Finding all rows based on previous rows column value diffrence

Question

I have an interesting requirement. There is a DATE column and a user in a table and I have to find all rows for all the user for which the previous row and current row has time difference of lets say more than 30 minutes. Rows are already sorted by time.
For example in the following table, we need ID 4 and 6 for user 1.
ID User date
1 1 today 1 hour 0 min 0 s
2 1 today 1 hour 1 min 0 s
3 1 today 1 hour 29 min 0 s
*4 1 today 1 hour 59 min 3 s*
5 1 today 2 hour 10 min 2 s
*6 1 today 2 hour 50 min 7 s*
Edited by: user733179 on Mar 5, 2009 12:00 PM

Answer 1

Hi,
To get all the users in the result set, outer-join to a table (or subquery, as shown bleow) that has one row per user:
WITH     got_dif     AS
     SELECT     id,     dt,     usr
     ,     (dt - LAG (dt)
                OVER ( PARTITION BY  usr
                          ORDER BY          dt
          )             -- difference in days
              * 24 * 60  AS minutes_dif
     FROM     table_x
--     WHERE     ...     -- if needed
,     all_users     AS
     SELECT DISTINCT  usr
     FROM              table_x
SELECT     gd.id,     gd.dt,     au.usr
,     NVL2 ( gd.usr
          , 'Okay'
          , 'Dif is never > 30'
          )    AS flag
FROM          all_users
LEFT OUTER JOIN     got_dif          ON  gd.usr         = au.usr
                            AND gd.minutes_dif > 30
;Notice that the test for a 30-minute difference is part of the join condition.
You don't need the flag column, and you may not want it.
If you would prefer a separate query of just the users that are not in the original result set, then you can use the original query (modified only slightly) as an EXISTS ior NOT IN sub-query. For example:
SELECT DISTINCT     usr
FROM    table_x
WHERE   usr  NOT IN
     WITH     got_dif     AS
          SELECT     id,     dt,     usr
          ,     (dt - LAG (dt)
                         OVER ( PARTITION BY  usr
                                  ORDER BY          dt
                )             -- difference in days
                    * 24 * 60  AS minutes_dif
           FROM     table_x
--          WHERE     ...     -- if needed
     SELECT  usr   -- Only 1 column in SELECT clause
     FROM     got_dif
     WHERE     minutes_dif     > 30
;