我需要编写一个 sql 查询，返回两个给定日期之间的工作日数（周一 - 周五）。
我想知道最有效的方法是什么？
SELECT --Start with total number of days including weekends
(DATEDIFF(dd,@StartDate,@EndDate)+1) --Subtact 2 days for each full weekend
(DATEDIFF(wk,@StartDate,@EndDate)*2) --If StartDate is a Sunday, Subtract 1
ELSE 0 END) --If EndDate is a Saturday, Subtract 1
FROM dual
然后，能够从该计数中删除假期（例如圣诞节和节礼日）也会很有帮助。
有任何想法吗？
计算两个日期之间的工作日数的简单方法是：
SELECT
date1,
date2,
((date2-date1)-2*FLOOR((date2-date1)/7)-DECODE(SIGN(TO_CHAR(date2,'D')-
TO_CHAR(date1,'D')),-1,2,0)+DECODE(TO_CHAR(date1,'D'),7,1,0)-
DECODE(TO_CHAR(date2,'D'),7,1,0))*24 as WorkDays
FROM
tablename
ORDER BY date1,date2
就是这么简单：
SQL> Select count(*)
2 from ( select rownum rnum
3 from all_objects
4 where rownum <= to_date('18-dec-2009','dd-mon-yyyy') -
to_date('16-nov-2009')+1 )
5 where to_char( to_date('16-nov-2009','dd-mon-yyyy')+rnum-1, 'DY' )
6 not in ( 'SAT', 'SUN' )
COUNT(*)
----------
25
SQL> Select to_char( to_date('16-nov-2009','dd-mon-yyyy')+rnum-1, 'DY dd-mon-yyyy' )
2 from ( select rownum rnum
3 from all_objects
4 where rownum <= to_date('18-dec-2009','dd-mon-yyyy') - to_date('16-nov-2009')+1 )
5 where to_char( to_date('16-nov-2009','dd-mon-yyyy')+rnum-1, 'DY' )
6 not in ( 'SAT', 'SUN' )
DAY_DATE
---------------
MON 16-nov-2009
TUE 17-nov-2009
WED 18-nov-2009
THU 19-nov-2009
FRI 20-nov-2009
MON 23-nov-2009
TUE 24-nov-2009
WED 25-nov-2009
THU 26-nov-2009
FRI 27-nov-2009
MON 30-nov-2009
TUE 01-dec-2009
WED 02-dec-2009
THU 03-dec-2009
FRI 04-dec-2009
MON 07-dec-2009
TUE 08-dec-2009
WED 09-dec-2009
THU 10-dec-2009
FRI 11-dec-2009
MON 14-dec-2009
TUE 15-dec-2009
WED 16-dec-2009
THU 17-dec-2009
FRI 18-dec-2009
25 rows selected.
这是一个例子
with given_days(d) as(
select <<start_date>> + level - 1
from dual
connect by level < = (<<end_date>> - <<start_date>>) + 1
)
select count(*)
from given_days
where to_char(d, 'DY', 'NLS_DATE_LANGUAGE=english') not in ('SUN', 'SAT')
示范
HR\XE> with given_days as(
2 select (to_date('&&1', 'dd.mm.yyyy') + level - 1) as g_day
3 from dual
4 connect by level < = (to_date('&2', 'dd.mm.yyyy') - to_date('&&1', 'dd.mm.yyyy')) + 1
5 )
6 select count(g_day) as cnt
7 from given_days
8 where to_char(g_day, 'DY', 'NLS_DATE_LANGUAGE=english') not in ('SUN', 'SAT');
Enter value for 1: 10.10.2012
old 2: select to_date('&&1', 'dd.mm.yyyy') + level - 1
new 2: select to_date('10.10.2012', 'dd.mm.yyyy') + level - 1
Enter value for 2: 17.10.2012
old 4: connect by level < = (to_date('&2', 'dd.mm.yyyy') - to_date('&&1', 'dd.mm.yyyy')) + 1
new 4: connect by level < = (to_date('17.10.2012', 'dd.mm.yyyy') - to_date('10.10.2012', 'dd.mm.yyyy')) + 1
cnt
----------
6
它可以通过以下方式实现：
select SUM(decode ( to_CHAR((sysdate-ROWNUM),'DY'),'SUN',0,'SAT',0,1)) from all_objects where rownum < sysdate - (sysdate -9)
我就是这样做的，假设您已经有一个带有列的日历表，该列指示一天是否是工作日：向您的日历表添加一个新列，例如 workday_num 并使用运行编号填充一次
sum(case when workingday then 1 else 0 end)
over (order by calendardate rows unbounded preceding)
现在它是您日历的两个连接，以及 p_start_date 和 p_end_date 的 workday_nums 的简单区别。
干得好...
获取两个日期之间的工作日（周一至周五），然后减去假期。
create or replace
FUNCTION calculate_business_days (p_start_date IN DATE, p_end_date IN DATE)
RETURN NUMBER IS
v_holidays NUMBER;
v_start_date DATE := TRUNC (p_start_date);
v_end_date DATE := TRUNC (p_end_date);
BEGIN
IF v_end_date >= v_start_date
THEN
SELECT COUNT (*)
INTO v_holidays
FROM holidays
WHERE day BETWEEN v_start_date AND v_end_date
AND day NOT IN (
SELECT hol.day
FROM holidays hol
WHERE MOD(TO_CHAR(hol.day, 'J'), 7) + 1 IN (6, 7)
);
RETURN GREATEST (NEXT_DAY (v_start_date, 'MON') - v_start_date - 2, 0)
+ ( ( NEXT_DAY (v_end_date, 'MON')
- NEXT_DAY (v_start_date, 'MON')
)
/ 7
)
* 5
- GREATEST (NEXT_DAY (v_end_date, 'MON') - v_end_date - 3, 0)
- v_holidays;
ELSE
RETURN NULL;
END IF;
END calculate_business_days;
之后，您可以对其进行测试，例如：
select
calculate_business_days('21-AUG-2013','28-AUG-2013') as business_days
from dual;