Question

我正在尝试提取在2个工作日内到期的活动的所有活动结束日期。因此，对于星期五和星期四，我希望日期在4天或更短时间内结束，星期六日期在3天内结束，太阳星期三日期在2天内结束，+ 3表示所有公共假日，列在日期列中的tempdate

以下脚本不起作用，只接受+2天的日期。我理解我需要使用case语句，但我真的不明白如何实现它。有人有什么想法吗？

select event_end_date
from mdmins.table

/*separating by date to find expiring dates*/ 
where  (  (dayofweek(c.event_end_date) in (5,6) /*Friday & Thursday*/ 
and    date(c.event_end_date) < current_date + 4 days 
and    date(c.event_end_date) >= current_date) 
or     (dayofweek(c.event_end_date) in (1,2,3,4) /*All other days of week*/ 
and    date(c.event_end_date) < current_date + 2 days 
and    date(c.event_end_date) >= current_date) 
or     (dayofweek(c.event_end_date) = 7 /*Saturday - not that this should be needed, but just in case*/ 
and    date(c.event_end_date) < current_date + 3 days 
and    date(c.event_end_date) >= current_date) 
or     (date(c.event_end_date) < (select dateval from mdmins.tempdate where dateval = (current date + 3 days)) /*a holiday is on in 3 days, so 2 working days before*/ 
and    date(c.event_end_date) >= current_date) )

Answer 1

从逻辑上讲，本周数据之间唯一不同的是你看多远，所以使用案例限制逻辑：

...
where date(c.event_end_date) between current_date and current_date +
  case
    when dayofweek(c.event_end_date) in (5, 6) then 4
    when dayofweek(c.event_end_date) in (1, 2, 3, 4) then 2
    else 3
  end days
or (date(c.event_end_date) < (select dateval
  from mdmins.tempdate
  where dateval = (current date + 3 days)) 
  and date(c.event_end_date) >= current_date)

Answer 2

where ((date(c.event_end_date) between current_date and current_date +
  case
    when dayofweek(current date) in (5,6) then 4
    when dayofweek(current date) in (1, 2, 3, 4) then 2
    else 3
  end days)
  or    date(c.event_end_date) < (select dateval
      from mdmins.tempdate
      where dateval = (current date + 3 days)) 
      and date(c.event_end_date) >= current_date )

Answer 3

这是一个典型的例子，说明为什么你想要一个包含所有日期的日历，以及一堆额外的信息。你能够做类似的事情：

WITH Desired_Dates AS (SELECT MIN(calendarDate) AS startRange,
                              MAX(calendarDate) + 1 DAY AS endRange
                       FROM(SELECT calendarDate
                            FROM Calendar
                            WHERE calendarDate >= CURRENT_DATE
                                  AND dayOfWeekISO IN (1, 2, 3, 4, 5)
                                  AND isHolday = '0'
                            ORDER BY calendarDate
                            FETCH FIRST 3 ROWS ONLY) D)
SELECT C.<column_list
FROM <Your_Table> C
JOIN Desired_Dates D
  ON C.event_end_date >= D.startRange
     AND C.event_end_Date < D.endRange

（请注意，我特别使用ISO星期几，其中星期一是1.星期日，因为1是美国的东西，并且函数来推导星期几从给定日期开始受到当前文化设置的影响。同样，如果event_end_date类似于时间戳，则需要使用专用结束范围
这实际上有可能使用仅索引信息来获取所需的日期（假设相关索引），并且在任何情况下都只会获取CTE中的数据一次。

如果您没有日历表，可以制作一个：

WITH Working_Dates AS (SELECT CURRENT_DATE - 1 DAY AS calendarDate, 
                              0 AS workingDayNum, '0' AS isWorkingDay
                       FROM SYSIBM/SYSDUMMY1
                       UNION ALL 
                       SELECT calendarDate, 
                              workingDayNum + CASE WHEN isWorkingDay = 1
                                                   THEN 1
                                                   ELSE 0 END AS workingDayNum,
                              isWorkingDay
                       FROM (SELECT calendarDate, workingDayNum,
                                    CASE WHEN DAYOFWEEK_ISO(WD.calendarDate) IN (6, 7)
                                              OR Holiday.dateval IS NOT NULL 
                                         THEN '0'
                                         ELSE '1' END AS isWorkingDay
                             FROM (SELECT calendarDate + 1 DAY, workingDayNum
                                   FROM Working_Dates
                                   WHERE workingDayNum < 2) WD
                             LEFT JOIN mdmins.tempdate Holiday
                                    ON Holiday.dateval = WD.calendarDate) WD),

     Desired_Dates AS (SELECT MIN(calendarDate) AS startRange,
                              MAX(calendarDate) + 1 DAY AS endRange
                       FROM Working_Date
                       WHERE isWorkingDay = '1')

SELECT C.<column_list
FROM <Your_Table> C
JOIN Desired_Dates D
  ON C.event_end_date >= D.startRange
     AND C.event_end_Date < D.endRange

这是使用递归CTE循环从当前日期（一个之前）到三个工作日过去的天数。（我必须提前一天开始，所以我只需要写＃34;它是一个工作日＆＃34;逻辑一次，这也是所有嵌套的原因。）

请注意，您的原始尝试或任何现有答案都不考虑这些：

如果周五或周一有假期（5天跳过，而不是4天）会怎样？
如果按顺序有多个假期（跳过未知天数）会怎样？

通过建立实际工作日的清单，我们可以将自己与问题隔离开来。

DB2 Sql如何使用case语句？

3 个答案: