甲骨文计算支付日期
Oracle calculate pay DATE
我们使用双月工资单，员工在每个月的第 15 天和最后一天领取工资。
如果这些日子是星期六或星期日，那么我们会在前一天收到付款（如果那一天没有定义为节假日）。
我设法让部分查询工作，我可以在其中排除周末和节假日，但我可以使用一些帮助来确定应该在什么日期向人们付款。这些是我想在输出中看到的唯一几天
提前感谢所有回答者以及您的耐心、帮助和专业知识。
以下是我目前的资料。
CREATE OR REPLACE TYPE nt_date IS TABLE OF DATE;
/
CREATE OR REPLACE FUNCTION generate_dates_pipelined(
p_from IN DATE,
p_to IN DATE
)
RETURN nt_date PIPELINED DETERMINISTIC
IS
v_start DATE := TRUNC(LEAST(p_from, p_to));
v_end DATE := TRUNC(GREATEST(p_from, p_to));
BEGIN
LOOP
PIPE ROW (v_start);
EXIT WHEN v_start >= v_end;
v_start := v_start + INTERVAL '1' DAY;
END LOOP;
RETURN;
END generate_dates_pipelined;
/
create table holidays(
holiday_date DATE not null,
holiday_name VARCHAR2(20),
constraint holidays_pk primary key (holiday_date),
constraint is_midnight check ( holiday_date = trunc ( holiday_date ) )
);
INSERT into holidays (HOLIDAY_DATE,HOLIDAY_NAME)
WITH dts as (
select to_date('15-APR-2022 00:00:00','DD-MON-YYYY HH24:MI:SS'), 'Passover 2022' from dual union all
select to_date('31-DEC-2022 00:00:00','DD-MON-YYYY HH24:MI:SS'), 'New Year Eve 2022' from dual
)
SELECT * from dts;
SELECT
COLUMN_VALUE
FROM
TABLE(generate_dates_pipelined(DATE '2022-01-01',
DATE '2022-12-31')) c
where
to_char(COLUMN_VALUE, 'DY') NOT IN ('SAT', 'SUN')
AND NOT EXISTS (
SELECT 1
FROM holidays h
WHERE c.COLUMN_VALUE = h.holiday_date
);
您可以使用下面的方法生成不包括星期六和星期日的日期日历。假期列表已建立并过滤。
主查询
with cte_h as
(
select to_date('15-APR-2022','DD-MON-YYYY') h_dt,
'PASSOVER' h_name from dual union all
select to_date('31-DEC-2022','DD-MON-YYYY') h_dt,
'New Year eve' h_name from dual
)
,cte_cal as (
select
decode(mod(level-1,2),
0,
case
when add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)+14
IN
(select h_dt from cte_h)
then add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)+13
else add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)+14 END,
case
when last_day(add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1))
IN
(select h_dt from cte_h)
then last_day(add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)) - 1
else last_day(add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)) END ) day
from dual
connect by level < 25),
cte_pd as
(select
DAY as date_orig_w_h_adj,
case
when instr(to_char(day,'DAY'),'SUNDAY')>0 then day-2
when instr(to_char(day,'DAY'),'SATURDAY')>0 then day-1
else day END as pay_date, to_char(day,'DAY') dow_orig
from cte_cal)
select * from cte_pd;
上面的输出-
DATE_ORIG_W_H_ADJ PAY_DATE DOW_ORIG
--------------------- --------- ------------
15-JAN-22 14-JAN-22 SATURDAY
31-JAN-22 31-JAN-22 MONDAY
15-FEB-22 15-FEB-22 TUESDAY
28-FEB-22 28-FEB-22 MONDAY
15-MAR-22 15-MAR-22 TUESDAY
31-MAR-22 31-MAR-22 THURSDAY
14-APR-22 14-APR-22 THURSDAY
30-APR-22 29-APR-22 SATURDAY
15-MAY-22 13-MAY-22 SUNDAY
31-MAY-22 31-MAY-22 TUESDAY
15-JUN-22 15-JUN-22 WEDNESDAY
30-JUN-22 30-JUN-22 THURSDAY
15-JUL-22 15-JUL-22 FRIDAY
31-JUL-22 29-JUL-22 SUNDAY
15-AUG-22 15-AUG-22 MONDAY
31-AUG-22 31-AUG-22 WEDNESDAY
15-SEP-22 15-SEP-22 THURSDAY
30-SEP-22 30-SEP-22 FRIDAY
15-OCT-22 14-OCT-22 SATURDAY
31-OCT-22 31-OCT-22 MONDAY
15-NOV-22 15-NOV-22 TUESDAY
30-NOV-22 30-NOV-22 WEDNESDAY
15-DEC-22 15-DEC-22 THURSDAY
30-DEC-22 30-DEC-22 FRIDAY
24 rows selected.
