计算每个用户登录之间的时间

Question

I have a table that contains a unique ID, start time, end time and user ID, we'll call this table User_Logged_In where the start time is when the user is logged in and the end time is when they are logged out.

I'm trying to display the latest 5 calculated "time since last activity" times, in days/hours/minutes/seconds, grouped by individual users (user IDs) and ordered from most recent to oldest.

User A may have 3 logins, user B may have 12 logins, user C may have 8 logins. For each user, I'm looking for the time offline between each login. Ordered by user, then start time DESC, I'd expect the most recent record for each user wouldn't have a value.

Example:

+------+---------------------+---------------------+---------+
|  id  | starttime           | endtime             | user_id |
+------+---------------------+---------------------+---------+
|  35  | 2018-12-19 15:20:45 | 2018-12-19 19:21:06 | 1       |
|  22  | 2018-12-19 14:51:39 | 2018-12-19 21:56:40 | 1       |
|  11  | 2018-01-05 14:28:17 | 2018-01-05 21:30:42 | 1       |
|  97  | 2019-09-25 13:17:34 | 2019-09-25 23:31:31 | 2       |
|  86  | 2019-07-31 20:42:16 | 2019-07-31 22:42:38 | 2       |
|  72  | 2019-05-21 17:49:21 | 2019-05-21 20:03:03 | 2       |
|  83  | 2019-07-31 14:30:39 | 2019-07-31 20:30:58 | 3       |
|  79  | 2019-07-30 13:49:23 | 2019-07-30 23:59:01 | 3       |
|  61  | 2019-04-12 15:03:25 | 2019-04-13 01:54:23 | 3       |
|  54  | 2019-02-03 06:45:33 | 2019-02-03 17:33:48 | 3       |
|  51  | 2019-01-23 09:12:15 | 2019-01-29 14:44:38 | 3       |
+------+---------------------+---------------------+---------+

http://sqlfiddle.com/#!9/0dba46/2

End Result:

+----+---------------------+---------------------+-------------------------------------------+---------+
| id | starttime           | endtime             | difference                                | user_id |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 35 | 2018-12-19 15:20:45 | 2018-12-19 19:21:06 | -                                         | 1       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 22 | 2018-12-19 14:51:39 | 2018-12-19 21:56:40 | 0 Days, 6 Hours, 35 Minutes, 55 Seconds   | 1       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 11 | 2018-01-05 14:28:17 | 2018-01-05 21:30:42 | 347 Days, 17 Hours, 57 Minutes, 23 Seconds | 1       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 97 | 2019-09-25 13:17:34 | 2019-09-25 23:31:31 | -                                         | 2       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 86 | 2019-07-31 20:42:16 | 2019-07-31 22:42:38 | 55 Days, 14 Hours, 56 Minutes, 15 Seconds  | 2       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 72 | 2019-05-21 17:49:21 | 2019-05-21 20:03:03 | 71 Days, 0 Hours, 39 Minutes, 13 Seconds  | 2       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 83 | 2019-07-31 14:30:39 | 2019-07-31 20:30:58 | -                                         | 3       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 79 | 2019-07-30 13:49:23 | 2019-07-30 23:59:01 | 1 Days, 6 Hours, 41 Minutes, 35 Seconds   | 3       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 61 | 2019-04-12 15:03:25 | 2019-04-13 01:54:23 | 108 Days, 11 Hours, 55 Minutes, 0 Seconds | 3       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 54 | 2019-02-03 06:45:33 | 2019-02-03 17:33:48 | 67 Days, 21 Hours, 29 Minutes, 37 Seconds  | 3       |
+----+---------------------+---------------------+-------------------------------------------+---------+
| 51 | 2019-01-29 09:12:15 | 2019-01-23 14:44:38 | 10 Days, 16 Hours, 0 Minutes, 55 Seconds   | 3       |
+----+---------------------+---------------------+-------------------------------------------+---------+

This value will be a column in a larger set of data. I've tried a number of different things, including joining back to the same table, but can't find a way of returning this value quickly, or at all.

Currently on MySQL v5.6.

Answer 1

Having taken the code from your SQLFiddle link, I created an output that seems desired. It all relies on the functions TIMESTAMPDIFF , SEC_TO_TIME and TIME_FORMAT , you may not need all the resulting columns, I have selected the difference in seconds just to have an intermediate result for the desired output.
The following query might meet the requirements:

SELECT
  user_id AS User,
  starttime AS Start,
  endtime AS End,
  TIMESTAMPDIFF(SECOND, starttime, endtime) AS "Difference in seconds",
  CONCAT(
    FLOOR(TIME_FORMAT(SEC_TO_TIME(TIMESTAMPDIFF(SECOND, starttime, endtime)), '%H') / 24), ' days ',
      MOD(
        TIME_FORMAT(SEC_TO_TIME(TIMESTAMPDIFF(SECOND, starttime, endtime)), '%H'), 24), ' hours, ',
        TIME_FORMAT(SEC_TO_TIME(TIMESTAMPDIFF(SECOND, starttime, endtime)), '%i minutes and %s seconds'
      )
   ) AS "readable Difference"
FROM
  Table1

