如果将其分为以下任务,应该很容易:
- 使用Row Generator方法生成两个给定日期之间的所有日期,如下所示。
- 忽略周末的日期,即周六和周日
- 检查范围内的日期是否与假期表中的任何匹配。
以下行生成器查询将为您提供工作日的总数,即不包括周六和周日:
SQL> WITH dates AS
2 (SELECT to_date('01/01/2014', 'DD/MM/YYYY') date1,
3 to_date('31/12/2014', 'DD/MM/YYYY') date2
4 FROM dual
5 )
6 SELECT SUM(weekday) weekday_count
7 FROM
8 (SELECT
9 CASE
10 WHEN TO_CHAR(date1+LEVEL-1, 'DY','NLS_DATE_LANGUAGE=AMERICAN')
11 NOT IN ('SAT', 'SUN')
12 THEN 1
13 ELSE 0
14 END weekday
15 FROM dates
16 CONNECT BY LEVEL <= date2-date1+1
17 )
18 /
WEEKDAY_COUNT
-------------
261
SQL>
现在,基于上面的行生成器查询,让我们看一个测试用例。
以下查询将计算2014 年 1 月 1 日至 2014年12 月 31 日之间的工作日数,不包括表中提到的节假日。
WITH子句仅将其用作表,在您的情况下,您可以简单地使用您的holiday table。
SQL> WITH dates
2 AS (SELECT To_date('01/01/2014', 'DD/MM/YYYY') date1,
3 To_date('31/12/2014', 'DD/MM/YYYY') date2
4 FROM dual),
5 holidays
6 AS (SELECT To_date('12.06.2011', 'DD.MM.YYYY') holiday FROM dual UNION ALL
7 SELECT To_date('19.06.2014', 'DD.MM.YYYY') holiday FROM dual UNION ALL
8 SELECT To_date('09.05.2013', 'DD.MM.YYYY') holiday FROM dual),
9 count_of_weekdays
10 AS (SELECT SUM(weekday) weekday_count
11 FROM (SELECT CASE
12 WHEN To_char(date1 + LEVEL - 1, 'DY',
13 'NLS_DATE_LANGUAGE=AMERICAN')
14 NOT IN (
15 'SAT',
16 'SUN' ) THEN 1
17 ELSE 0
18 END weekday
19 FROM dates
20 CONNECT BY LEVEL <= date2 - date1 + 1)),
21 count_of_holidays
22 AS (SELECT Count(*) holiday_count
23 FROM holidays
24 WHERE holiday NOT BETWEEN To_date('01/01/2015', 'DD/MM/YYYY') AND
25 To_date('31/03/2015', 'DD/MM/YYYY'))
26 SELECT weekday_count - holiday_count as working_day_count
27 FROM count_of_weekdays,
28 count_of_holidays
29 /
WORKING_DAY_COUNT
-----------------
258
SQL>
共有261个工作日,其中假期表中有3个假期。因此,输出中的工作日总数为。261 - 3 = 258