获取每年两个日期之间天数的平均值、最大值和最小值的查询

Question

我有一张租船的桌子：
出租：
- 租期PK
- 返回日期，
- 船号
以及此处不需要的其他字段

我需要进行一个返回四列的查询：
- 年
- 今年平均租房天数
- 今年最多租用天数
- 今年最少租用天数

现在我有这个查询：

SELECT 
YEAR(RentDate), 
AVG(DATEDIFF(DD, RentDate, ReturnDate))  AS AVERAGE, 
MAX(DATEDIFF(DD, RentDate, ReturnDate)) AS MAXIMUM, 
MIN(DATEDIFF(DD, RentDate, ReturnDate)) AS MINIMUM  
FROM RENTING
WHERE YEAR(RentDate) = YEAR(ReturnDate)
GROUP BY YEAR(RentDate)

问题是我正在考虑租金在年底开始的可能性，并且在另一年结束 - RentDate Year ！= ReturnDate。我认为这个查询不包括这种可能性。

Answer 1

我认为这个查询不包括这种可能性

不，它没有，但它很容易修复 - 只需删除您添加的 WHERE 子句：

SELECT 
YEAR(RentDate), 
AVG(DATEDIFF(DD, RentDate, ReturnDate))  AS AVERAGE, 
MAX(DATEDIFF(DD, RentDate, ReturnDate)) AS MAXIMUM, 
MIN(DATEDIFF(DD, RentDate, ReturnDate)) AS MINIMUM  
FROM RENTING
/*WHERE YEAR(RentDate) = YEAR(ReturnDate)*/
GROUP BY YEAR(RentDate)

DATEDIFF 工作正常，即使年份发生变化。

Answer 2

根据您对 D Stanley 回答的评论，您希望拆分任何跨越两年（或更多）年的租金，并将部分租金归于每年。

为此，您需要计算每年的租金天数。我知道的最简单的方法是使用Calendar Table。在您的情况下，您最感兴趣的是一年中的每一天的租金。

给定一个日历表：

CREATE TABLE Calendar
    ([CalendarDate] date, [CalendarYear] char(4))

您将租赁表加入日历，日历会将每个租赁期扩展到与租赁天数相等的行数。您可以COUNT 按年份分组的天数，按日历年拆分租金。

SELECT RentDate, COUNT(*) AS DayCount, CalendarYear
FROM Renting INNER JOIN Calendar ON CalendarDate >= RentDate 
  AND CalendarDate < ReturnDate
GROUP BY RentDate, CalendarYear

包含 PK RentDate 以区分每个不同的租金。否则，您每年都会得到一个数字，并且无法计算您的最小值、最大值和平均值。

要获取聚合值，请将第一个查询包装在另一个查询中：

SELECT CalendarYear,
       AVG(DayCount) AS AVERAGE,
       MAX(DayCount) AS MAXIMUM, 
       MIN(DayCount) AS MINIMUM  
FROM (
    SELECT RentDate, COUNT(*) AS DayCount, CalendarYear
    FROM Renting INNER JOIN Calendar ON CalendarDate >= RentDate 
      AND CalendarDate < ReturnDate
    GROUP BY RentDate, CalendarYear
) AS T

这是SQL Fiddle，您可以看到它的实际效果。

Answer 3

从表面上看，您需要在年底拆分覆盖年终的租金。我会使用我喜欢称之为测试驱动查询设计 (TDQD) 的方式来构建查询。

一年的租金

这可能涵盖了大部分数据：

SELECT YEAR(RentalDate) AS RentalYear,
       DATEDIFF(dd, RentalDate, ReturnDate) AS RentalDays
  FROM Renting
 WHERE YEAR(RentalDate) = YEAR(ReturnDate)

跨年的租金

此查询处理起始年份的租金部分：

SELECT YEAR(RentalDate) AS RentalYear,
       DATEDIFF(dd, RentalDate, DATEFROMPARTS(YEAR(RentalDate), 12, 31)) AS RentalDays
  FROM Renting
 WHERE YEAR(RentalDate) + 1 = YEAR(ReturnDate)

这个查询处理的是下一年的租金部分：

SELECT YEAR(ReturnDate) AS RentalYear,
       DATEDIFF(dd, DATEFROMPARTS(YEAR(ReturnDate), 1, 1), ReturnDate) AS RentalDays
  FROM Renting
 WHERE YEAR(RentalDate) + 1 = YEAR(ReturnDate)

租期超过两年

这还是比较棘手的。我要指出的是，理论上，单次租金可以从 2011-04-14 开始，到 2014-09-30 结束（为了争论），在这种情况下，2011 年有部分年租金，在 2012 年和 2013 年租了两次全年租金（但一个租了 366 天，另一个租了 365 天），然后是 2014 年的部分年租。但我不会解决这部分问题。

不带聚合的查询

前面三个查询需要结合 UNION ALL 来创建应该运行聚合的原始数据：

SELECT YEAR(RentalDate) AS RentalYear,
       DATEDIFF(dd, RentalDate, ReturnDate) AS RentalDays
  FROM Renting
 WHERE YEAR(RentalDate) = YEAR(ReturnDate)
UNION ALL
SELECT YEAR(RentalDate) AS RentalYear,
       DATEDIFF(dd, RentalDate, DATEFROMPARTS(YEAR(RentalDate), 12, 31)) AS RentalDays
  FROM Renting
 WHERE YEAR(RentalDate) + 1 = YEAR(ReturnDate)
UNION ALL
SELECT YEAR(ReturnDate) AS RentalYear,
       DATEDIFF(dd, DATEFROMPARTS(YEAR(ReturnDate), 1, 1), ReturnDate) AS RentalDays
  FROM Renting
 WHERE YEAR(RentalDate) + 1 = YEAR(ReturnDate)

聚合数据

SELECT RentalYear,
       AVG(RentalDays) AS Average,
       MIN(RentalDays) AS Minimum,
       MAX(RentalDays) AS Maximum
  FROM (SELECT YEAR(RentalDate) AS RentalYear,
               DATEDIFF(dd, RentalDate, ReturnDate) AS RentalDays
          FROM Renting
         WHERE YEAR(RentalDate) = YEAR(ReturnDate)
        UNION ALL
        SELECT YEAR(RentalDate) AS RentalYear,
               DATEDIFF(dd, RentalDate, DATEFROMPARTS(YEAR(RentalDate), 12, 31)) AS RentalDays
          FROM Renting
         WHERE YEAR(RentalDate) + 1 = YEAR(ReturnDate)
        UNION ALL
        SELECT YEAR(ReturnDate) AS RentalYear,
               DATEDIFF(dd, DATEFROMPARTS(YEAR(ReturnDate), 1, 1), ReturnDate) AS RentalDays
          FROM Renting
         WHERE YEAR(RentalDate) + 1 = YEAR(ReturnDate)
       ) AS Rentals
 GROUP BY RentalYear