查找两个日期之间的第五个工作日sql

Question

我正在尝试查找两个日期之间的第五个工作日，不包括节假日。

这是我写的查询：

with s_date as (select TO_DATE('21-04-2020','DD-MM-YYYY') d from dual),
     e_date as (select TO_DATE('01-05-2020','DD-MM-YYYY') d from dual),
     no_of_days as (select abs(trunc(s_date.d -  e_date.d))+1 no from s_date,e_date),
    cal as (select d+rownum-1 dt
            from s_date
            connect by level <= (select no from no_of_days) )
select cal.dt  from cal where to_char(cal.dt, 'DY', 'NLS_DATE_LANGUAGE=ENGLISH') NOT IN ('SAT', 'SUN')
and cal.dt not in (select day from T_PUBLIC_HOLIDAYS) order by cal.dt desc;

输出：

DT
1:30-04-2020
2:29-04-2020
3:28-04-2020
4:27-04-2020
5:24-04-2020
6:23-04-2020
7:22-04-2020
8:21-04-2020

如何将输出限制为： DT 1： 24-04-2020

Answer 1

对于第五个，您只需在请求的末尾添加：

fetch first 5 rows only

您可以只使用一行（然后结束...）：

fetch first row only

您首先选择第 5 行（按日期降序排列），然后选择唯一的第一行，按日期升序排列前一个结果。

你的 SQL 变成了：

select * from (
with s_date as (select TO_DATE('21-04-2020','DD-MM-YYYY') d from dual),
     e_date as (select TO_DATE('01-05-2020','DD-MM-YYYY') d from dual),
     no_of_days as (select abs(trunc(s_date.d -  e_date.d))+1 no from s_date,e_date),
    cal as (select d+rownum-1 dt
            from s_date
            connect by level <= (select no from no_of_days) )
select cal.dt  from cal where to_char(cal.dt, 'DY', 'NLS_DATE_LANGUAGE=ENGLISH') NOT IN ('SAT', 'SUN')
and cal.dt not in (select day from T_PUBLIC_HOLIDAYS)
order by cal.dt desc
fetch first 5 rows only
) x
order by x.dt
fetch first row only
;

Answer 2

请检查以下查询，让我知道您的问题是否已解决。

with s_date as (select TO_DATE('21-04-2020','DD-MM-YYYY') d from dual),
     e_date as (select TO_DATE(DT_EXPR_PLDG,'DD-MM-YYYY') d from T_ANCRDT_FNNCL_C_FR where DT_EXPR_PLDG is not null ),
     no_of_days as (select abs(trunc(s_date.d -  e_date.d))+1 no from s_date,e_date),
    cal as (select d+rownum-1 dt
            from s_date
            connect by level <= (select no from no_of_days) )
select cal.dt  from cal where to_char(cal.dt, 'DY', 'NLS_DATE_LANGUAGE=ENGLISH') NOT IN ('SAT', 'SUN')
AND to_char (date '21-04-2020', 'D')=5 -- you can add weekday() function as well 
and cal.dt not in (select day from T_PUBLIC_HOLIDAYS) 
order by cal.dt desc;

Answer 3

将它放到另一个 SELECT 语句中并使用 ROWNUM。

select date_value FROM (
select cal.dt as date_value, ROWNUM as row_num from cal where to_char(cal.dt, 'DY', 'NLS_DATE_LANGUAGE=ENGLISH') NOT IN ('SAT', 'SUN')
and cal.dt not in (select day from T_PUBLIC_HOLIDAYS) order by cal.dt desc
)) WHERE row_num = 5

顺便说一句，你的整个 SQL 语句很丑，你不应该使用它:)

Answer 4

首先创建附加列以对结果中的日期进行排序，然后选择所需的任何日期，在您的情况下为 day_order = 6

select 
    cal.dt 
    , rank() over (order by cal.dt) as day_order    
from cal 
where to_char(cal.dt, 'dy', 'nls_date_language=english') not in ('sat', 'sun') 
and cal.dt not in (select day from T_PUBLIC_HOLIDAYS)

代码如下：

with 
s_date as
(
select to_date('21-04-2020','dd-mm-yyyy') d from dual
)
, e_date as 
(
select to_date('01-05-2020','dd-mm-yyyy') d from dual
)
, no_of_days as 
(
select abs(trunc(s_date.d -  e_date.d))+1 no from s_date,e_date
)
, cal as
(   
select d+rownum-1 dt
from s_date
connect by level <= (select no from no_of_days) 
)
, cal2 as 
(select 
    cal.dt 
    , rank() over (order by cal.dt) as day_order    
from cal 
where to_char(cal.dt, 'dy', 'nls_date_language=english') not in ('sat', 'sun') 
and cal.dt not in (select day from T_PUBLIC_HOLIDAYS)
)
select dt from cal2 
where day_order = 5 -- day_order = 6 if you want to exclude the start day

Answer 5

这将在一个语句中计算多个开始日期和偏移日期的值：

Oracle 设置：

CREATE TABLE test_data ( start_date, days ) AS
SELECT DATE '2019-01-01' + date_offset, days
FROM   ( SELECT LEVEL - 1 AS date_offset FROM DUAL CONNECT BY LEVEL <= 7 )
       CROSS JOIN
       ( SELECT LEVEL - 1 AS days FROM DUAL CONNECT BY LEVEL <= 7 );

CREATE TABLE holidays ( holiday_date ) AS
SELECT DATE '2019-01-01' FROM DUAL UNION ALL
SELECT DATE '2019-01-09' FROM DUAL UNION ALL
SELECT DATE '2019-01-10' FROM DUAL;

