甲骨文获取上个工作日周一至周五

Question

我想获得最后一个工作日。 如果是周二到周六，那就是前一天。如果是周日或周一，那就是周五。

到目前为止，我已经尝试过这个，但我正在努力获得所需的输出。

SELECT
    level AS dow,
    trunc(sysdate, 'D') + level day,
    to_char(trunc(sysdate, 'D') + level, 'Day') AS day_week,
    CASE
        WHEN to_char(trunc(sysdate, 'D') + level, 'Day') IN (
            'Sunday',
            'Monday'
        ) THEN
            trunc(sysdate - 2, 'IW') + 4
        ELSE
            sysdate - 1
    END calculation
FROM
    dual
CONNECT BY
    level <= 7;

Answer 1

此解决方案不受语言和地域的影响：

SELECT date_value,
       date_value - CASE TRUNC(date_value) - TRUNC(date_value, 'IW')
                    WHEN 0 THEN 3 -- Monday
                    WHEN 6 THEN 2 -- Sunday
                    ELSE 1        -- Tuesday to Saturday
                    END AS previous_weekday
FROM   table_name;

其中，对于样本数据：

CREATE TABLE table_name (date_value) AS
SELECT TRUNC(sysdate - LEVEL + 1)
FROM   DUAL
CONNECT BY LEVEL <= 7;

输出（日期格式YYYY-MM-DD HH24:MI:SS (DY)）：

DATE_VALUE	PREVIOUS_WEEKDAY
2021-07-20 00:00:00 (TUE)	2021-07-19 00:00:00 (MON)
2021-07-19 00:00:00 (MON)	2021-07-16 00:00:00 (FRI)
2021-07-18 00:00:00 (SUN)	2021-07-16 00:00:00 (FRI)
2021-07-17 00:00:00 (SAT)	2021-07-16 00:00:00 (FRI)
2021-07-16 00:00:00 (FRI)	2021-07-15 00:00:00 (THU)
2021-07-15 00:00:00 (THU)	2021-07-14 00:00:00 (WED)
2021-07-14 00:00:00 (WED)	2021-07-13 00:00:00 (TUE)

db小提琴here