如何在oracle中查找不包括周六、周日和节假日的48个工作时间日期
How to find 48 working hours date excluding saturday,sundays and holidays in oracle
某模块要求求第48、24个工作时间
要求：
假设如果我将 2nd May 作为参数传递给函数，则 48 hours 的输出应为 27th April，24 hours 的输出应为 28th APRIL（因为5月1日是假期，4月29日和30日属于周六和周日）
问题出在两个连续的假期上。例如，要创建一个 dummy 数据，我们插入 5 月 2 日作为假期，运行 3rd May 上的代码应该为 48 hours 检索 27th April 和 28th APRIL 对于 24 hours.
但我的功能似乎不适用于连续的假期。某处计数器增量似乎在错误的位置。
考虑因素：
周末：周六和周日
需要排除的天数：给定节假日日历中的周六、周日和节假日：
假期table创作：
CREATE TABLE HOLIDAY_TAB
(
HOL_DATE DATE,
DESCRIPTION VARCHAR2 (100) DEFAULT NULL
);
insert into HOLIDAY_TAB values (TO_DATE ('26-Jan-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('29-Mar-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('14-Apr-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('01-May-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('02-Jun-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('26-Jun-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('15-Aug-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('25-Aug-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('28-Sep-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('02-Oct-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('19-Oct-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('25-Dec-2017', 'DD-MON-YYYY'),NULL);
commit;
写入捕获假期的函数：
函数：
CREATE OR REPLACE FUNCTION CSE.F_HOL_CHECK_ABC (i_hol_date DATE)
RETURN DATE
AS
valid_working_day DATE := i_hol_date;
day_C holiday_nvs%ROWTYPE;
CURSOR c_hol
IS
SELECT *
FROM HOLIDAY_TAB
WHERE TRUNC (hol_date) = TRUNC (i_hol_date);
CURSOR c_hol_24
IS
SELECT *
FROM HOLIDAY_TAB
WHERE TRUNC (hol_date) = TRUNC (i_hol_date + 3);
CURSOR c_hol_48
IS
SELECT *
FROM HOLIDAY_NVS
WHERE TRUNC (hol_date) = TRUNC (i_hol_date + 2);
BEGIN
-- FOR rec24 IN c_hol_24
-- LOOP
-- IF (rec24.hol_date IS NOT NULL)
-- THEN
-- valid_working_day := i_hol_date - 1;
-- END IF;
-- END LOOP;
OPEN c_hol;
FETCH c_hol INTO day_C;
IF c_hol%FOUND
THEN
SELECT DECODE (TO_CHAR (i_hol_date - 1, 'D'),
1, i_hol_date - 3,
i_hol_date - )
INTO valid_working_day
FROM DUAL;
END IF;
CLOSE c_hol;
RETURN (valid_working_day);
END;
/
不确定功能是否正确。但是有一种奇怪的情况，即与日期文字相比，我的查询在尝试使用 SYSDATE 时没有给出相同的结果。
运行 1 手动设置日期：
SELECT TRUNC (
DECODE (
TO_CHAR (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY'), 'D'),
2, F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 4),
DECODE (
TO_CHAR (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY'), 'D'),
3, F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 4),
F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 2))))
AS "48HOURS",
TRUNC (
DECODE (
TO_CHAR (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY'), 'D'),
2, F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 3),
F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 1)))
AS "24HOURS"
FROM DUAL;
运行 2 系统日期：
SELECT TRUNC (
DECODE (
TO_CHAR (SYSDATE, 'D'),
2, F_HOL_CHECK_ABC (SYSDATE - 4),
DECODE (TO_CHAR (SYSDATE, 'D'),
3, F_HOL_CHECK_ABC (SYSDATE - 4),
F_HOL_CHECK_ABC (SYSDATE -2)))) as "48 hour" ,
TRUNC (
DECODE (
TO_CHAR (SYSDATE,
'D'),
2, F_HOL_CHECK_ABC (
SYSDATE - 3),
F_HOL_CHECK_ABC (
SYSDATE -1))) as "24 hour" from dual;
非常感谢您提供这方面的帮助。我所需要的只是跳过节假日、星期日和星期六，即；非工作时间，给我每天 48 小时工作时间和 24 小时工作时间
这是使用计数器对代码 2 进行的另一次尝试：
CREATE OR REPLACE FUNCTION CSE.F_HOL_CHECK_S_NS (i_hol_date DATE,
i_S_NS NUMBER)
RETURN DATE
AS
valid_working_day DATE := i_hol_date;
counter NUMBER := 0;
day_number NUMBER := 0;
hol_count NUMBER := 0;
day_C holiday_nvs%ROWTYPE;
CURSOR c_hol (hol_date_c DATE)
IS
SELECT *
FROM HOLIDAY_TAB
WHERE TRUNC (hol_date) = TRUNC (TO_DATE (hol_date_c, 'DD-MON-YYYY'));
BEGIN
IF i_S_NS = 0
THEN
LOOP
IF c_hol%ISOPEN
THEN
CLOSE c_hol;
END IF;
OPEN c_hol (valid_working_day);
IF c_hol%FOUND
THEN
valid_working_day := valid_working_day - 1;
SELECT TO_CHAR (TO_DATE (valid_working_day, 'DD-MON-YYYY'), 'D')
INTO day_number
FROM DUAL;
SELECT COUNT (*)
INTO hol_count
FROM HOLIDAY_TAB
WHERE TRUNC (hol_date) =
TRUNC (TO_DATE (valid_working_day, 'DD-MON-YYYY'));
--valid_working_day:=valid_working_day-1;
-- SELECT DECODE (TO_CHAR (valid_working_day - 1, 'D'),
-- 1, valid_working_day - 3,
-- valid_working_day - 1)
-- INTO valid_working_day
-- FROM DUAL;
IF (hol_count > 0)
THEN
valid_working_day := valid_working_day - 1;
-- counter := counter + 1;
ELSIF (day_number = 1 OR day_number = 7)
THEN
valid_working_day := valid_working_day - 1;
END IF;
ELSIF ( TO_CHAR (TO_DATE (valid_working_day, 'DD-MON-YYYY'), 'D') =
1
OR TO_CHAR (TO_DATE (valid_working_day, 'DD-MON-YYYY'), 'D') =
7)
THEN
valid_working_day := valid_working_day - 1;
ELSE
counter := counter + 1;
valid_working_day := valid_working_day - 1;
END IF;
EXIT WHEN counter >= 3;
END LOOP;
--elsif (i_S_NS <> 0) then
--null;
END IF;
--valid_working_day := valid_working_day - 1;
RETURN (valid_working_day);
END;
这是一个纯粹的 SQL 答案。诀窍是生成一系列涵盖所有可能性的先前日期。在我所知道的国家/地区，连续的假期不超过两个 public（英国的圣诞节和复活节，苏格兰的除夕节）。也考虑到周末，这意味着最多可以排除四天的时间。
正如评论者指出的那样，在其他国家/地区可能会有更长的 public 假期，因此您可能需要相应地调整偏移量。
无论如何，有两天的目标，我们需要一个可以追溯到六天的范围（加上运气）。这将为我们提供目标日期前 7 天的结果集和 :
select (tgt_date - 7) + (level-1)
from dual
connect by level <= 7
现在我们准备好了。我们可以使用带有 'IW' 日期掩码的技巧来将星期几确定为数字，并以文化中立的方式排除星期六和星期日。我们可以左加入 holiday_tab 以排除 public 假期。然后我们对剩下的和 select 最近的两个日期进行排名：
SQL> with hdr as (
2 select dr.dt
3 , case
4 when (1 + dt - trunc(dr.dt, 'IW') in (6,7) then 1
5 when h.hol_date is not null then 1
6 else 0
7 end as hol
8 from ( select trunc(date '2017-05-02' - 7) + (level-1) as dt
9 from dual
10 connect by level <= 7
11 ) dr
12 left join holiday_tab h
13 on h.hol_date = dr.dt
14 )
15 , rhdr as (
16 select hdr.dt
17 , row_number() over (order by hdr.dt desc) rn
18 from hdr
19 where hdr.hol = 0
20 )
21 select rhdr.dt
22 , decode( rhdr.rn, 1, '24hr', '48hr') as cat
23 , to_char(rhdr.dt, 'DY') as dy
24 from rhdr
25 where rn <= 2;
DT CAT DY
--------- ---- ------------
28-APR-17 24hr FRI
27-APR-17 48hr THU
SQL>
鉴于昨天 02-MAY-2017 作为目标日期，这将跳过星期一（五一假期）和周末来识别前两个工作日。
如果你需要一个函数，你可以这样做：
create or replace type dt_nt as table of date;
create or replace function prior_working_days
( p_target_date in date
, p_no_of_days in number := 2)
return dt_nt
is
return_value dt_nt;
offset pls_integer := (p_no_of_days+4+1);
begin
with hdr as (
select dr.dt
, case
when to_char(1 + dt - trunc(dr.dt, 'IW') in (6,7) then 1
when h.hol_date is not null then 1
else 0
end as hol
from ( select (trunc(p_target_date) - offset) + (level-1) as dt
from dual
connect by level <= offset
) dr
left join holiday_tab h
on h.hol_date = dr.dt
)
, rhdr as (
select hdr.dt
, row_number() over (order by hdr.dt desc) rn
from hdr
where hdr.hol = 0
)
select rhdr.dt
bulk collect into return_value
from rhdr
where rn <= p_no_of_days;
return return_value;
end prior_working_days;
/
returns SQL table 个日期：
SQL> select * from table( prior_working_days(sysdate));
COLUMN_VA
---------
02-MAY-17
28-APR-17
SQL>
我的建议是这样的函数：
CREATE OR REPLACE FUNCTION F_HOL_CHECK_S_NS (i_hol_date DATE, i_S_NS NUMBER) RETURN DATE AS
TYPE DATE_TABLE_TYPE is TABLE OF DATE;
Holidays DATE_TABLE_TYPE;
the_date DATE := i_hol_date;
duration INTEGER := 0;
BEGIN
ID i_hol_date IS NULL OR i_S_NS IS NULL THEN
-- Avoid infinite loop
RETURN NULL;
END IF;
-- Just for performance reason
SELECT HOL_DATE
BULK COLLECT INTO Holidays
FROM HOLIDAY_TAB
WHERE HOL_DATE < i_hol_date;
LOOP
the_date := the_date - 1;
IF TO_CHAR(the_date, 'fmDy', 'NLS_DATE_LANGUAGE = american') NOT IN ('Sat', 'Sun') AND TRUNC(the_date) NOT MEMBER OF Holidays THEN
duration := duration + 24;
END IF;
EXIT WHEN duration >= i_S_NS;
END LOOP;
RETURN the_date;
END;
SELECT F_HOL_CHECK_S_NS(DATE '2017-05-02', 24) FROM dual;
SELECT F_HOL_CHECK_S_NS(DATE '2017-05-02', 48) FROM dual;
SELECT F_HOL_CHECK_S_NS(SYSDATE, 24) FROM dual;
SELECT F_HOL_CHECK_S_NS(SYSDATE, 48) FROM dual;
您可以在 SQL 中完成所有操作：
WITH dates ( dt, lvl ) AS (
SELECT CAST( TRUNC( :your_date ) AS DATE ), 0 FROM DUAL
UNION ALL
SELECT CAST( dt - INTERVAL '1' DAY AS DATE ),
CASE
WHEN ( dt - INTERVAL '1' DAY ) - TRUNC( dt - INTERVAL '1' DAY, 'IW' ) >= 5
OR hol_date IS NOT NULL
THEN lvl
ELSE lvl + 1
END
FROM dates d
LEFT OUTER JOIN
holidays h
ON ( d.dt - INTERVAL '1' DAY = h.hol_date )
WHERE lvl < 2
)
SELECT *
FROM dates
PIVOT ( MAX( dt ) FOR lvl IN ( 1 AS DATE24, 2 AS DATE48 ) );
（注意：使用 CAST 不是必需的，但没有我得到 ORA-01790: expression must have same datatype as corresponding expression）
或者，作为函数：
CREATE OR REPLACE FUNCTION F_HOL_CHECK_S_NS (
i_hol_date DATE,
i_S_NS NUMBER
) RETURN DATE
AS
p_date DATE;
BEGIN
WITH dates ( dt, lvl ) AS (
SELECT CAST( TRUNC( i_hol_date ) AS DATE ), 0 FROM DUAL
UNION ALL
SELECT CAST( dt - INTERVAL '1' DAY AS DATE ),
CASE
WHEN ( dt - INTERVAL '1' DAY ) - TRUNC( dt - INTERVAL '1' DAY, 'IW' ) >= 5
OR hol_date IS NOT NULL
THEN lvl
ELSE lvl + 1
END
FROM dates d
LEFT OUTER JOIN
holidays h
ON ( d.dt - INTERVAL '1' DAY = h.hol_date )
WHERE lvl < i_s_ns
)
SELECT dt
INTO p_date
FROM dates
WHERE lvl = i_s_ns;
RETURN p_date;
END;
/
某模块要求求第48、24个工作时间
要求：
假设如果我将 2nd May 作为参数传递给函数，则 48 hours 的输出应为 27th April，24 hours 的输出应为 28th APRIL（因为5月1日是假期，4月29日和30日属于周六和周日）
问题出在两个连续的假期上。例如，要创建一个 dummy 数据，我们插入 5 月 2 日作为假期，运行 3rd May 上的代码应该为 48 hours 检索 27th April 和 28th APRIL 对于 24 hours.