创建一个辅助函数来添加工作日。这不是必需的，因为您可以将函数展开到查询中，但它会使正在发生的事情更加明显：

CREATE FUNCTION add_weekdays(
  start_date IN DATE,
  weekdays   IN NUMBER
) RETURN DATE DETERMINISTIC
IS
BEGIN
  RETURN start_date
         -- add full weeks
         + FLOOR( weekdays / 5 ) * 7
         -- check to see if the remaining days go over a weekend, if so add 2 extra days
         + CASE WHEN ( start_date - TRUNC( start_date, 'IW' ) ) + MOD( weekdays, 5 ) >= 5
           THEN MOD( weekdays, 5 ) + 2
           ELSE MOD( weekdays, 5 )
           END;
END;
/

查询：

WITH offsets ( start_date, days, offset_date, holiday_days ) AS (
  SELECT start_date,
         days,
         add_weekdays( start_date, days ),
         ( SELECT COUNT(*)
           FROM   holidays
           WHERE  holiday_date BETWEEN start_date
                               AND add_weekdays( start_date, days )
         )
  FROM   test_data
UNION ALL
  -- recurse and add the previously found holiday days and look to see if any more holidays overlap
  SELECT start_date,
         days,
         add_weekdays( offset_date, holiday_days ),
         ( SELECT COUNT(*)
           FROM   holidays
           WHERE  holiday_date BETWEEN offset_date + 1
                               AND add_weekdays( offset_date, holiday_days )
         )
  FROM   offsets
  -- stop recursing when no holiday days are found
  WHERE  holiday_days > 0
)
CYCLE start_date, days, offset_date SET is_cycle TO 1 DEFAULT 0
SELECT TO_CHAR( start_date, 'DY YYYY-MM-DD' ) AS start_dt,
       days,
       TO_CHAR( offset_date, 'DY YYYY-MM-DD' ) AS offset_dt,
       ( offset_date - start_date ) - days AS skipped_days
FROM   offsets
WHERE  holiday_days = 0
ORDER BY start_date, days;

输出：

（测试的节假日日期为2019-01-01、2019-01-09 和2019-01-10，跳过的天数包括节假日和周末）

START_DT |天 | OFFSET_DT | SKIPPED_DAYS :------------- | ---: | :------------- | ------------: 周二 2019-01-01 | 0 |周三 2019-01-02 | 1 周二 2019-01-01 | 1 |星期四 2019-01-03 | 1 周二 2019-01-01 | 2 |周五 2019-01-04 | 1 周二 2019-01-01 | 3 |星期一 2019-01-07 | 3 周二 2019-01-01 | 4 |周二 2019-01-08 | 3 周二 2019-01-01 | 5 |周五 2019-01-11 | 5 周二 2019-01-01 | 6 |星期一 2019-01-14 | 7 周三 2019-01-02 | 0 |周三 2019-01-02 | 0 周三 2019-01-02 | 1 |星期四 2019-01-03 | 0 周三 2019-01-02 | 2 |周五 2019-01-04 | 0 周三 2019-01-02 | 3 |星期一 2019-01-07 | 2 周三 2019-01-02 | 4 |周二 2019-01-08 | 2 周三 2019-01-02 | 5 |周五 2019-01-11 | 4 周三 2019-01-02 | 6 |星期一 2019-01-14 | 6 星期四 2019-01-03 | 0 |星期四 2019-01-03 | 0 星期四 2019-01-03 | 1 |周五 2019-01-04 | 0 星期四 2019-01-03 | 2 |星期一 2019-01-07 | 2 星期四 2019-01-03 | 3 |周二 2019-01-08 | 2 星期四 2019-01-03 | 4 |周五 2019-01-11 | 4 星期四 2019-01-03 | 5 |星期一 2019-01-14 | 6 星期四 2019-01-03 | 6 |周二 2019-01-15 | 6 周五 2019-01-04 | 0 |周五 2019-01-04 | 0 周五 2019-01-04 | 1 |星期一 2019-01-07 | 2 周五 2019-01-04 | 2 |周二 2019-01-08 | 2 周五 2019-01-04 | 3 |周五 2019-01-11 | 4 周五 2019-01-04 | 4 |星期一 2019-01-14 | 6 周五 2019-01-04 | 5 |周二 2019-01-15 | 6 周五 2019-01-04 | 6 |周三 2019-01-16 | 6 SAT 2019-01-05 | 0 |星期一 2019-01-07 | 2 SAT 2019-01-05 | 1 |周二 2019-01-08 | 2 SAT 2019-01-05 | 2 |周五 2019-01-11 | 4 SAT 2019-01-05 | 3 |星期一 2019-01-14 | 6 SAT 2019-01-05 | 4 |周二 2019-01-15 | 6 SAT 2019-01-05 | 5 |周三 2019-01-16 | 6 SAT 2019-01-05 | 6 |周四 2019-01-17 | 6 孙 2019-01-06 | 0 |周二 2019-01-08 | 2 孙 2019-01-06 | 1 |周五 2019-01-11 | 4 孙 2019-01-06 | 2 |星期一 2019-01-14 | 6 孙 2019-01-06 | 3 |周二 2019-01-15 | 6 孙 2019-01-06 | 4 |周三 2019-01-16 | 6 孙 2019-01-06 | 5 |周四 2019-01-17 | 6 孙 2019-01-06 | 6 |周五 2019-01-18 | 6 星期一 2019-01-07 | 0 |星期一 2019-01-07 | 0 星期一 2019-01-07 | 1 |周二 2019-01-08 | 0 星期一 2019-01-07 | 2 |周五 2019-01-11 | 2 星期一 2019-01-07 | 3 |星期一 2019-01-14 | 4 星期一 2019-01-07 | 4 |周二 2019-01-15 | 4 星期一 2019-01-07 | 5 |周三 2019-01-16 | 4 星期一 2019-01-07 | 6 |周四 2019-01-17 | 4

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