如果两排相隔20分钟，你似乎对它们都感兴趣。以下尝试模拟超前和滞后：

SELECT * FROM (
    SELECT *, (
        SELECT CASE WHEN EVENT = 'LOGOUT' AND main.Event = 'LOGIN' THEN TimeDIFF(main.Time, Time) END
        FROM t AS prev
        WHERE User = main.User AND Time < main.Time
        ORDER BY Time DESC
        LIMIT 1
    ) AS diff_lag, (
        SELECT CASE WHEN EVENT = 'LOGIN' AND main.Event = 'LOGOUT' THEN TimeDIFF(Time, main.Time) END
        FROM t AS next
        WHERE User = main.User AND Time > main.Time
        ORDER BY Time ASC
        LIMIT 1
    ) AS diff_lead
    FROM t AS main
) x
WHERE diff_lag > '00:20:00' OR diff_lead > '00:20:00'

或者，尝试以下方法，除了将两行放在一起之外，其他方法似乎有效：

SELECT *
FROM t AS o
INNER JOIN t AS i ON o.User = i.User AND o.Time < i.Time                    -- join logouts with potential logins
LEFT JOIN t AS x ON o.User = x.User AND o.Time < x.Time AND x.Time < i.Time -- any row present between logout and login
WHERE o.Event = 'LOGOUT' AND i.Event = 'LOGIN' AND x.ID IS NULL AND TIMEDIFF(i.Time, o.Time) > '00:20:00'

这回答了一个问题：
我需要选择所有用户记录谁没有重新登录后20分钟内注销。

select lo.*
from (select l.*,
             (select min(l2.time)
              from logs l2
              where l2.user = l.user and l2.time > l.time and
                    l2.event = 'LOGIN'
             ) as next_login_time
      from logs l
      where l.event  = 'LOGOUT'
     ) lo
where next_login_time > time + interval 20 minute;

您的示例结果包括登录结果。根据你提出的问题，目前还不清楚这些是如何定义的。