Oracle 查询排除周末和下午 6 点到晚上 9 点

Posted 2023-03-27

【中文标题】Oracle 查询排除周末和下午 6 点到晚上 9 点

【英文标题】：Oracle query to Exclude weekends, and 6PM to 9PM

【发布时间】：2020-12-30 21:07:34

【问题描述】：

我正在尝试实现一个查询，该查询返回两个日期之间的时差，不包括周末（周六和周日）和不包括时间（下午 6 点至上午 9 点）。 目前，我有一个排除周末的功能，但我无法从查询中排除时间。有人能帮忙吗？ 我求助的文章是this

CREATE OR REPLACE FUNCTION get_bus_minutes_between(

                                   p_start_date DATE,

                                   p_end_date DATE

                                  )

    RETURN NUMBER
    DETERMINISTIC -- ***** Can't hurt
    IS
        days_diff     NUMBER  := 0;
        end_date      DATE    := p_end_date;
        minutes_diff  NUMBER;
        start_date    DATE    := p_start_date;
        weeks_diff    NUMBER;
BEGIN
    IF  start_date <= end_date
    THEN
        --  Move start_date and end_date away from weekends
        IF  start_date > TRUNC (start_date, 'IW') + 5
        THEN     -- Use next Monday for start_date
            start_date := TRUNC (start_date, 'IW') + 7;
        END IF;

        IF  end_date > TRUNC (end_date, 'IW') + 5
        THEN     -- Use Friday quitting time
            end_date := TRUNC (end_date, 'IW') + 4 + (16.5 / 24);
        END IF;

       -- Move start_date into the same weeek as end_date
       -- (Remember how many weeks we had to move it)
       weeks_diff := ( TRUNC (end_date,   'IW')
                     - TRUNC (start_date, 'IW')
                     ) / 7;
       IF  weeks_diff > 0
       THEN
           start_date := start_date + (7 * weeks_diff);
       END IF;

       -- Make start_date the same day as end_date
       -- (Remember how many days we had to move it)
       days_diff := TRUNC (end_date) - TRUNC (start_date);
       IF  days_diff > 0
       THEN
           start_date := start_date + days_diff;
       END IF;

       -- Move start_date up to starting time
       start_date := GREATEST ( start_date
                              , TRUNC (start_date) + (8.75 / 24)
                              );

       -- Move end_date back to quitting time
       end_date := LEAST ( end_date
                         , TRUNC (end_date) + ( CASE
                                                    WHEN  TO_CHAR ( end_date
                                                                  , 'DY'
                                                                  , 'NLS_DATE_LANGUAGE=ENGLISH'
                                                                  ) = 'FRI'
                                                    THEN  16.5
                                                    ELSE  17
                                                END
                                              / 24
                                              )
                         );

         minutes_diff := ( GREATEST ( 0
                                    , end_date - start_date
                                    )
                         * 24 * 60
                         )
                     +  (days_diff * 495)     --  495 minutes per full day (Mon.-Thu.)
                     +  (weeks_diff * 2445);  -- 2445 minutes per full week
    ELSIF  start_date > end_date
    THEN
        minutes_diff := -get_bus_minutes_between (end_date, start_date);
    ELSE    -- One of the arguments was NULL
        minutes_diff := NULL;
    END IF;

   RETURN  ROUND(minutes_diff);
END  get_bus_minutes_between;

【问题讨论】：

要尝试的几件事：验证传递给您的函数的DATE 值是否具有时间组件。许多数据库应用程序经常丢弃插入的时间部分。第二件事：我会尝试设置一个测试SELECT 查询，并将每个逻辑设置为输出，以查看逻辑中的每个“步骤”是否正在执行预期的操作。还要验证TRUNC 函数是否返回了您所期望的；见docs.oracle.com/en/database/oracle/oracle-database/21/sqlrf/…

也许这个：***.com/questions/41936398/…

【参考方案1】：

可以直接计算天差（改编自我的回答here）：

SELECT start_date,
       end_date,
       ROUND(
         (
           -- Calculate the full weeks difference from the start of ISO weeks.
           ( TRUNC( end_date, 'IW' ) - TRUNC( start_date, 'IW' ) ) * (9/24) * (5/7)
           -- Add the full days for the final week.
           + LEAST( TRUNC( end_date ) - TRUNC( end_date, 'IW' ), 5 ) * (9/24)
           -- Subtract the full days from the days of the week before the start date.
           - LEAST( TRUNC( start_date ) - TRUNC( start_date, 'IW' ), 5 ) * (9/24)
           -- Add the hours of the final day
           + LEAST( GREATEST( end_date - TRUNC( end_date ) - 9/24, 0 ), 9/24 )
           -- Subtract the hours of the day before the range starts.
           - LEAST( GREATEST( start_date - TRUNC( start_date ) - 9/24, 0 ), 9/24 )
         )
         -- Multiply to give minutes rather than fractions of full days.
         * 24 * 60
       ) AS work_day_mins_diff
FROM   table_name;

其中，对于样本数据：

CREATE TABLE table_name ( start_date, end_date ) AS
SELECT DATE '2020-12-30' + INTERVAL '00' HOUR, DATE '2020-12-30' + INTERVAL '12' HOUR FROM DUAL UNION ALL
SELECT DATE '2020-12-30' + INTERVAL '18' HOUR, DATE '2020-12-30' + INTERVAL '20' HOUR FROM DUAL UNION ALL
SELECT DATE '2020-12-30' + INTERVAL '17:30' HOUR TO MINUTE, DATE '2020-12-30' + INTERVAL '21:30' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT DATE '2021-01-01' + INTERVAL '00' HOUR, DATE '2021-01-04' + INTERVAL '00' HOUR FROM DUAL UNION ALL
SELECT DATE '2021-01-02' + INTERVAL '00' HOUR, DATE '2021-01-04' + INTERVAL '00' HOUR FROM DUAL UNION ALL
SELECT DATE '2020-12-28' + INTERVAL '00' HOUR, DATE '2021-01-04' + INTERVAL '00' HOUR FROM DUAL UNION ALL
SELECT DATE '2020-12-28' + INTERVAL '00' HOUR, DATE '2020-12-29' + INTERVAL '00' HOUR FROM DUAL;

输出：

（使用ALTER SESSION SET NLS_DATE_FORMAT = 'YYYY-MM-DD HH24:MI:SS (DY)';）

START_DATE | END_DATE | WORK_DAY_MINS_DIFF :------------------------ | :------------------------ | -----------------: 2020-12-30 00:00:00 (周三) | 2020-12-30 12:00:00 (周三) | 180 2020-12-30 18:00:00 (周三) | 2020-12-30 20:00:00 (周三) | 0 2020-12-30 17:30:00 (周三) | 2020-12-30 21:30:00 (周三) | 30 2021-01-01 00:00:00 (周五) | 2021-01-04 00:00:00 (周一) | 540 2021-01-02 00:00:00 (周六) | 2021-01-04 00:00:00 (周一) | 0 2020-12-28 00:00:00 (周一) | 2021-01-04 00:00:00 (周一) | 2700 2020-12-28 00:00:00 (周一) | 2020-12-29 00:00:00 (星期二) | 540

db小提琴here