Question

我正在使用票务系统，并且需要编写一个查询来检查代理商是否在交付时迟到了。

这就是我现在所拥有的以及我需要的东西：

我有：ticket_id time_created
我需要选择所有相同的ticket_id，其中两个之间的time_created是大于24小时。

例如：

ticket_id time_created 
1         2013-08-19 12:11
1         2013-08-17 12:07
1         2013-08-17 12:00
2         2013-08-19 12:11
2         2013-08-19 12:10

结果

ticket_id time_difference
1             48:04

我只需要检查具有相同ticket_id号码的票证。

EDIT 让我们说也有和id列我需要1,2和2,3之间的检查值（在这种情况下 - 所有ticket_id = 1）和4,5（ticket_id = 2）

Answer 1

像这样的东西可以解决这个问题：

select ticket_id, min(a.time_created), max(b.time_created),
       TIMESTAMPDIFF(HOUR,min(a.time_created), max(b.time_created))
  from tbl AS a JOIN tbl AS b USING(ticket_id)
  where a.time_created < (b.time_created - INTERVAL 1 DAY)
  group by(a.ticket_id);

产：

+------------+--------------------------------+--------------------------------+--------------------------------------------------------------+
| TICKET_ID  |      MIN(A.TIME_CREATED)       |      MAX(B.TIME_CREATED)       | TIMESTAMPDIFF(HOUR,MIN(A.TIME_CREATED), MAX(B.TIME_CREATED)) |
+------------+--------------------------------+--------------------------------+--------------------------------------------------------------+
|         1  | August, 17 2013 12:00:00+0000  | August, 19 2013 12:11:00+0000  |                                                           48 |
+------------+--------------------------------+--------------------------------+--------------------------------------------------------------+

请参阅http://sqlfiddle.com/#!2/4af76e/7

请注意，我在此处使用了TIMESTAMPDIFF，请求HOUR中的结果。由于TIMEDIFF值的范围有限（≈34天），我在此处未使用TIME。

Answer 2

为了获得差异，您需要相当于MySQL不支持的lag()函数。您可以使用相关子查询获得相同的效果：

select ticket_id, timediff(nextTimeCreated, time_created)
from (select t.*,
             (select time_created
              from t t2
              where t2.ticket_id = t.ticket_id and
                    t2.time_created > t.time_created
             ) as nextTimeCreated
      from t
     ) t
where nextTimeCreated > time_created + interval 1 day;

Answer 3

此查询会检查上次日期和最低日期之间的差异，您希望这个或最后一个日期和第二个日期之间的差异

SELECT ticket_id, 
DATEDIFF(MAX(TIME_CREATED),MIN(TIME_CREATED)) AS DiffDate 
FROM TABLE1 
    group by ticket_id

此查询应查找两张最新票证之间的差异

SELECT ticket_id, TIMEDIFF(MAX(A.TIME_CREATED),

             (SELECT MAX(TIME_CREATED) FROM TICKETS B 
              where B.TIME_CREATED <MAX(A.TIME_CREATED)))  AS DiffDate 
    FROM TICKETS A
        group by ticket_id

想想看，我现在认为这可能是

的重复

Difference between dates in two consecutive rows

你可以在那里试试......

SELECT ticketid，extract（来自tdiff的epoch）FROM（选择 ticketid， ticketdate - lag（ticketdate）OVER（按照ticketid ORDER BY ticketdate）AS tdiff， dense_rank（）OVER（PARTITION BY ticketid ORDER BY ticketdate）AS等级从表1 按顺序排列）x 等级= 2;

并在

处小提琴

http://sqlfiddle.com/#!1/0e213/1

同一个ID票务系统之间的时间

3 个答案: