SQL查询以查找日期之间差异的平均值

Question

我有一个列，它的数据类似于

Date
13/8/2011
2/9/2011
10/9/2011
20/9/2011

我需要编写一个SQL查询/过程，它将帮助我获得日期之间差异的平均值。对于上面的例子，它将是(19+8+10)/3=12.33 --请帮助解决这个问题。

谢谢你，盖萨

Answer 1

我不知道您的RDBMS，但这是来自Server的。另外，你的一个记录是错误的-02/09/2011-13/08/2011是20，而不是19。

create table dates (
    myDate  date
)

insert into dates
values ({d '2011-08-13'}), 
    ({d '2011-09-02'}), 
    ({d '2011-09-10'}), 
    ({d '2011-09-20'})

;with orderedDates as (
    select ROW_NUMBER() OVER (order by myDate) as row, myDate
    from dates
), datediffs as (
    select DATEDIFF(dd, o2.myDate, o1.myDate) as diff
    from orderedDates o1 left outer join
        orderedDates o2 on o1.row = o2.row + 1
)
select AVG(cast(diff as decimal))
from datediffs
where diff is NOT NULL

结果为12.667：(20 +8+ 10)/3。

Answer 2

试试这个：

SELECT DATEDIFF(
    day, 
    MIN(date_col), 
    MAX(date_col)) / (COUNT(date_col)-1) AS mean_val
FROM your_table

感谢ypercube的建议：

SELECT
  CASE
    WHEN COUNT(date_col) < 2 THEN 0
    ELSE DATEDIFF(
        day, 
        MIN(date_col), 
        MAX(date_col)) / (COUNT(date_col)-1)
  END
  as mean_val
FROM your_table