Question

我有一个表，其中包含以下示例之类的任务：

id user_id  project_id      done_at    duration type

1    13        15         2014-08-11   03:00:00  1
2    13        15         2014-08-11   04:00:00  1
10   13        15         2014-08-12   04:00:00  1
3    13        15         2014-08-12   08:00:00  2
5    13        13         2014-08-13   08:00:00  1
7    13        15         2014-08-14   04:00:00  1
8    13        15         2014-08-18   08:00:00  1
9    13        15         2014-08-19   08:00:00  2

如何按用户，项目和类型获取连续done_at列的开始/结束序列，如下所示：

 user_id project_id type start_done_at end_done_at   duration
  13        15       1     2014-08-11   2014-08-12   11:00:00
  13        15       2     2014-08-12   2014-08-12   08:00:00
  13        13       1     2014-08-13   2014-08-14   12:00:00
  13        15       1     2014-08-18   2014-08-18   08:00:00
  13        15       2     2014-08-19   2014-08-19   08:00:00

我已经尝试过以下示例：http://www.artfulsoftware.com/infotree/qrytip.php?id=76但是我没有做正确的持续时间。

我目前的疑问是：

SELECT task.user_id as user_id, task.done_at as start_done_at, MIN(task2.done_at) as end_done_at, task.project_id as project_id, task.timeoff_type_id as timeoff_type, SEC_TO_TIME(SUM(TIME_TO_SEC(task.duration))) as duration    
FROM tasks task 
LEFT JOIN tasks task1 ON task.done_at = task1.done_at + 1       
    AND task.user_id = task1.user_id 
    AND task.project_id = task1.project_id 
    AND task.timeoff_type_id = task1.timeoff_type_id 
LEFT JOIN tasks task2 ON task.done_at <= task2.done_at 
    AND task.user_id = task2.user_id 
    AND task.project_id = task2.project_id 
    AND task.timeoff_type_id = task2.timeoff_type_id
LEFT JOIN tasks task3 ON task2.done_at = task3.done_at -1 
    AND task2.user_id = task3.user_id 
    AND task2.project_id = task3.project_id  
    AND task2.timeoff_type_id = task3.timeoff_type_id       
WHERE task1.done_at IS NULL 
    AND task2.done_at IS NOT NULL 
    AND task3.done_at IS NULL 
GROUP BY task.user_id, task.project_id, task.timeoff_type_id, task.done_at

Answer 1

select 
user_id, project_id,
min(done_at) as start_done_at,
max(done_at) as end_done_at,
sec_to_time(sum(time_to_sec(duration))) as duration 
from (
    select
    t.*,
    @gn := if(timestampdiff(day, @prev_date, done_at) between 0 and 1 and @prev_type = type and @prev_project = project_id, @gn, @gn + 1) as group_number,
    @prev_date := done_at,
    @prev_type := type,
    @prev_project := project_id
    from
    t
    , (select @prev_project := null, @prev_date := null, @prev_type := null, @gn := 0) var_init
    order by project_id, done_at, duration
) sq
group by group_number

看到它在 sqlfiddle

连续日期按标准排序

1 个答案: