Question

计算一行中的连续（特定）匹配，但如果其他内容在中间，则从头开始。

我的想法是查看某人在一段时间内（在我的情况下为15分钟）内进行了多少次登录尝试并采取相应措施。例如（简化表）：

select count(Loginstatus) Result  from Userlog
where User_Id = 5
and Loginstatus = 'ATTEMPT'
and Timestamp > Systimestamp - Interval '15' Minute;

我的选择现在应该产生：2

到目前为止我得到了什么：

  File "C:\Users\X\AppData\Local\Programs\Python\Python35\lib\site-packages\pyassimp\helper.py", line 234, in search_library
    raise AssimpError("assimp library not found")
pyassimp.errors.AssimpError: assimp library not found

但不幸的是，这计算了所有的尝试。我很高兴得到任何帮助，并提前感谢你

Answer 1

您需要使用GROUP BY：

SELECT [LoginStatus]=MAX(Loginstatus), [Count]=count(Loginstatus) 
FROM Userlog
WHERE User_Id = 5
AND Loginstatus = 'ATTEMPT'
AND Timestamp > Systimestamp - Interval '15' Minute;
GROUP BY (Loginstatus)

Answer 2

您可以使用行号的差异来识别相邻序列：

select user_id, count(*) as num_consecutive
from (select ul.*,
             row_number() over (partition by user_id order by timestamp) as seqnum_u,
             row_number() over (partition by user_id, loginstatus order by timestamp) as seqnum_us

      from Userlog ul
      where User_Id = 5 and Timestamp > Systimestamp - Interval '15' Minute;
     ) ul
where Loginstatus = 'ATTEMPT'
group by user_id, (seqnum_u - seqnum_us);

如果您想要一个值，可以添加：

order by num_consecutive desc
fetch first 1 row only

计算数据库行中的匹配条目，除非该行为空

2 个答案: