平均TIMESTAMP（6）与TIME ZONE次

Question

我在类型TIMESTAMP（6）与时区数据库有2列的列表。我已经从另一个中减去一个来获得两个时间戳之间的时间。

select lastprocesseddate-importeddate 
from feedqueueitems 
where eventid = 2213283 
order by written desc; 

我怎样才能得到我有的时间差异列表的平均值？

这里有时间差的小样本：

+00 00:00:00.488871  
+00 00:00:00.464286 
+00 00:00:00.477107 
+00 00:00:00.507042 
+00 00:00:00.369144 
+00 00:00:00.488918 
+00 00:00:00.354797 
+00 00:00:00.378801 
+00 00:00:00.320040 
+00 00:00:00.361242 
+00 00:00:00.302327 
+00 00:00:00.331441 
+00 00:00:00.324065 

编辑：我也注意到 - 我已经尝试AVG函数，它只是返回

ORA-00932: inconsistent datatypes: expected NUMBER got INTERVAL DAY TO SECOND 
00932. 00000 - "inconsistent datatypes: expected %s got %s" 
*Cause:  
*Action: 
Error at Line: 3 Column: 29 

EDIT2：只是澄清上述片段。第3行是我的SQL查询，其格式如下：

select AVG(lastprocesseddate-importeddate) from feedqueueitems where eventid = 2213283; 

EDIT3：非常感谢Matt和Alex Poole。你俩都有助于大规模和我感谢您抽出宝贵的时间来帮助这一点，双方始终与更新的帮助响应反馈/其他问题包退！多谢你们！

Answer 1

使用AVG功能

SELECT avg(cast(lastprocesseddate as date)-cast(importeddate as date)) 
FROM feedqueueitems 
WHERE eventid = 2213283 
ORDER BY written DESC; 

在数据库与+1时区为importeddate和lastprocesseddate是UTC

SELECT avg(cast(cast(lastprocesseddate as timestamp with time zone) at time zone '+01:00' as date)-cast(importeddate as date)) 
FROM feedqueueitems 
WHERE eventid = 2213283 
ORDER BY written DESC; 

Answer 2

你可以提取每一个间隙值，这是一个时间的组件间隔数据类型，所以你最终以秒的图（包括小数部分），然后平均那些：

select avg(extract(second from gap) 
    + extract(minute from gap) * 60 
    + extract(hour from gap) * 60 * 60 
    + extract(day from gap) * 60 * 60 * 24) as avg_gap 
from (
    select lastprocesseddate-importeddate as gap 
    from feedqueueitems 
    where eventid = 2213283 
); 

使用CTE提供间隔值的演示你表明：

with cte as (
    select interval '+00 00:00:00.488871' day to second as gap from dual 
    union all select interval '+00 00:00:00.464286' day to second from dual 
    union all select interval '+00 00:00:00.477107' day to second from dual 
    union all select interval '+00 00:00:00.507042' day to second from dual 
    union all select interval '+00 00:00:00.369144' day to second from dual 
    union all select interval '+00 00:00:00.488918' day to second from dual 
    union all select interval '+00 00:00:00.354797' day to second from dual 
    union all select interval '+00 00:00:00.378801' day to second from dual 
    union all select interval '+00 00:00:00.320040' day to second from dual 
    union all select interval '+00 00:00:00.361242' day to second from dual 
    union all select interval '+00 00:00:00.302327' day to second from dual 
    union all select interval '+00 00:00:00.331441' day to second from dual 
    union all select interval '+00 00:00:00.324065' day to second from dual 
) 
select avg(extract(second from gap) 
    + extract(minute from gap) * 60 
    + extract(hour from gap) * 60 * 60 
    + extract(day from gap) * 60 * 60 * 24) as avg_gap 
from cte; 

    AVG_GAP 
---------- 
.397544692 

或者，如果你想让它作为间隔：

select numtodsinterval(avg(extract(second from gap) 
    + extract(minute from gap) * 60 
    + extract(hour from gap) * 60 * 60 
    + extract(day from gap) * 60 * 60 * 24), 'SECOND') as avg_gap 
... 

这给

AVG_GAP    
-------------------- 
0 0:0:0.397544692 

SQL Fiddle with answer in seconds。 （它似乎不喜欢显示时间间隔，所以不能演示）。

Answer 3

这个查询应该可以解决这个问题。

WITH t AS 
    (SELECT 
     TIMESTAMP '2015-04-23 12:00:00.5 +02:00' AS lastprocesseddate, 
     TIMESTAMP '2015-04-23 12:05:10.21 UTC' AS importeddate 
    FROM dual) 
SELECT 
    AVG(
     EXTRACT(SECOND FROM SYS_EXTRACT_UTC(lastprocesseddate) - SYS_EXTRACT_UTC(importeddate)) 
     + EXTRACT(MINUTE FROM SYS_EXTRACT_UTC(lastprocesseddate) - SYS_EXTRACT_UTC(importeddate)) * 60 
     + EXTRACT(HOUR FROM SYS_EXTRACT_UTC(lastprocesseddate) - SYS_EXTRACT_UTC(importeddate)) * 60 * 60 
     + EXTRACT(DAY FROM SYS_EXTRACT_UTC(lastprocesseddate) - SYS_EXTRACT_UTC(importeddate)) * 60 * 60 * 24 
    ) AS average_gap 
FROM t;