 'DAY')
END,
'THURSDAY') > END_DATE
THEN
END_DATE
ELSE
NEXT_DAY (
CASE
WHEN LEVEL = 1
THEN
CASE
WHEN TRIM (TO_CHAR (START_DATE, 'DAY')) = 'THURSDAY'
THEN
START_DATE - 7
ELSE
START_DATE
END
ELSE
TRUNC (START_DATE + LEVEL + 7, 'DAY')
END,
'THURSDAY')
END
END_DATE
FROM DUAL CROSS JOIN t
CONNECT BY LEVEL < END_DATE - START_DATE;
答案 2 :(得分：0)
您可以使用递归子查询分解子句：
WITH input_dates ( start_date, end_date ) AS (
SELECT DATE '2018-01-03', DATE '2018-01-31'
FROM DUAL
),
valid_start_date ( start_date, end_date ) AS (
SELECT CASE
WHEN start_date - TRUNC( start_date, 'IW' )
IN (
0, -- Monday
1, -- Tuesday
2, -- Wednesday
3, -- Thursday
6 -- Sunday
)
THEN start_date
ELSE NEXT_DAY( start_date, 'SUNDAY' )
END,
end_date
FROM input_dates
),
dates ( start_week, end_week, end_date ) AS (
SELECT start_date,
LEAST( NEXT_DAY( start_date, 'THURSDAY' ), end_date ),
end_date
FROM valid_start_date
WHERE start_date <= end_date
UNION ALL
SELECT NEXT_DAY( start_week, 'SUNDAY' ),
LEAST( end_week + INTERVAL '7' DAY, end_date ),
end_date
FROM dates
WHERE NEXT_DAY( start_week, 'SUNDAY' ) <= end_date
)
SELECT start_week, end_week
FROM dates;
输出：
START_DATE END_DATE
---------- ----------
2018-01-03 2018-01-04
2018-01-07 2018-01-11
2018-01-14 2018-01-18
2018-01-21 2018-01-25
2018-01-28 2018-01-31
答案 3 :(得分：0)
类似的东西：Oracle 11及更高版本（也许是Oracle 9和10 ......）
如果你的星期从星期日开始，也许你应该在内部sql中减去1天。 （“选择截断（dt，'IW'） - 1作为beg_period”从星期日开始的一周 - 而不是“select trunc（dt，'IW'）作为beg_period”从星期一开始的一周）
with periods as
(
select greatest(to_date(&date_from, 'dd.mm.yyyy'), beg_period) beg_period -- date_from or first day of week
, least(to_date(&date_to, 'dd.mm.yyyy'), lead(beg_period) over (order by beg_period)-1 ) end_period -- date_to or last day of week (and =NULL if week begins AFTER date_to)
from
(select trunc(dt, 'IW')-1 as beg_period -- calc the beginning of week for every day generated, taking week beginning in Sunday ("-1")
from
(-- generate all the days between &date_from and FIRST DAY OF NEXT WEEK JUST AFTER &date_to (including)
select to_date(&date_from, 'dd.mm.yyyy') + (level-1) as dt
from dual
connect by level <= (trunc(to_date(&date_to, 'dd.mm.yyyy'), 'IW') + 7 -- also calc the first day of next week just AFTER date_to; it needed
-- for lead(...) in further (just to calc end-of-week for date_to, by lead(...))
- to_date(&date_from, 'dd.mm.yyyy')
+ 1
)
)
group by trunc(dt, 'IW') -- take only 1 beg_period for every week (may use "group by", or "distinct")
)
)
select beg_period
, end_period - case to_char(end_period, 'DAY','NLS_DATE_LANGUAGE=''numeric date language''') when '7' then 2 when '6' then 1 else 0 end
as end_period -- correction for weekend (Friday and Saturday) for week beginning in Sunday
from periods
where end_period is not null -- exclude next week just after date_to (which has end_period=NULL)
order by 1
;
查看最后一个“案例”：它取决于NLS。在我的表达中，我的意思是星期六='7'和星期五='6'（我想这是正确的，如果一周从星期日开始）。