Question

经过一些搜索，我能够使用此查询获得结果。

SELECT 
    DATEDIFF(HOUR, LOGINTIME, LOGOUTTIME) 
FROM 
    [SESSION]
GROUP BY 
    DATEPART(YEAR, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
    DATEPART(MONTH, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
    DATEPART(DAY, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
    DATEPART(HOUR, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
    (DATEPART(MINUTE, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)) / 10),
    LoginTime, LogoutTime

这是一个包含用户登录和注销时间会话的表。我想要的是计算有多少用户有一个小时两个或更多。我想得到点数。

此查询的示例结果是。

Answer 1

尝试将其添加到子查询中并对结果进行分组和过滤，如下所示：

SELECT Hours, COUNT(T1.Hours) HourCount
FROM 
    (SELECT DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME) as Hours
    FROM [SESSION]
    GROUP BY 
        DATEPART(YEAR, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
        DATEPART(MONTH, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
        DATEPART(DAY, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
        DATEPART(HOUR, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
        (DATEPART(MINUTE, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)) / 10),
        LoginTime,LogoutTime) T1
GROUP BY T1.Hours
HAVING T1.Hours > 1

这将按Hours分割和计数，大于1.

Answer 2

试试这个：

SELECT MAX(Total) AS Total
FROM
    (SELECT 
        ROW_MUMBER() OVER (ORDER BY DATEDIFF(HOUR, LOGINTIME, LOGOUTTIME)) AS Total 
    FROM 
        [SESSION]
    WHERE
        DATEDIFF(HOUR, LOGINTIME, LOGOUTTIME) > 1
    GROUP BY 
        DATEPART(YEAR, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
        DATEPART(MONTH, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
        DATEPART(DAY, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
        DATEPART(HOUR, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)),
        (DATEPART(MINUTE, DATEDIFF(HOUR,LOGINTIME,LOGOUTTIME)) / 10),
        LoginTime, LogoutTime
) AS T

Answer 3

在添加＆＃34; SessionDuration＆＃34;后，我必须使用此查询。作为计算列，使用＆＃34; LogoutTime-LoginTime＆＃34;。

SELECT COUNT(DATEPART(HOUR,SessionDuration)) AS [Count],
       DATEPART(HOUR,SessionDuration) AS SessionDuration 
FROM [SESSION]
GROUP BY DATEPART(HOUR,SessionDuration)

SQL Server组按持续时间

3 个答案: