Total hours worked per employee per day including overnight sessions

I have a table like this:

CREATE TABLE Table1
(ID int, empid int, time datetime, state int);

+--------------+---------------------+-----------------+
|    empid     |       time          |      state      | 
+--------------+---------------------+-----------------+
(      4       | 2014-03-01 11:12:00 |         0       )  
(      5       | 2014-03-01 12:28:06 |         0       )
(      4       | 2014-03-01 12:50:07 |         1       )
(      4       | 2014-03-01 13:38:00 |         0       )
(      5       | 2014-03-01 13:28:06 |         1       )
(      4       | 2014-03-01 18:42:15 |         1       )
(      4       | 2014-03-02 08:11:08 |         0       ) 
(      4       | 2014-03-02 13:26:11 |         1       ) 
(      5       | 2014-03-02 14:16:15 |         0       ) 
(      4       | 2014-03-02 16:16:15 |         0       ) 
(      5       | 2014-03-02 17:48:21 |         1       ) 
(      4       | 2014-03-02 19:39:03 |         1       ) 
(      5       | 2014-03-02 20:16:15 |         0       )
(      5       | 2014-03-03 04:16:15 |         1       )
+--------------+---------------------+-----------------+

primary key omitted for brevity - please see http://www.sqlfiddle.com/#!9/30503/1

0 means login - 1 means logout. There can be several logins/logouts per user per day or period of days if for example an employee works overnight (as employee 5 does between 2014-03-02 and 2014-03-03 )

I am trying to get the total sum of hours worked per employee per day/shift (e.g. some shifts will span over two days):

1      2014-03-02   total hours worked    08:32:00

I came up with this query:

SELECT CONCAT(
MOD(TIMEDIFF(MAX(CASE WHEN state = '1' THEN time END),
MIN(CASE WHEN state =    '0' THEN time END)), 24 ), ' hours ',
MINUTE(TIMEDIFF(MAX(CASE WHEN state = '1' THEN time END),
MIN(CASE WHEN state    = '0' THEN time END))), ' minutes') as HoursWorked, 
empid, 
Date(time)
FROM emplog T
GROUP BY empid, Date(time)

Which only gave me the time e

Please check if the performance is sufficient:

SELECT empid
      ,SEC_TO_TIME(SUM(TIMESTAMPDIFF(SECOND, time, (SELECT IFNULL(MIN(time),NOW())
                                                      FROM emplog b
                                                     WHERE b.empid = a.empid
                                                       AND b.time  > a.time
                                                       AND b.state = 1
                                                    )))) date_worked
  FROM emplog a
 WHERE state=0
 GROUP BY empid;

Edit

To aggregate per day:

SELECT empid
      ,DATE(time) day
      ,SEC_TO_TIME(SUM(TIMESTAMPDIFF(SECOND, time, (SELECT IFNULL(MIN(time),NOW())
                                                      FROM emplog b
                                                     WHERE b.empid = a.empid
                                                       AND b.time  > a.time
                                                       AND b.state = 1
                                                    )))) date_worked
  FROM emplog a
 WHERE state=0
 GROUP BY empid, DATE(time);

StackExchange Database Administrators Q#130676, answer score: 5