没有周末和节假日的两个日期的区别sql查询ORACLE

Question

我有 2 个表格：第一个包含采购订单的开始日期和结束日期， 第二张表包含年假

-采购订单

-假期

我正在尝试计算不包括周末和节假日的两个日期之间的工作日数。

输出应该是这样的：

Start Date | End Date | Business Days

你能帮帮我吗

Answer 1

由于您已经有一个假期表，您可以计算开始日期和结束日期之间的假期，然后从结束日期和开始日期之间的天数差中减去。对于周末，您需要一个包含类似于您的假期表的周末天数的表，或者您可以按如下方式生成它们。

with sample_data(id, start_date, end_date) as (
  select 1, date '2015-03-06', date '2015-03-7' from dual union all
  select 2, date '2015-03-07', date '2015-03-8' from dual union all
  select 3, date '2015-03-08', date '2015-03-9' from dual union all
  select 4, date '2015-02-07', date '2015-06-26' from dual union all
  select 5, date '2015-04-17', date '2015-08-16' from dual
)
, holidays(holiday) as (
  select date '2015-01-01' from dual union all -- New Years
  select date '2015-01-19' from dual union all -- MLK Day
  select date '2015-02-16' from dual union all -- Presidents Day
  select date '2015-05-25' from dual union all -- Memorial Day
  select date '2015-04-03' from dual union all -- Independence Day (Observed)
  select date '2015-09-07' from dual union all -- Labor Day
  select date '2015-11-11' from dual union all -- Veterans Day
  select date '2015-11-26' from dual union all -- Thanks Giving
  select date '2015-11-27' from dual union all -- Black Friday
  select date '2015-12-25' from dual           -- Christmas
)
-- If your calendar table doesn't already hold weekends you can generate
-- the weekends with these next two subfactored queries (common table Expressions)
, firstweekend(weekend, end_date) as (
  select next_day(min(start_date),'saturday'), max(end_date) from sample_data
  union all
  select next_day(min(start_date),'sunday'), max(end_date) from sample_data
)
, weekends(weekend, last_end_date) as (
  select weekend, end_date from firstweekend
  union all
  select weekend + 7, last_end_date from weekends where weekend+7 <= last_end_date
)
-- if not already in the same table combine distinct weekend an holiday days
-- to prevent double counting (in case a holiday is also a weekend).
, days_off(day_off) as (
  select weekend from weekends
  union
  select holiday from holidays
)
select id
     , start_date
     , end_date
     , end_date - start_date + 1
     - (select count(*) from days_off where day_off between start_date and end_date) business_days
  from sample_data;

        ID START_DATE  END_DATE    BUSINESS_DAYS
---------- ----------- ----------- -------------
         1 06-MAR-2015 07-MAR-2015             1
         2 07-MAR-2015 08-MAR-2015             0
         3 08-MAR-2015 09-MAR-2015             1
         4 07-FEB-2015 26-JUN-2015            98
         5 17-APR-2015 16-AUG-2015            85

Answer 2

此查询应为purchase 表中的每个范围生成准确的工作日数：

with days as (
  select rn, sd + level - 1 dt, sd, ed 
    from (select row_number() over (order by start_date) rn, 
                 start_date sd, end_date ed from purchase_order)
    connect by prior rn = rn and sd + level - 1 <= ed
           and prior dbms_random.value is not null)
select sd start_date, ed end_date, count(1) business_days
  from days d left join holidays h on holiday_date = d.dt
  where dt - trunc(dt, 'iw') not in (5, 6) and h.holiday_date is null
  group by rn, sd, ed

^{SQLFiddle demo}

对于purchase_orders 中的每一行，查询从这个范围内生成日期（这是由子查询dates 完成的）。 主查询检查这是周末还是节假日，并计算其余日期。

如果purchase_orders 中有大量数据，用于生成日期的分层查询可能会导致速度变慢 或经期较长。在这种情况下，首选方法是创建日历表，正如 cmets 中已建议的那样。

