Question

我在TIMESTAMP（6）WITH TIME ZONE类型的数据库中有2列。我从另一个中减去一个以获得两个时间戳之间的时间。

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

如何获得我所拥有的时差列表的平均值？

以下是时差的一小部分样本：

+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：我也应该注意到 - 我已经尝试了AVG功能，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：只是为了澄清上面的代码片段。第3行是我的SQL查询全部在一行上，格式如下：

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

EDIT3：非常感谢Matt和Alex Poole。您已经大量帮助了，我感谢您花时间来帮助解决这个问题，并且一致地回复最新的帮助以回应反馈/进一步的问题！谢谢你们！

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importdate和lastprocesseddate的+1时区的数据库上是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

您可以从每个间隙值中提取时间分量，这是一种间隔数据类型，因此您最终得到一个以秒为单位的数字（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提供显示的间隔值的演示：

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

或者如果你想要它作为间隔：

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
...

给出了

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

SQL Fiddle with answer in seconds。（它似乎不喜欢显示间隔，因此无法演示）。

Answer 3

此查询应解决此问题。

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;

使用TIME ZONE时间平均TIMESTAMP（6）列表

3 个答案: