计算重叠日期范围的价格

Question

我有一个名为 HotelRate 的简单表

HID  |  START_DATE  |   END_DATE    |   PRICE_PER_DAY
--------------------------------------
1        01/1/2015       10/1/2015       100
1        11/1/2015       20/1/2015       75
1        21/1/2015       30/1/2015       110

如果用户查询5/1/2015 到25/1/2015 之间的总价格，计算酒店房间价格的最简单方法是什么。

我已经检查过：

• How can I query for overlapping date ranges?
• Overlapping date range MySQL

但这对我来说没有多大意义。

我尝试了几个查询，但这些查询似乎是在盲目地击中箭头。有人可以建议我一个简单而优雅的方法吗？

---

@JamesZ

在运行第一个查询时我得到

start_date end_date   duration    price_per_day
---------- ---------- ----------- -------------
2015-01-01 2015-01-10 5           100
2015-01-11 2015-01-20 9           75
2015-01-21 2015-01-30 4           110

第一个范围5 可以，第二个范围应该是10，第三个应该是5

如何计算天数：start 和 end 日期之间的总夜数，与天数差异相同

05-Jan-15   06-Jan-15   1 Night
06-Jan-15   07-Jan-15   1 Night
07-Jan-15   08-Jan-15   1 Night
08-Jan-15   09-Jan-15   1 Night
09-Jan-15   10-Jan-15   1 Night
10-Jan-15   11-Jan-15   1 Night
11-Jan-15   12-Jan-15   1 Night
12-Jan-15   13-Jan-15   1 Night
13-Jan-15   14-Jan-15   1 Night
14-Jan-15   15-Jan-15   1 Night
15-Jan-15   16-Jan-15   1 Night
16-Jan-15   17-Jan-15   1 Night
17-Jan-15   18-Jan-15   1 Night
18-Jan-15   19-Jan-15   1 Night
19-Jan-15   20-Jan-15   1 Night
20-Jan-15   21-Jan-15   1 Night
21-Jan-15   22-Jan-15   1 Night
22-Jan-15   23-Jan-15   1 Night
23-Jan-15   24-Jan-15   1 Night
24-Jan-15   25-Jan-15   1 Night
               Count : 20 Night

Answer 1

这样的事情应该可以解决问题：

declare @startdate date, @enddate date

set @startdate = '20150105'
set @enddate = '20150125'

select
  start_date,
  end_date,
  datediff(
    day, 
    case when @startdate > start_date then @startdate else start_date end, 
    case when @enddate < end_date then @enddate else end_date end) as duration,
  price_per_day
from
  reservation
where
  end_date >= @startdate and
  start_date <= @enddate

这只是处理带有大小写的重叠范围，因此如果预订开始是正确的，则使用它，否则搜索条件和结束日期相同。此处的天数和价格是分开的，但您可以将它们相乘得到结果。

SQL 小提琴：http://sqlfiddle.com/#!3/4027b3/1

编辑，这样得到总和：

declare @startdate date, @enddate date

set @startdate = '20150105'
set @enddate = '20150125'

select
  sum(datediff(
    day, 
    case when @startdate > start_date then @startdate else start_date end, 
    case when @enddate < end_date then @enddate else end_date end)  
  * price_per_day)
from
  reservation
where
  end_date >= @startdate and
  start_date <= @enddate

Answer 2

您将需要一个日历表，但每个数据库都应该有一个。 实际实现总是特定于用户和 DBMS（例如MS SQL Server），因此搜索“日历表”+ yourDBMS 可能会显示您系统的一些源代码。

select HID, sum(PRICE_PER_DAY)
from calendar_table as c
join HotelRate
  on calendar_date between START_DATE and END_DATE
group by HID

Answer 3

如果您有可用的现有日期表，这很容易处理。还没有？您将在下面找到两个可帮助您入门的功能。这就是你使用它们的方式：

-- Arguments can be passed in any order
SELECT * FROM dbo.RangeDate('2015-12-31', '2015-01-01');
SELECT * FROM dbo.RangeSmallInt(10, 0);

SELECT A.HID, SUM(A.PRICE_PER_DAY)
FROM dbo.RangeDate('2000-01-01', '2020-12-31') Calendar
JOIN HotelRate A
    ON Calendar.D BETWEEN A.START_DATE and A.END_DATE
GROUP BY A.HID;

