MYSQL每个问题在每个用户下的处理时长

题目二：数据统计。有一张任务处理日志表，记录了每个任务的流转日志，如系统在某个时点分配给A用户一个任务，A用户在某个时点流转到B用户，B用户在某个时点流转到处理完成节点。按问题统计出每个用户处理任务的时长。设计一张表，并且写出sql，计算出每个问题在每个用户下的处理时长。

Table_name : project_circulation

按项目归纳。

Select pro_id, user_id, sec_to_time(sum(timestampdiff(second, start_time, end_time))) as time from project_circulation group by user_id,pro_id;

创建Mysql数据表：

获取每个用户完成每个工程的结束时间：

(Select a.pro_id, a.user_id, a.start_time, b.start_time as end_time

from  project_circulation a, project_circulation b

where a.pro_id = b.pro_id and a.user_id = b.post_user_id and b.start_time > a.start_time  

and b.start_time = (select min(start_time) from project_circulation c where a.pro_id = c.pro_id and a.user_id = c.post_user_id and c.start_time > a.start_time)

)

运行结果图：

完整代码--获取每个用户完成每个工程所需的时间：

Select pro_id, user_id, sum(datediff(end_time, start_time)) as time

from (Select a.pro_id, a.user_id, a.start_time, b.start_time as end_time

from  project_circulation a, project_circulation b

where a.pro_id = b.pro_id and a.user_id = b.post_user_id and b.start_time > a.start_time and

b.start_time = (select min(start_time) from project_circulation c where a.pro_id = c.pro_id and a.user_id = c.post_user_id and c.start_time > a.start_time))as c

group by user_id,pro_id;

运行结果图：

补充知识点：

-- 获取多列最大最小值 

SELECT GREATEST(1,2,3,4);结果：4

SELECT LEAST(1,2,3,4);结果：1

-- 获取一列中最大最小值 

SELECT MAX(col1); 

SELECT MIN(col1);