但我的功能似乎不适用于连续的假期。某处计数器增量似乎在错误的位置。
考虑因素： 周末：周六和周日 需要排除的天数：给定节假日日历中的周六、周日和节假日：
假期table创作：
CREATE TABLE HOLIDAY_TAB
(
HOL_DATE DATE,
DESCRIPTION VARCHAR2 (100) DEFAULT NULL
);
insert into HOLIDAY_TAB values (TO_DATE ('26-Jan-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('29-Mar-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('14-Apr-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('01-May-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('02-Jun-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('26-Jun-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('15-Aug-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('25-Aug-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('28-Sep-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('02-Oct-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('19-Oct-2017', 'DD-MON-YYYY'),NULL);
insert into HOLIDAY_TAB values (TO_DATE ('25-Dec-2017', 'DD-MON-YYYY'),NULL);
commit;
写入捕获假期的函数：
函数：
CREATE OR REPLACE FUNCTION CSE.F_HOL_CHECK_ABC (i_hol_date DATE)
RETURN DATE
AS
valid_working_day DATE := i_hol_date;
day_C holiday_nvs%ROWTYPE;
CURSOR c_hol
IS
SELECT *
FROM HOLIDAY_TAB
WHERE TRUNC (hol_date) = TRUNC (i_hol_date);
CURSOR c_hol_24
IS
SELECT *
FROM HOLIDAY_TAB
WHERE TRUNC (hol_date) = TRUNC (i_hol_date + 3);
CURSOR c_hol_48
IS
SELECT *
FROM HOLIDAY_NVS
WHERE TRUNC (hol_date) = TRUNC (i_hol_date + 2);
BEGIN
-- FOR rec24 IN c_hol_24
-- LOOP
-- IF (rec24.hol_date IS NOT NULL)
-- THEN
-- valid_working_day := i_hol_date - 1;
-- END IF;
-- END LOOP;
OPEN c_hol;
FETCH c_hol INTO day_C;
IF c_hol%FOUND
THEN
SELECT DECODE (TO_CHAR (i_hol_date - 1, 'D'),
1, i_hol_date - 3,
i_hol_date - )
INTO valid_working_day
FROM DUAL;
END IF;
CLOSE c_hol;
RETURN (valid_working_day);
END;
/
不确定功能是否正确。但是有一种奇怪的情况，即与日期文字相比，我的查询在尝试使用 SYSDATE 时没有给出相同的结果。
运行 1 手动设置日期：
SELECT TRUNC (
DECODE (
TO_CHAR (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY'), 'D'),
2, F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 4),
DECODE (
TO_CHAR (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY'), 'D'),
3, F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 4),
F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 2))))
AS "48HOURS",
TRUNC (
DECODE (
TO_CHAR (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY'), 'D'),
2, F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 3),
F_HOL_CHECK_ABC (TO_DATE ('03-MAY-2017', 'DD-MON-YYYY') - 1)))
AS "24HOURS"
FROM DUAL;
运行 2 系统日期：
SELECT TRUNC (
DECODE (
TO_CHAR (SYSDATE, 'D'),
2, F_HOL_CHECK_ABC (SYSDATE - 4),
DECODE (TO_CHAR (SYSDATE, 'D'),
3, F_HOL_CHECK_ABC (SYSDATE - 4),
F_HOL_CHECK_ABC (SYSDATE -2)))) as "48 hour" ,
TRUNC (
DECODE (
TO_CHAR (SYSDATE,
'D'),
2, F_HOL_CHECK_ABC (
SYSDATE - 3),
F_HOL_CHECK_ABC (
SYSDATE -1))) as "24 hour" from dual;
非常感谢您提供这方面的帮助。我所需要的只是跳过节假日、星期日和星期六，即；非工作时间，给我每天 48 小时工作时间和 24 小时工作时间
这是使用计数器对代码 2 进行的另一次尝试：
CREATE OR REPLACE FUNCTION CSE.F_HOL_CHECK_S_NS (i_hol_date DATE,
i_S_NS NUMBER)
RETURN DATE
AS
valid_working_day DATE := i_hol_date;
counter NUMBER := 0;
day_number NUMBER := 0;
hol_count NUMBER := 0;
day_C holiday_nvs%ROWTYPE;
CURSOR c_hol (hol_date_c DATE)
IS
SELECT *
FROM HOLIDAY_TAB
WHERE TRUNC (hol_date) = TRUNC (TO_DATE (hol_date_c, 'DD-MON-YYYY'));
BEGIN
IF i_S_NS = 0
THEN
LOOP
IF c_hol%ISOPEN
THEN
CLOSE c_hol;
END IF;
OPEN c_hol (valid_working_day);
IF c_hol%FOUND
THEN
valid_working_day := valid_working_day - 1;
SELECT TO_CHAR (TO_DATE (valid_working_day, 'DD-MON-YYYY'), 'D')
INTO day_number
FROM DUAL;
SELECT COUNT (*)
INTO hol_count
FROM HOLIDAY_TAB
WHERE TRUNC (hol_date) =
TRUNC (TO_DATE (valid_working_day, 'DD-MON-YYYY'));
--valid_working_day:=valid_working_day-1;
-- SELECT DECODE (TO_CHAR (valid_working_day - 1, 'D'),
-- 1, valid_working_day - 3,
-- valid_working_day - 1)
-- INTO valid_working_day
-- FROM DUAL;
IF (hol_count > 0)
THEN
valid_working_day := valid_working_day - 1;
-- counter := counter + 1;
ELSIF (day_number = 1 OR day_number = 7)
THEN
valid_working_day := valid_working_day - 1;
END IF;
ELSIF ( TO_CHAR (TO_DATE (valid_working_day, 'DD-MON-YYYY'), 'D') =
1
OR TO_CHAR (TO_DATE (valid_working_day, 'DD-MON-YYYY'), 'D') =
7)
THEN
valid_working_day := valid_working_day - 1;
ELSE
counter := counter + 1;
valid_working_day := valid_working_day - 1;
END IF;
EXIT WHEN counter >= 3;
END LOOP;
--elsif (i_S_NS <> 0) then
--null;
END IF;
--valid_working_day := valid_working_day - 1;
RETURN (valid_working_day);
END;
这是一个纯粹的 SQL 答案。诀窍是生成一系列涵盖所有可能性的先前日期。在我所知道的国家/地区，连续的假期不超过两个 public（英国的圣诞节和复活节，苏格兰的除夕节）。也考虑到周末，这意味着最多可以排除四天的时间。
正如评论者指出的那样，在其他国家/地区可能会有更长的 public 假期，因此您可能需要相应地调整偏移量。
无论如何，有两天的目标，我们需要一个可以追溯到六天的范围（加上运气）。这将为我们提供目标日期前 7 天的结果集和 :
select (tgt_date - 7) + (level-1)
from dual
connect by level <= 7
现在我们准备好了。我们可以使用带有 'IW' 日期掩码的技巧来将星期几确定为数字，并以文化中立的方式排除星期六和星期日。我们可以左加入 holiday_tab 以排除 public 假期。然后我们对剩下的和 select 最近的两个日期进行排名：
SQL> with hdr as (
2 select dr.dt
3 , case
4 when (1 + dt - trunc(dr.dt, 'IW') in (6,7) then 1
5 when h.hol_date is not null then 1
6 else 0
7 end as hol
8 from ( select trunc(date '2017-05-02' - 7) + (level-1) as dt
9 from dual
10 connect by level <= 7
11 ) dr
12 left join holiday_tab h
13 on h.hol_date = dr.dt
14 )
15 , rhdr as (
16 select hdr.dt
17 , row_number() over (order by hdr.dt desc) rn
18 from hdr
19 where hdr.hol = 0
20 )
21 select rhdr.dt
22 , decode( rhdr.rn, 1, '24hr', '48hr') as cat
23 , to_char(rhdr.dt, 'DY') as dy
24 from rhdr
25 where rn <= 2;
DT CAT DY
--------- ---- ------------
28-APR-17 24hr FRI
27-APR-17 48hr THU
SQL>
鉴于昨天 02-MAY-2017 作为目标日期，这将跳过星期一（五一假期）和周末来识别前两个工作日。
如果你需要一个函数，你可以这样做：
create or replace type dt_nt as table of date;
create or replace function prior_working_days