This will output the following in the mentioned fiddle-link:

User    Start                   End                     Difference in seconds   readable Difference
1       2018-12-19T15:20:45Z    2018-12-19T19:21:06Z    14421                   0 days 4 hours, 00 minutes and 21 seconds
1       2018-12-19T14:51:39Z    2018-12-19T21:56:40Z    25501                   0 days 7 hours, 05 minutes and 01 seconds
1       2018-01-05T14:28:17Z    2018-01-05T21:30:42Z    25345                   0 days 7 hours, 02 minutes and 25 seconds
2       2019-09-25T13:17:34Z    2019-09-25T23:31:31Z    36837                   0 days 10 hours, 13 minutes and 57 seconds
2       2019-07-31T20:42:16Z    2019-07-31T22:42:38Z    7222                    0 days 2 hours, 00 minutes and 22 seconds
2       2019-05-21T17:49:21Z    2019-05-21T20:03:03Z    8022                    0 days 2 hours, 13 minutes and 42 seconds
3       2019-07-31T14:30:39Z    2019-07-31T20:30:58Z    21619                   0 days 6 hours, 00 minutes and 19 seconds
3       2019-07-30T13:49:23Z    2019-07-30T23:59:01Z    36578                   0 days 10 hours, 09 minutes and 38 seconds
3       2019-04-12T15:03:25Z    2019-04-13T01:54:23Z    39058                   0 days 10 hours, 50 minutes and 58 seconds
3       2019-02-03T06:45:33Z    2019-02-03T17:33:48Z    38895                   0 days 10 hours, 48 minutes and 15 seconds
3       2019-01-29T09:12:15Z    2019-01-23T14:44:38Z    -498457                 -6 days -18 hours, -27 minutes and 37 seconds

As you can see, the last row of the results seems undesired due to invalid data.

Answer 2

the solution to your question is to use window functions. you need to compare the begin of each session with the end of each previous session which can be achieved using the lag or lead function which allow to access the preceeding or following row within a window.

to show only the last 5 logins of a user we use the row_number function to generate a row count for each window (the window spans across all records of the same userid) and use an outer select to filter it by values <= 5.

see the example below:

select *
from (
         select l.userid,
                l.starttime,
                l.endtime,
                datediff(starttime, lead(l.endtime, 1)
                                         over (partition by l.userid order by l.endtime desc)) date_difference,
                timediff(starttime, lead(l.endtime, 1)
                                         over (partition by l.userid order by l.endtime desc)) time_difference,
                row_number() over (partition by l.userid)                                      rn
         from User_Logged_In l
         order by l.userid desc, l.starttime desc
     ) x
where x.rn <= 5

this results in the following result:

the date_format and time_format functions can be used to format the dates and times according to your needs.

i used the following sql to create and fill the table:

drop table if exists User_Logged_In;
create table User_Logged_In
(
    id        int      not null unique,
    starttime datetime not null,
    endtime   datetime not null,
    userid    int      not null
);

insert into User_Logged_In(id, starttime, endtime, userid)
values (5, '2018-12-19 15:20:45', '2018-12-19 19:21:06', 1);
insert into User_Logged_In(id, starttime, endtime, userid)
values (22, '2018-12-19 14:51:39', '2018-12-19 21:56:40', 1);
insert into User_Logged_In(id, starttime, endtime, userid)
values (11, '2018-01-05 14:28:17', '2018-01-05 21:30:42', 1);
insert into User_Logged_In(id, starttime, endtime, userid)
values (97, '2019-09-25 13:17:34', '2019-09-25 23:31:31', 2);
insert into User_Logged_In(id, starttime, endtime, userid)
values (86, '2019-07-31 20:42:16', '2019-07-31 22:42:38', 2);
insert into User_Logged_In(id, starttime, endtime, userid)
values (72, '2019-05-21 17:49:21', '2019-05-21 20:03:03', 2);
insert into User_Logged_In(id, starttime, endtime, userid)
values (83, '2019-07-31 14:30:39', '2019-07-31 20:30:58', 3);
insert into User_Logged_In(id, starttime, endtime, userid)
values (79, '2019-07-30 13:49:23', '2019-07-30 23:59:01', 3);
insert into User_Logged_In(id, starttime, endtime, userid)
values (61, '2019-04-12 15:03:25', '2019-04-13 01:54:23', 3);
insert into User_Logged_In(id, starttime, endtime, userid)
values (54, '2019-02-03 06:45:33', '2019-02-03 17:33:48', 3);
insert into User_Logged_In(id, starttime, endtime, userid)
values (51, '2019-01-29 09:12:15', '2019-01-23 14:44:38', 3);