ticket_number | datetime | status
---------------+----------------------------+---------
1 | 2020-08-03 03:52:58.048196 | replied
1 | 2020-08-02 19:55:49.121455 | new
2 | 2020-07-30 03:52:58.048196 | pending
2 | 2020-07-28 20:15:41.213842 | replied
2 | 2020-07-26 03:52:58.482911 | new
3 | 2020-06-17 19:55:49.394628 | closed
3 | 2020-06-14 03:52:58.513141 | replied
3 | 2020-06-11 19:55:49.242859 | new
4 | 2020-05-14 07:13:50.527481 | new
5 | 2020-05-13 11:24:38.558921 | new
上表按datetime. 订购后，我需要分组，ticket_number以便每张票只出现一次，并选择status最新的datetime每张票。然后我需要计算每个status.
工单 1 的状态为已回复 - reply_count = 1，工
单 2 的状态为待处理 - pending_count = 1，
工单 3 的状态为已关闭 - closed_count = 1，工
单 4 的状态为 new - new_count = 1，
票 5 的状态为 mew - new_count = 2
预期结果：
replied_count | pending_count | closed_count | new_count
---------------+-----------------+----------------+------------
1 | 1 | 1 | 2