( p_target_date in date
, p_no_of_days in number := 2)
return dt_nt
is
return_value dt_nt;
offset pls_integer := (p_no_of_days+4+1);
begin
with hdr as (
select dr.dt
, case
when to_char(1 + dt - trunc(dr.dt, 'IW') in (6,7) then 1
when h.hol_date is not null then 1
else 0
end as hol
from ( select (trunc(p_target_date) - offset) + (level-1) as dt
from dual
connect by level <= offset
) dr
left join holiday_tab h
on h.hol_date = dr.dt
)
, rhdr as (
select hdr.dt
, row_number() over (order by hdr.dt desc) rn
from hdr
where hdr.hol = 0
)
select rhdr.dt
bulk collect into return_value
from rhdr
where rn <= p_no_of_days;
return return_value;
end prior_working_days;
/
returns SQL table 个日期：
SQL> select * from table( prior_working_days(sysdate));
COLUMN_VA
---------
02-MAY-17
28-APR-17
SQL>
我的建议是这样的函数：
CREATE OR REPLACE FUNCTION F_HOL_CHECK_S_NS (i_hol_date DATE, i_S_NS NUMBER) RETURN DATE AS
TYPE DATE_TABLE_TYPE is TABLE OF DATE;
Holidays DATE_TABLE_TYPE;
the_date DATE := i_hol_date;
duration INTEGER := 0;
BEGIN
ID i_hol_date IS NULL OR i_S_NS IS NULL THEN
-- Avoid infinite loop
RETURN NULL;
END IF;
-- Just for performance reason
SELECT HOL_DATE
BULK COLLECT INTO Holidays
FROM HOLIDAY_TAB
WHERE HOL_DATE < i_hol_date;
LOOP
the_date := the_date - 1;
IF TO_CHAR(the_date, 'fmDy', 'NLS_DATE_LANGUAGE = american') NOT IN ('Sat', 'Sun') AND TRUNC(the_date) NOT MEMBER OF Holidays THEN
duration := duration + 24;
END IF;
EXIT WHEN duration >= i_S_NS;
END LOOP;
RETURN the_date;
END;
SELECT F_HOL_CHECK_S_NS(DATE '2017-05-02', 24) FROM dual;
SELECT F_HOL_CHECK_S_NS(DATE '2017-05-02', 48) FROM dual;
SELECT F_HOL_CHECK_S_NS(SYSDATE, 24) FROM dual;
SELECT F_HOL_CHECK_S_NS(SYSDATE, 48) FROM dual;
您可以在 SQL 中完成所有操作：
WITH dates ( dt, lvl ) AS (
SELECT CAST( TRUNC( :your_date ) AS DATE ), 0 FROM DUAL
UNION ALL
SELECT CAST( dt - INTERVAL '1' DAY AS DATE ),
CASE
WHEN ( dt - INTERVAL '1' DAY ) - TRUNC( dt - INTERVAL '1' DAY, 'IW' ) >= 5
OR hol_date IS NOT NULL
THEN lvl
ELSE lvl + 1
END
FROM dates d
LEFT OUTER JOIN
holidays h
ON ( d.dt - INTERVAL '1' DAY = h.hol_date )
WHERE lvl < 2
)
SELECT *
FROM dates
PIVOT ( MAX( dt ) FOR lvl IN ( 1 AS DATE24, 2 AS DATE48 ) );
（注意：使用 CAST 不是必需的，但没有我得到 ORA-01790: expression must have same datatype as corresponding expression）
或者，作为函数：
CREATE OR REPLACE FUNCTION F_HOL_CHECK_S_NS (
i_hol_date DATE,
i_S_NS NUMBER
) RETURN DATE
AS
p_date DATE;
BEGIN
WITH dates ( dt, lvl ) AS (
SELECT CAST( TRUNC( i_hol_date ) AS DATE ), 0 FROM DUAL
UNION ALL
SELECT CAST( dt - INTERVAL '1' DAY AS DATE ),
CASE
WHEN ( dt - INTERVAL '1' DAY ) - TRUNC( dt - INTERVAL '1' DAY, 'IW' ) >= 5
OR hol_date IS NOT NULL
THEN lvl
ELSE lvl + 1
END
FROM dates d
LEFT OUTER JOIN
holidays h
ON ( d.dt - INTERVAL '1' DAY = h.hol_date )
WHERE lvl < i_s_ns
)
SELECT dt
INTO p_date
FROM dates
WHERE lvl = i_s_ns;
RETURN p_date;
END;
/