根据您目前的情况：
WITH days AS
(
SELECT days.column_value
FROM TABLE(generate_dates_pipelined(DATE '2022-01-01', DATE '2022-12-31')) days
LEFT JOIN holidays holy ON days.column_value = holy.holiday_date
WHERE holy.holiday_date IS NULL
WHERE TO_CHAR(days.column_value, 'DY') NOT IN ('SAT', 'SUN')
)
SELECT payment_day
FROM (SELECT MAX(days.column_value) AS payment_day, TRUNC(days.column_value,'MM')
FROM days
WHERE days.column_value <= TRUNC(days.column_value,'MM') + 14
GROUP BY TRUNC(days.column_value,'MM')
UNION
SELECT MAX(days.column_value) , TRUNC(days.column_value,'MM')
FROM days
WHERE days.column_value <= ADD_MONTHS(TRUNC(days.column_value,'MM'),1) - 1
GROUP BY TRUNC(days.column_value,'MM')
)
我用我的这个变体做了一些测试：
WITH holidays AS
(
SELECT TO_DATE('15-APR-2022 00:00:00', 'DD-MON-YYYY HH24:MI:SS') AS dt,
'Passover 2022' AS NAME
FROM DUAL
UNION ALL
SELECT TO_DATE('31-DEC-2022 00:00:00', 'DD-MON-YYYY HH24:MI:SS'),
'New Year Eve 2022'
FROM DUAL
),
workdays AS
(
SELECT TRUNC(wknd.dt) AS dt
FROM (SELECT SYSDATE + ROWNUM - 1 dt
FROM dual
WHERE 1 = 1 /* you can limit those such as in your pipelined function here */
CONNECT BY LEVEL <= 366) wknd
WHERE TO_CHAR(wknd.dt, 'FMDAY', 'NLS_DATE_LANGUAGE=AMERICAN') NOT IN ('SUNDAY','SATURDAY')
)
SELECT payment_day
FROM (SELECT MAX(dt) AS payment_day, TRUNC(dt,'MM')
FROM (SELECT wd.*
FROM workdays wd
LEFT JOIN holidays hd ON wd.dt = hd.dt
WHERE hd.dt IS NULL) first_payroll
WHERE first_payroll.dt <= TRUNC(first_payroll.dt,'MM') + 14
GROUP BY TRUNC(dt,'MM')
UNION
SELECT MAX(dt), TRUNC(dt,'MM')
FROM (SELECT wd.*
FROM workdays wd
LEFT JOIN holidays hd ON wd.dt = hd.dt
WHERE hd.dt IS NULL) second_payroll
WHERE second_payroll.dt <= ADD_MONTHS(TRUNC(second_payroll.dt,'MM'),1) - 1
GROUP BY TRUNC(dt,'MM'));
希望它能给你一些有用的想法。
建立日历和假期 table 后，您需要参考它们两次，一次是确定 'nominal' 发薪日（每个月的第 15 天和最后一天）并确定如果这是一个假期或周末，另一个时间来查找该日期之前的最近工作日：
With Cal as
(SELECT TRUNC (SYSDATE - ROWNUM) dt
FROM DUAL CONNECT BY ROWNUM < 366),
PayCal as (
select c.dt as THIS_DATE
, CASE WHEN h.holiday_date is not null then 'Y' else 'N' end as IS_HOLIDAY
, CASE WHEN to_char(dt, 'DY') IN ('SAT', 'SUN') then 'Y' else 'N' end as IS_SAT_OR_SUN
from Cal c
left join
-- determine if c is a public holiday
holidays h
on c.dt = h.holiday_date
)
select
c.THIS_DATE
, case when IS_HOLIDAY='Y' or IS_SAT_OR_SUN='Y' then
(select max(c1.THIS_DATE)
from PayCal c1
where c1.THIS_DATE < c.THIS_DATE
and c1.IS_HOLIDAY='N'
and c1.IS_SAT_OR_SUN='N')
else c.THIS_DATE
end as PAY_ON_THIS_DATE
from PayCal c
where EXTRACT(DAY FROM c.THIS_DATE)=15
OR
LAST_DAY(c.THIS_DATE)=c.THIS_DATE
我使用了在 SO 中找到的日历数据生成逻辑版本，但您可以用您的 looping-function.
