mysql - Mysql中时间戳的总和

Question

I'm trying to get the average of Completed time, so at first, I'm getting the completed time of individual processOrders.

下面的查询将获取单个 processOrders 的完成时间。

select processorder_prod_number,ifnull(time(round( timediff(max(completed_time), 
min(start_time)))) ,'00:00:00') as `completedtime`
from processorder_prod where week(start_time)=week(curdate())
group by processorder_prod_number;

输出是这样的

+--------------------+----------------+
| processorder_number| completedtime  | 
+--------------------+----------------+
|   1002135897       |  01:42:15      |
|   1002135898       |  01:39:43      |
|   1002135900       |  05:31:52      |
|   1002135901       |  02:31:52      |
+--------------------+----------------+

当我尝试获取总时间，即 sum(completedtime) 时，它得到了错误的输出。

select count(distinct processorder_prod_number),time(round(sum(completedtime))) as totaltime from
(select processorder_prod_number,ifnull(time(round( timediff(max(completed_time), 
min(start_time)))) ,'00:00:00') as `completedtime`
from processorder_prod where week(start_time)=week(curdate())
group by processorder_prod_number) as t;

上述查询的输出

+---------------------------+----------------+
| count(processorder_number)| completedtime  | 
+---------------------------+----------------+
|   4                       |   01:42:15     |
+---------------------------+----------------+

请在我做错的地方帮助我。欢迎任何建议和更正，在此先感谢。

Answer 1

在这里，您将获得已完成时间的总和。

SELECT SEC_TO_TIME( SUM( TIME_TO_SEC( `time` ) ) ) AS `completedtime` FROM dual;

更新的答案

 SELECT COUNT(DISTINCT processorder_prod_number), TIME(ROUND(SUM(completedtime))) AS  totaltime FROM (SELECT processorder_prod_number, IFNULL(SEC_TO_TIME( SUM( TIME_TO_SEC( `completed_time` ) ) ),'00:00:00') AS  `completedtime`  FROM processorder_prod WHERE WEEK(start_time)=WEEK(CURDATE())  GROUP BY processorder_prod_number) AS t;