Mysql 登录时间和注销时间之间的总时差

输入

输出：

 userid           login time                               logout time
----------------------------------------------------------------------------
   1          2012-08-01 16:08:26                     2012-08-01 16:08:29
   2          2012-08-01 16:22:49                     2012-08-01 16:25:44
   3          2012-08-01 16:08:26                     2012-08-01 16:08:29
   3          2012-08-01 16:22:49                     2012-08-01 16:25:44
   3          2012-08-01 16:08:26                     2012-08-01 16:08:29
   4          2012-08-01 16:22:49                     2012-08-01 16:25:44

我尝试过的查询是：

 userid        date         total time difference b/w login time and logout time
---------------------------------------------------------------------------------
   1        2012-08-01                         00:08:29
   2        2012-08-01                         1:25:44
   3        2012-08-01                         00;55;5
   4        2012-08-01                         1:25:44

---

我认为您只需要将函数用作：

SELECT distinct t.user_id, DATE_FORMAT(t.login_time,\'%d %b %Y\') AS datez,
       SEC_TO_TIME(SUM(TIME_TO_SEC(t.logout_time) - TIME_TO_SEC(t.login_time))) AS timediffe
from login_log t 
where user_id=5 
  AND login_time between '2012-08-01' AND '2012-08-2' 
GROUP BY t.user_id,datez

---

我认为您只需要将函数用作：

SELECT distinct t.user_id, DATE_FORMAT(t.login_time,\'%d %b %Y\') AS datez,
       SEC_TO_TIME(SUM(TIME_TO_SEC(t.logout_time) - TIME_TO_SEC(t.login_time))) AS timediffe
from login_log t 
where user_id=5 
  AND login_time between '2012-08-01' AND '2012-08-2' 
GROUP BY t.user_id,datez

---

请使用此方法查找时间差异：

SELECT t.user_id,
       DATE_FORMAT(t.login_time,'%d %b %Y') AS datez,
       SEC_TO_TIME(SUM(TIME_TO_SEC(TIMEDIFF(t.logout_time, t.login_time)))) AS timediffe
FROM tic_login_log t
WHERE user_id = 5
      AND login_time between '2012-08-01' AND '2012-08-2'
GROUP BY t.user_id,datez;

---

请使用此方法查找时间差异：

SELECT t.user_id,
       DATE_FORMAT(t.login_time,'%d %b %Y') AS datez,
       SEC_TO_TIME(SUM(TIME_TO_SEC(TIMEDIFF(t.logout_time, t.login_time)))) AS timediffe
FROM tic_login_log t
WHERE user_id = 5
      AND login_time between '2012-08-01' AND '2012-08-2'
GROUP BY t.user_id,datez;

您可以使用

TIMEDIFF（）

方法

试试这个：

select TIMEDIFF(NOW(), '2012-06-20 06:24:43')

例如：-

SELECT timediff(your_logout_time,your_login_time);

输出：-

'24:00:03'

您可以使用

TIMEDIFF（）

方法

试试这个：

select TIMEDIFF(NOW(), '2012-06-20 06:24:43')

例如：-

SELECT timediff(your_logout_time,your_login_time);

---

输出：-

'24:00:03'

根据“2012-08-01”的相关要求输出

SELECT timediff('2012-08-02 16:08:29','2012-08-01 16:08:26');

每个用户的日期vise总登录时间

select userid, date(login_time) as Date, sum(timediff(logout_time-login_time) as
'Logged in Time' where date(login_time)='2012-08-01' group by userid;

select userid, date(login_time) as Date, sum(timediff(logout_time-login_time) as
'Logged in Time'    group by userid,date(login_time);

每个用户在历史记录中的总登录时间

select userid, date(login_time) as Date, sum(timediff(logout_time-login_time) as
'Logged in Time' where date(login_time)='2012-08-01' group by userid;

select userid, date(login_time) as Date, sum(timediff(logout_time-login_time) as
'Logged in Time'    group by userid,date(login_time);

---

根据“2012-08-01”的相关要求输出

SELECT timediff('2012-08-02 16:08:29','2012-08-01 16:08:26');

每个用户的日期vise总登录时间

select userid, date(login_time) as Date, sum(timediff(logout_time-login_time) as
'Logged in Time' where date(login_time)='2012-08-01' group by userid;

select userid, date(login_time) as Date, sum(timediff(logout_time-login_time) as
'Logged in Time'    group by userid,date(login_time);

每个用户在历史记录中的总登录时间

select userid, date(login_time) as Date, sum(timediff(logout_time-login_time) as
'Logged in Time' where date(login_time)='2012-08-01' group by userid;

select userid, date(login_time) as Date, sum(timediff(logout_time-login_time) as
'Logged in Time'    group by userid,date(login_time);

---

但是val将采用整数格式，而不是时间格式

SEC\u TO\u time

再次将其转换为

time

格式。只需尝试此查询，应该对您有效。但是val将以整数格式出现，而不是时间格式

SEC\u TO_time

再次将其转换为

time

格式。试试这个查询，应该对你有用。问题是每个用户登录了多少时间？对吧，伙计？问题是1。每个用户在特定日期登录的时间是多少？或2。每个用户登录date vise的时间有多长？或3。每个用户登录的总时间是多少？现在我认为最好的答案是回答这三个问题。问题是每个用户登录的时间是多少？对吧，伙计？问题是1。每个用户在特定日期登录的时间是多少？或2。每个用户登录date vise的时间有多长？或3。每个用户登录的总时间是多少？现在我认为最好的答案是回答这三个问题。