替换我的 Cal table
create table holidays(
holiday_date DATE not null,
holiday_name VARCHAR2(20),
constraint holidays_pk primary key (holiday_date),
constraint is_midnight check ( holiday_date = trunc ( holiday_date ) )
);
INSERT into holidays (HOLIDAY_DATE,HOLIDAY_NAME)
WITH dts as (
select to_date('15-APR-2022 00:00:00','DD-MON-YYYY HH24:MI:SS'), 'Passover 2022' from dual union all
select to_date('31-DEC-2022 00:00:00','DD-MON-YYYY HH24:MI:SS'), 'New Year Eve 2022' from dual
)
SELECT * from dts;
WITH parameters (first_month, num_months) AS
(
SELECT TO_DATE ('JAN-2022', 'MON-YYYY') -- First month wanted
, 4 -- Total number of months
FROM dual
)
, all_scheduled_paydays AS
(
SELECT ADD_MONTHS (first_month + 14, LEVEL - 1) AS payday_1
, ADD_MONTHS (LAST_DAY (first_month), LEVEL - 1) AS payday_2
FROM parameters
CONNECT BY LEVEL <= num_months
)
SELECT c.actual_payday
, a.scheduled_payday -- For debugging
FROM all_scheduled_paydays
UNPIVOT ( scheduled_payday
FOR col IN (payday_1, payday_2)
) a
CROSS APPLY (
SELECT a.scheduled_payday + 1 - LEVEL AS actual_payday
FROM dual
WHERE CONNECT_BY_ISLEAF = 1
CONNECT BY TO_CHAR ( a.scheduled_payday + 2 - LEVEL
, 'DY'
, 'NLS_DATE_LANGUAGE=ENGLISH'
) IN ('SAT', 'SUN')
OR a.scheduled_payday + 2 - LEVEL
IN (
SELECT holiday_date
FROM holidays
)
) c
ORDER BY scheduled_payday
;
我们使用双月工资单，员工在每个月的第 15 天和最后一天领取工资。
如果这些日子是星期六或星期日，那么我们会在前一天收到付款（如果那一天没有定义为节假日）。
我设法让部分查询工作，我可以在其中排除周末和节假日，但我可以使用一些帮助来确定应该在什么日期向人们付款。这些是我想在输出中看到的唯一几天
提前感谢所有回答者以及您的耐心、帮助和专业知识。
以下是我目前的资料。
CREATE OR REPLACE TYPE nt_date IS TABLE OF DATE;
/
CREATE OR REPLACE FUNCTION generate_dates_pipelined(
p_from IN DATE,
p_to IN DATE
)
RETURN nt_date PIPELINED DETERMINISTIC
IS
v_start DATE := TRUNC(LEAST(p_from, p_to));
v_end DATE := TRUNC(GREATEST(p_from, p_to));
BEGIN
LOOP
PIPE ROW (v_start);
EXIT WHEN v_start >= v_end;
v_start := v_start + INTERVAL '1' DAY;
END LOOP;
RETURN;
END generate_dates_pipelined;
/
create table holidays(
holiday_date DATE not null,
holiday_name VARCHAR2(20),
constraint holidays_pk primary key (holiday_date),
constraint is_midnight check ( holiday_date = trunc ( holiday_date ) )
);
INSERT into holidays (HOLIDAY_DATE,HOLIDAY_NAME)
WITH dts as (
select to_date('15-APR-2022 00:00:00','DD-MON-YYYY HH24:MI:SS'), 'Passover 2022' from dual union all
select to_date('31-DEC-2022 00:00:00','DD-MON-YYYY HH24:MI:SS'), 'New Year Eve 2022' from dual
)
SELECT * from dts;
SELECT
COLUMN_VALUE
FROM
TABLE(generate_dates_pipelined(DATE '2022-01-01',
DATE '2022-12-31')) c
where
to_char(COLUMN_VALUE, 'DY') NOT IN ('SAT', 'SUN')
AND NOT EXISTS (
SELECT 1
FROM holidays h
WHERE c.COLUMN_VALUE = h.holiday_date
);
您可以使用下面的方法生成不包括星期六和星期日的日期日历。假期列表已建立并过滤。
主查询
with cte_h as
(
select to_date('15-APR-2022','DD-MON-YYYY') h_dt,
'PASSOVER' h_name from dual union all