您可以将 RangeDate 函数用作日历，也可以使用它来构建自己的日历函数/表格。

-- Generate a range of up to 65,536 contiguous DATES
CREATE FUNCTION dbo.RangeDate (   
    @date1 DATE = NULL
  , @date2 DATE = NULL
)   
RETURNS TABLE   
AS   
RETURN (
    SELECT D = DATEADD(DAY, A.N, CASE WHEN @date1 <= @date2 THEN @date1 ELSE @date2 END)
    FROM dbo.RangeSmallInt(
        CASE WHEN @date1 IS NOT NULL AND @date2 IS NOT NULL THEN 0 END
      , ABS(DATEDIFF(DAY, @date1, @date2))
    ) A
);

-- Generate a range of up to 65,536 contiguous BIGINTS
CREATE FUNCTION dbo.RangeSmallInt (
    @n1 BIGINT = NULL
  , @n2 BIGINT = NULL
)
RETURNS TABLE
AS
RETURN (
    WITH Numbers AS (
        SELECT N FROM(VALUES
            (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 16
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 32
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 48
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 64
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 80
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 96
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 112
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 128
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 144
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 160
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 176
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 192
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 208
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 224
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 240
          , (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1), (1) -- 256
        ) V (N)
    )    
    SELECT TOP (
               CASE
                   WHEN @n1 IS NOT NULL AND @n2 IS NOT NULL THEN ABS(@n2 - @n1) + 1
                   ELSE 0
               END
           )
           N = ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) - 1 + CASE WHEN @n1 <= @n2 THEN @n1 ELSE @n2 END
    FROM Numbers A, Numbers B
    WHERE ABS(@n2 - @n1) + 1 < 65537
);

Answer 4

您可以使用它来计算每个期间的价格，然后将其总结为总成本。它使用案例语句来计算每个时期有多少天，因此在您的示例中这是 5,9 和 4：

Declare @startdate date = '2015-01-05',
        @todate date = '2015-01-25'

Select sum(price_per_period) as TotalPrice -- The cost for all periods is summed to give a total
from
-- First it works out the number of days in the period with a case statement and then
-- multiplies this by the daily rate to get the total for that period
(Select price_per_day * case when Start_date <= @startdate then DATEDIFF(day, @startdate,end_date) else  
    case when Start_date > @startdate and end_date < @todate then DATEDIFF(day, start_date,end_date) else 
        case when Start_date > @startdate and end_date >= @todate then  DATEDIFF(day, start_date, @todate) end
        end
    end price_per_period

 from pricetable
 where (Start_date between @Startdate and @todate) or 
      (end_date between @Startdate and @todate)
) a

这消除了对单独日历表的需要

SQL 小提琴：http://www.sqlfiddle.com/#!6/25e63/4/0

Answer 5

这应该足够快，因为您首先生成日历，然后只使用联接。每家酒店的总价也可以通过分组来实现：

数据定义：

create table HotelRate(HID int, START_DATE date, END_DATE date, PRICE_PER_DAY int);

insert into HotelRate values
(1, '20150101', '20150110', 100),
(1, '20150111', '20150120', 75),
(1, '20150121', '20150130', 110),
(2, '20150101', '20150110', 10),
(2, '20150111', '20150120', 5),
(2, '20150121', '20150130', 50)

查询：

declare @sd date = '20150105' , @ed date = '20150125'

;with c as(select @sd d union all select dateadd(dd, 1, d) from c where d < @ed)
select h.HID, h.START_DATE, h.END_DATE, sum(PRICE_PER_DAY) PRICE
from c join HotelRate h on c.d >= h.START_DATE and c.d < h.END_DATE
group by grouping sets((h.HID, h.START_DATE, h.END_DATE),(h.HID))

输出：

HID START_DATE  END_DATE    PRICE
1   2015-01-01  2015-01-10  500
1   2015-01-11  2015-01-20  675
1   2015-01-21  2015-01-30  550
1   (null)      (null)      1725
2   2015-01-01  2015-01-10  50
2   2015-01-11  2015-01-20  45
2   2015-01-21  2015-01-30  250
2   (null)      (null)      345

这可以通过计数表进一步优化。更重要的是，如果您在数据库中创建日历表，它将是即时的。

这是小提琴http://sqlfiddle.com/#!3/25e7bc/1

假设您创建了一些日历表Calendar(d date)，其中包含从例如1900-01-01 到结束2100-01-01 的日期。在日期列的Calendar 和HotelRange 表上添加索引。那么上面的查询可以改写为：

select h.HID, h.START_DATE, h.END_DATE, sum(PRICE_PER_DAY) PRICE
from Calendar c join HotelRate h on c.d >= h.START_DATE and c.d < h.END_DATE
where c.d between @sd and @ed
group by grouping sets((h.HID, h.START_DATE, h.END_DATE),(h.HID))