Answer 3

编辑：我忽略了 Oracle 标记的存在，并开始为 SQL Server 编写脚本。但是这个概念并没有改变。

为了超级准确，我会创建一个如下格式的表格。

Year int, month int, DaysInMonth int, firstOccuranceOfSunday int

创建一个过程以从该表上的特定年份和月份中提取周末。

CREATE FUNCTION [dbo].[GetWeekendsForMonthYear]
(
    @year int,
    @month int
)
RETURNS @weekends TABLE
(
[Weekend] date
)
AS
BEGIN

declare @firstsunday int = 0
Declare @DaysInMonth int = 0
Select @DaysInMonth = DaysInMonth, @firstsunday = FirstSunday from Months
Where [Year] = @year and [month] = @month
Declare @FirstSaterday int = @firstsunday - 1

declare @CurrentDay int = 0
Declare @CurrentDayIsSunday bit = 0

if @FirstSaterday !< 1
 Begin
    insert into @Weekends values(DATEADD(year, @year -1900, DATEADD(month, @month -1, DATEADD(day, @Firstsaterday -1, 0))))

    insert into @Weekends values(DATEADD(year, @year -1900, DATEADD(month, @month -1, DATEADD(day, @FirstSunday -1, 0))))
    set @CurrentDayIsSunday = 1
    set @CurrentDay = @firstsunday
 END
else
    begin
     insert into @Weekends values(DATEADD(year, @year -1900, DATEADD(month, @month -1, DATEADD(day, @FirstSunday -1, 0))))
     set @FirstSaterday = @firstsunday + 6
     insert into @Weekends values(DATEADD(year, @year -1900, DATEADD(month, @month -1, DATEADD(day, @Firstsaterday -1, 0))))
     set @CurrentDayIsSunday = 0
    set @CurrentDay = @FirstSaterday
    end 

 declare @done bit = 0

 while @done = 0
 Begin
    if @CurrentDay <= @DaysInMonth
        Begin
            If @CurrentDayIsSunday = 1
            begin
                set @CurrentDay = @CurrentDay + 6
                set @CurrentDayIsSunday = 0
                if @CurrentDay <= @DaysInMonth
                begin
                    insert into @Weekends Values(DATEADD(year, @year -1900, DATEADD(month, @month -1, DATEADD(day, @CurrentDay -1, 0))))
                end
            end
            else
                begin
                    set @CurrentDay = @CurrentDay + 1
                    set @CurrentDayIsSunday = 1
                    if @CurrentDay <= @DaysInMonth
                    begin
                        insert into @Weekends Values(DATEADD(year, @year -1900, DATEADD(month, @month -1, DATEADD(day, @CurrentDay -1, 0))))
                    end
                end
            end
    ELSE
    begin
        Set @done = 1
    end
 end
    RETURN
END

当调用并提供年份和月份时，这将返回代表周末的日期列表。

现在，使用该函数，创建一个过程，为特定日期范围内的每个适用行调用一次该函数，并在临时表中返回值。

请注意，我现在发布此内容，以便您查看发生了什么，但我将继续处理代码。当它们出现时，我会发布更新。

更多内容：获取特定日期范围的周末列表（格式化），从该列表中删除可以在您的假期表中找到的所有日期。

很遗憾，我明天还要上班，我要睡觉了。

Answer 4

您可以使用如下查询删除非周末假期：

select (t.end_date - t.start_date) - count(c.date)
from table1 t left join
     calendar c
     on c.date between t1.start_date and t1.end_date and
        to_char(c.date, 'D') not in ('1', '7')
group by t.end_date, t.start_date;

删除周末的日子会更加复杂。整周有两个周末，所以这很容易。所以一个很好的近似值是：

select (t.end_date - t.start_date) - (count(c.date) +
       2 * floor((t.end_date - t.start_date) / 7))
from table1 t left join
     calendar c
     on c.date between t1.start_date and t1.end_date and
        to_char(c.date, 'D') not in ('1', '7')
group by t.end_date, t.start_date;

这没有得到星期几，这实际上是如果结束日期在开始日期之前，那么它在下一周。然而，这个逻辑在 Oracle 处理星期几的方式中变得相当复杂，所以也许上面的近似值就足够了。