select to_date('31-DEC-2022','DD-MON-YYYY') h_dt,
'New Year eve' h_name from dual
)
,cte_cal as (
select
decode(mod(level-1,2),
0,
case
when add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)+14
IN
(select h_dt from cte_h)
then add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)+13
else add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)+14 END,
case
when last_day(add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1))
IN
(select h_dt from cte_h)
then last_day(add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)) - 1
else last_day(add_months(trunc(sysdate,'YEAR'),ceil(level/2)-1)) END ) day
from dual
connect by level < 25),
cte_pd as
(select
DAY as date_orig_w_h_adj,
case
when instr(to_char(day,'DAY'),'SUNDAY')>0 then day-2
when instr(to_char(day,'DAY'),'SATURDAY')>0 then day-1
else day END as pay_date, to_char(day,'DAY') dow_orig
from cte_cal)
select * from cte_pd;
上面的输出-
DATE_ORIG_W_H_ADJ PAY_DATE DOW_ORIG
--------------------- --------- ------------
15-JAN-22 14-JAN-22 SATURDAY
31-JAN-22 31-JAN-22 MONDAY
15-FEB-22 15-FEB-22 TUESDAY
28-FEB-22 28-FEB-22 MONDAY
15-MAR-22 15-MAR-22 TUESDAY
31-MAR-22 31-MAR-22 THURSDAY
14-APR-22 14-APR-22 THURSDAY
30-APR-22 29-APR-22 SATURDAY
15-MAY-22 13-MAY-22 SUNDAY
31-MAY-22 31-MAY-22 TUESDAY
15-JUN-22 15-JUN-22 WEDNESDAY
30-JUN-22 30-JUN-22 THURSDAY
15-JUL-22 15-JUL-22 FRIDAY
31-JUL-22 29-JUL-22 SUNDAY
15-AUG-22 15-AUG-22 MONDAY
31-AUG-22 31-AUG-22 WEDNESDAY
15-SEP-22 15-SEP-22 THURSDAY
30-SEP-22 30-SEP-22 FRIDAY
15-OCT-22 14-OCT-22 SATURDAY
31-OCT-22 31-OCT-22 MONDAY
15-NOV-22 15-NOV-22 TUESDAY
30-NOV-22 30-NOV-22 WEDNESDAY
15-DEC-22 15-DEC-22 THURSDAY
30-DEC-22 30-DEC-22 FRIDAY
24 rows selected.
根据您目前的情况：
WITH days AS
(
SELECT days.column_value
FROM TABLE(generate_dates_pipelined(DATE '2022-01-01', DATE '2022-12-31')) days
LEFT JOIN holidays holy ON days.column_value = holy.holiday_date
WHERE holy.holiday_date IS NULL
WHERE TO_CHAR(days.column_value, 'DY') NOT IN ('SAT', 'SUN')
)
SELECT payment_day
FROM (SELECT MAX(days.column_value) AS payment_day, TRUNC(days.column_value,'MM')
FROM days
WHERE days.column_value <= TRUNC(days.column_value,'MM') + 14
GROUP BY TRUNC(days.column_value,'MM')
UNION
SELECT MAX(days.column_value) , TRUNC(days.column_value,'MM')
FROM days
WHERE days.column_value <= ADD_MONTHS(TRUNC(days.column_value,'MM'),1) - 1
GROUP BY TRUNC(days.column_value,'MM')
)
我用我的这个变体做了一些测试：
WITH holidays AS
(
SELECT TO_DATE('15-APR-2022 00:00:00', 'DD-MON-YYYY HH24:MI:SS') AS dt,
'Passover 2022' AS NAME
FROM DUAL
UNION ALL
SELECT TO_DATE('31-DEC-2022 00:00:00', 'DD-MON-YYYY HH24:MI:SS'),
'New Year Eve 2022'
FROM DUAL
),
workdays AS
(
SELECT TRUNC(wknd.dt) AS dt
FROM (SELECT SYSDATE + ROWNUM - 1 dt
FROM dual
WHERE 1 = 1 /* you can limit those such as in your pipelined function here */
CONNECT BY LEVEL <= 366) wknd
