0

我想为字符串列赋值。下面是包含一个月的休假和工作日数据的列。

Holiday_list (Column name)
--------------------------------------------------
WHHHHHHHWHHHHHHHWHHHHHHHWHHHHH

我想知道列表中的所有日期是工作日和假期。

4

2 回答 2

1

您可以使用递归子查询因式分解子句和SUBSTR：

WITH holidays ( column_name, dt, is_working, lvl ) AS (
  SELECT column_name, 
         TRUNC( SYSDATE, 'MM' ), -- Replace with the start date for your list
         SUBSTR( column_name, 1, 1 ),
         1
  FROM   holiday_list
UNION ALL
  SELECT column_name, 
         dt + INTERVAL '1' DAY,
         SUBSTR( column_name, lvl + 1, 1 ),
         lvl + 1
  FROM   holidays
  WHERE  lvl < LENGTH( column_name )
)
SELECT dt,
       is_working
FROM   holidays;

其中，对于您的示例数据：

CREATE TABLE Holiday_list (Column_name) AS
SELECT 'WHHHHHHHWHHHHHHHWHHHHHHHWHHHHH' FROM DUAL;

输出：

DT | 正在工作中
:----------------- | :---------
2021-03-01 00:00:00 | W         
2021-03-02 00:00:00 | H         
2021-03-03 00:00:00 | H         
2021-03-04 00:00:00 | H         
2021-03-05 00:00:00 | H         
2021-03-06 00:00:00 | H         
2021-03-07 00:00:00 | H         
2021-03-08 00:00:00 | H         
2021-03-09 00:00:00 | W         
2021-03-10 00:00:00 | H         
2021-03-11 00:00:00 | H         
2021-03-12 00:00:00 | H         
2021-03-13 00:00:00 | H         
2021-03-14 00:00:00 | H         
2021-03-15 00:00:00 | H         
2021-03-16 00:00:00 | H         
2021-03-17 00:00:00 | W         
2021-03-18 00:00:00 | H         
2021-03-19 00:00:00 | H         
2021-03-20 00:00:00 | H         
2021-03-21 00:00:00 | H         
2021-03-22 00:00:00 | H         
2021-03-23 00:00:00 | H         
2021-03-24 00:00:00 | H         
2021-03-25 00:00:00 | W         
2021-03-26 00:00:00 | H         
2021-03-27 00:00:00 | H         
2021-03-28 00:00:00 | H         
2021-03-29 00:00:00 | H         
2021-03-30 00:00:00 | H

或者，如果您希望它们都在一行中：

SELECT SUBSTR( column_name,  1, 1 ) AS day01,
       SUBSTR( column_name,  2, 1 ) AS day02,
       SUBSTR( column_name,  3, 1 ) AS day03,
       SUBSTR( column_name,  4, 1 ) AS day04,
       SUBSTR( column_name,  5, 1 ) AS day05,
       SUBSTR( column_name,  6, 1 ) AS day06,
       SUBSTR( column_name,  7, 1 ) AS day07,
       SUBSTR( column_name,  8, 1 ) AS day08,
       SUBSTR( column_name,  9, 1 ) AS day09,
       SUBSTR( column_name, 10, 1 ) AS day10,
       SUBSTR( column_name, 11, 1 ) AS day11,
       SUBSTR( column_name, 12, 1 ) AS day12,
       SUBSTR( column_name, 13, 1 ) AS day13,
       SUBSTR( column_name, 14, 1 ) AS day14,
       SUBSTR( column_name, 15, 1 ) AS day15,
       SUBSTR( column_name, 16, 1 ) AS day16,
       SUBSTR( column_name, 17, 1 ) AS day17,
       SUBSTR( column_name, 18, 1 ) AS day18,
       SUBSTR( column_name, 19, 1 ) AS day19,
       SUBSTR( column_name, 20, 1 ) AS day20,
       SUBSTR( column_name, 21, 1 ) AS day21,
       SUBSTR( column_name, 22, 1 ) AS day22,
       SUBSTR( column_name, 23, 1 ) AS day23,
       SUBSTR( column_name, 24, 1 ) AS day24,
       SUBSTR( column_name, 25, 1 ) AS day25,
       SUBSTR( column_name, 26, 1 ) AS day26,
       SUBSTR( column_name, 27, 1 ) AS day27,
       SUBSTR( column_name, 28, 1 ) AS day28,
       SUBSTR( column_name, 29, 1 ) AS day29,
       SUBSTR( column_name, 30, 1 ) AS day30,
       SUBSTR( column_name, 31, 1 ) AS day31
FROM   holiday_list

哪个输出：

DAY01 | DAY02 | DAY03 | DAY04 | DAY05 | DAY06 | DAY07 | DAY08 | DAY09 | DAY10 | DAY11 | 第十二天 | 第十三天 | DAY14 | DAY15 | DAY16 | DAY17 | DAY18 | DAY19 | DAY20 | DAY21 | DAY22 | DAY23 | DAY24 | DAY25 | DAY26 | DAY27 | DAY28 | DAY29 | DAY30 | DAY31
:---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :----
W | H | H | H | H | H | H | H | W | H | H | H | H | H | H | H | W | H | H | H | H | H | H | H | W | H | H | H | H | H | 无效的

db<>在这里摆弄

于 2021-03-29T09:27:15.560 回答
0

如果唯一的输入是像您在问题中显示的那样由 30 个字母组成的字符串，那么它是没有意义的。第一个字母是 W，但它对应的日期是什么？

在下面的示例中，我假设您的输入有两列 - 一列显示“月”，另一列显示 W 和 H 字符串，如您的问题中所示。我故意选择 2021 年 3 月来说明您的安排存在问题：该月有 31 天，但您的字符串只有 30 个字母。那么，3 月 31 日是工作日还是假期？

以下查询的输出与 MT0 的答案相同。查询本身是不同的。该with子句模拟您的数据（您的表） - 它不是您将运行的查询的一部分。相反，主select语句将引用您的实际表和该表中的列。

with holidays as (
  select 'March 2021' as mth, 'WHHHHHHHWHHHHHHHWHHHHHHHWHHHHH' as holiday_list
  from   dual
)
select  to_date(mth, 'Month yyyy', 'nls_date_language=English') + level - 1 as dt,
        substr(holiday_list, level, 1) as is_working_day
from    holidays
connect by level <= length(holiday_list)
;
于 2021-03-29T14:47:43.813 回答