MySQL lag / lead function?

Question

I want to calculate the difference between user start time and stop time from a previous session.

The table contains: user number, start / stop unix timestamp. It is sorted after user start and start.

user  start   stop
1    150000 150010
1    160011 160050
1    160100 160200
2    150011 150023
2    155001 155055
2    160001 160500
3    159001 159550

I expect the result to look like this:

user start   stop  diff
1   150000 150010  160011-150010=10001
1   160011 160050  160100-160050
1   160100 160200  0 (new user on next row)
2   150011 150023  155001-150023
2   155001 155055  160001-155055
2   160001 160500  0 (new user on next row)
3   159001 159550  and so on..

Can this be done, if so, how?

Answer 1

This is best done at the application level, but just for fun, here it is at the database level:

select `user`, `start`, `stop`, diff from (
    select
    t.*
    , if(@prev_user = `user`, (`stop` - @prev) * -1, 0) as diff
    , @prev := `start`
    , @prev_user := `user`
    from
    t
    , (select @prev := null, @prev_user := null) var_init
    order by `user`, `start` desc
) sq
order by `user`, `start`

• See how it works in sqlfiddle mode

, MySQL /. , , . SELECT . , SELECT , , " " SELECT.
, ORDER BY . . , ORDER BY.

EDIT:

UPDATE inactivitytmp
JOIN (
SELECT
inactivitytmp.*
, if(@prev_user_id = `user_id`, (`end_ts` - @prev) * -1, 0) as diff2
, @prev := `start_ts`
, @prev_user_id := `user_id`
FROM
inactivitytmp
, (SELECT @prev := null, @prev_user_id := null) var_init
ORDER BY `user_id`, `start_ts` DESC
) query_alias
ON inactivitytmp.user_id=query_alias.user_id AND inactivitytmp.start_ts=q uery_alias.start_ts AND inactivitytmp.end_ts=query_alias.end_ts
SET inactivitytmp.diff=query_alias.diff2;