WHERE TO_CHAR(wknd.dt, 'FMDAY', 'NLS_DATE_LANGUAGE=AMERICAN') NOT IN ('SUNDAY','SATURDAY')
)
SELECT payment_day
FROM (SELECT MAX(dt) AS payment_day, TRUNC(dt,'MM')
FROM (SELECT wd.*
FROM workdays wd
LEFT JOIN holidays hd ON wd.dt = hd.dt
WHERE hd.dt IS NULL) first_payroll
WHERE first_payroll.dt <= TRUNC(first_payroll.dt,'MM') + 14
GROUP BY TRUNC(dt,'MM')
UNION
SELECT MAX(dt), TRUNC(dt,'MM')
FROM (SELECT wd.*
FROM workdays wd
LEFT JOIN holidays hd ON wd.dt = hd.dt
WHERE hd.dt IS NULL) second_payroll
WHERE second_payroll.dt <= ADD_MONTHS(TRUNC(second_payroll.dt,'MM'),1) - 1
GROUP BY TRUNC(dt,'MM'));
希望它能给你一些有用的想法。
建立日历和假期 table 后，您需要参考它们两次，一次是确定 'nominal' 发薪日（每个月的第 15 天和最后一天）并确定如果这是一个假期或周末，另一个时间来查找该日期之前的最近工作日：
With Cal as
(SELECT TRUNC (SYSDATE - ROWNUM) dt
FROM DUAL CONNECT BY ROWNUM < 366),
PayCal as (
select c.dt as THIS_DATE
, CASE WHEN h.holiday_date is not null then 'Y' else 'N' end as IS_HOLIDAY
, CASE WHEN to_char(dt, 'DY') IN ('SAT', 'SUN') then 'Y' else 'N' end as IS_SAT_OR_SUN
from Cal c
left join
-- determine if c is a public holiday
holidays h
on c.dt = h.holiday_date
)
select
c.THIS_DATE
, case when IS_HOLIDAY='Y' or IS_SAT_OR_SUN='Y' then
(select max(c1.THIS_DATE)
from PayCal c1
where c1.THIS_DATE < c.THIS_DATE
and c1.IS_HOLIDAY='N'
and c1.IS_SAT_OR_SUN='N')
else c.THIS_DATE
end as PAY_ON_THIS_DATE
from PayCal c
where EXTRACT(DAY FROM c.THIS_DATE)=15
OR
LAST_DAY(c.THIS_DATE)=c.THIS_DATE
我使用了在 SO 中找到的日历数据生成逻辑版本，但您可以用您的 looping-function.
替换我的 Cal table
create table holidays(
holiday_date DATE not null,
holiday_name VARCHAR2(20),
constraint holidays_pk primary key (holiday_date),
constraint is_midnight check ( holiday_date = trunc ( holiday_date ) )
);
INSERT into holidays (HOLIDAY_DATE,HOLIDAY_NAME)
WITH dts as (
select to_date('15-APR-2022 00:00:00','DD-MON-YYYY HH24:MI:SS'), 'Passover 2022' from dual union all
select to_date('31-DEC-2022 00:00:00','DD-MON-YYYY HH24:MI:SS'), 'New Year Eve 2022' from dual
)
SELECT * from dts;
WITH parameters (first_month, num_months) AS
(
SELECT TO_DATE ('JAN-2022', 'MON-YYYY') -- First month wanted
, 4 -- Total number of months
FROM dual
)
, all_scheduled_paydays AS
(
SELECT ADD_MONTHS (first_month + 14, LEVEL - 1) AS payday_1
, ADD_MONTHS (LAST_DAY (first_month), LEVEL - 1) AS payday_2
FROM parameters
CONNECT BY LEVEL <= num_months
)
SELECT c.actual_payday
, a.scheduled_payday -- For debugging
FROM all_scheduled_paydays
UNPIVOT ( scheduled_payday
FOR col IN (payday_1, payday_2)
) a
CROSS APPLY (
SELECT a.scheduled_payday + 1 - LEVEL AS actual_payday
FROM dual
WHERE CONNECT_BY_ISLEAF = 1
CONNECT BY TO_CHAR ( a.scheduled_payday + 2 - LEVEL
, 'DY'
, 'NLS_DATE_LANGUAGE=ENGLISH'
) IN ('SAT', 'SUN')
OR a.scheduled_payday + 2 - LEVEL
IN (
SELECT holiday_date
FROM holidays
)
) c
ORDER BY scheduled_payday
;