如何获得 datediff 的平均时间？

Question

通过下面的查询，我得到了每个的 DATEDIFF 记录在 table.

如何获取所有记录的datediff的AVG？

我不关心日期，只关心时间。

SELECT 
right('0' + convert(float,datediff 
                    (second, callstartdt,endstartdt)/ 60 % 60), 2) + ':' +
right('0' + convert(float,datediff 
                    (second, callstartdt,endstartdt)% 60),2)  as total_time
FROM times

http://sqlfiddle.com/#!6/f01c6/2

我有 sql 2012.

Answer 1

with avg_s as (
    select avg(datediff(ss, callstartdt, endstartdt)) as ss from times
)
select
    ss / 86400 as days,
    ss % 86400 / 3600 as hours,
    ss % 3600 / 60 as minutes,
    ss % 60 as seconds,
    concat(
        format(ss / 3600, '00:'), /* could be greater than 24 */
        format(ss % 3600 / 60, '00:'),
        format(ss % 60, '00')
    )
from avg_s

较旧的日期时间 hack 会让你做这样的事情，但在 Sql Server 的更高版本中，很多都被弃用了。

select cast(avg(
        cast(cast(endstartdt as datetime) -
        cast(callstartdt as datetime)
    as float)) as datetime)
from times

Answer 2

你可以在里面放一个 AVG():

SELECT 
    right('0' + convert(float,AVG(datediff (second, callstartdt,endstartdt))/ 60 % 60), 2) + ':' 
  + right('0' + convert(float,AVG(datediff (second, callstartdt,endstartdt))% 60),2)  as total_time
FROM times

演示：SQL Fiddle

如何获得 datediff 的平均时间？

How to get average time of datediff?

sql

sql-server

sql-server-2012