如何在SQL Query中获取此逻辑？

Question

我有两个表来跟踪用户活动：

我需要编写一个SQL查询来获取格式为

的数据

： 用户ID |的UserRole | LoginTime | LogOutTime

注意：LoginTime / LogOutTime是ActivityAuditTrail表中的Activity_Time_Stamp。我无法弄清楚这个逻辑。

Answer 1

;with LoginTime AS(
SELECT User_Id, Activity_Time_Stamp 
  FROM Activity_Audit_Trail
 WHERE Activity_Id = 01
),LogOutTime AS(
SELECT ROW_NUMBER() OVER(PARTITION BY a.User_Id ORDER BY Sa.User_Id, a.Activity_Time_Stamp DESC) AS RowNumber
       a.User_Id, a.Activity_Time_Stamp AS LogOutTimeValue, li.Activity_Time_Stamp AS LoginTimeValue
  FROM Activity_Audit_Trail a
  JOIN LoginTime li
    ON li.User_Id =  a.User_Id
 WHERE Activity_Id = 02
   AND a.Activity_Time_Stamp > li.Activity_Time_Stamp
     ) 
  Select User_Id
  ,User_Role
  ,LoginTimeValue
  ,LogOutTimeValue
FROM LogOutTime
WHERE RowNumber = 1

请试试这个..因为我没有表格的架构脚本，没有经过测试。您可能会遇到一些语法错误。请遵循逻辑..

Answer 2

尝试对ActivityAuditTrail用户，角色和会话令牌中的行进行分组，然后分别选择最低和最高活动时间戳作为登录和注销时间戳：

SELECT User_Id, User_Role, MIN(Activity_Time_Stamp) AS LoginTime, CASE COUNT(*) WHEN 2 THEN MAX(Activity_Time_Stamp) ELSE NULL END AS LogOutTime
FROM ActivityAuditTrail
GROUP BY User_Id, User_Role, User_Session

此查询将为您提供每个用户会话的登录和注销时间戳。

Answer 3

我认为你不需要你的活动表。 如果用户仅登录和退出系统一次，此查询将为您提供所需的信息：

Select AT.UserId. AT.UserRole, MIN(Activity_Time_Stamp) as LoginTime, MAX(Activity_Time_Stamp) as LogoutTime
from ActivityAuditTrail AT
group by AT.UserId. AT.UserRole

当然这还不够，所以你需要一种方法来通过注销对登录进行分组。 我假设这是Activity_ID字段的用途。因此，每个活动都由几个活动审计跟踪组成，其中第一个是登录，最后一个是注销 在这种情况下，你可以运行：

Select AT.UserId. AT.UserRole, AT.Activity_ID, MIN(Activity_Time_Stamp) as LoginTime, MAX(Activity_Time_Stamp) as LogoutTime
from ActivityAuditTrail AT
group by AT.UserId. AT.UserRole, AT.Activity_ID

编辑：

select distinct AT.UserId, AT.UserRole,
(select Activity_Time_Stamp from ActivityAuditTrail where UserId=AT.UserId and Activity_ID=1) as LoginTime,
(select Activity_Time_Stamp from ActivityAuditTrail where UserId=AT.UserId and Activity_ID=2) as LogoutTime
from ActivityAuditTrail AT

Answer 4

SELECT User_Id, User_Role, Activity_Time_Stamp AS LoginTime, 
( SELECT Activity_Time_Stamp 
  FROM ActivityAuditTrail aud2
  WHERE aud2.Activity_Id=2 
  AND ActivityAuditTrail.User_Id=aud2.User_Id 
  AND ActivityAuditTrail.User_Session=aud2.User_Session 
) AS LogOutTime
FROM ActivityAuditTrail
WHERE Activity_Id=1

我是安全的，并假设会话不是用户唯一的，但如果他们是，你可以完全从WHERE中省略user_id。

Answer 5

SELECT b.UserId, b.User_Role MIN(b.Activity_Time_Stamp) AS LoginTime, MAX(b.Activity_Time_Stamp) AS LogoutTime  
FROM `Activity` a 
  INNER JOIN `Activity Audit Trail` b ON a.Id = b.Activity_Id 
GROUP BY b.User_Id

Answer 6

如何识别整个会话，我的意思是，用户可能已多次登录并注销，如何知道哪个LogOut匹配哪个登录。