Oracle: Select rows when value in one column changes - sql

Question

I have the following table:
PLACE USER_ID Date
---------- ---------- -----------------------------
ABC 4 14/04/20 12:05:29,255000000
ABC 4 14/04/20 15:42:28,389000000
ABC 4 14/04/20 18:33:20,202000000
ABC 4 14/04/20 22:51:28,339000000
XYZ 4 14/04/20 11:07:23,335000000
XYZ 2 14/04/20 12:15:12,123000000
ABC 4 13/04/20 22:09:33,255000000
QWE 4 13/04/20 10:18:29,144000000
XYZ 2 14/04/20 10:05:47,255000000
And I need to get the rows when the place changes order by date for the user_id that I select.
So the desired result should be this (for user_id 4):
PLACE USER_ID DATE
---------- ---------- -----------------------------
ABC 4 14/04/20 12:05:29,255000000
XYZ 4 14/04/20 11:07:23,335000000
ABC 4 13/04/20 22:09:33,255000000
QWE 4 13/04/20 10:18:29,144000000
I tried with min date but in my example, I lose data if the user goes back to that place:
SELECT MIN(DATE), PLACE FROM user_places WHERE USER_ID=4 GROUP BY PLACE
Result I get (missing one row):
PLACE USER_ID DATE
---------- ---------- -----------------------------
XYZ 4 14/04/20 11:07:23,335000000
ABC 4 13/04/20 22:09:33,255000000
QWE 4 13/04/20 10:18:29,144000000
Thanks in advance!

Answer 1

In Oracle 12.1 and higher, gaps-and-islands problems like this one are an easy task for the match_recognize clause. For example:
Table setup
alter session set nls_timestamp_format = 'dd/mm/rr hh24:mi:ss,ff';
create table user_places (place, user_id, date_) as
select 'ABC', 4, to_timestamp('14/04/20 12:05:29,255000000') from dual union all
select 'ABC', 4, to_timestamp('14/04/20 15:42:28,389000000') from dual union all
select 'ABC', 4, to_timestamp('14/04/20 18:33:20,202000000') from dual union all
select 'ABC', 4, to_timestamp('14/04/20 22:51:28,339000000') from dual union all
select 'XYZ', 4, to_timestamp('14/04/20 11:07:23,335000000') from dual union all
select 'XYZ', 2, to_timestamp('14/04/20 12:15:12,123000000') from dual union all
select 'ABC', 4, to_timestamp('13/04/20 22:09:33,255000000') from dual union all
select 'QWE', 4, to_timestamp('13/04/20 10:18:29,144000000') from dual union all
select 'XYZ', 2, to_timestamp('14/04/20 10:05:47,255000000') from dual
;
commit;
Query and output
select place, user_id, date_
from (select * from user_places where user_id = 4)
match_recognize (
order by date_
all rows per match
pattern (a {- b* -} )
define b as place = a.place
)
order by date_ desc -- if needed
;
PLACE USER_ID DATE_
----- ------- ---------------------------
ABC 4 14/04/20 12:05:29,255000000
XYZ 4 14/04/20 11:07:23,335000000
ABC 4 13/04/20 22:09:33,255000000
QWE 4 13/04/20 10:18:29,144000000
A few things to note here:
DATE is a reserved keyword. Not a good column name. I used DATE_
instead; notice the trailing underscore.
I hardcoded the value 4. Of course, the better practice is to make that into a bind variable.
If you really only need to do this for one user_id at a time, it is most efficient to do what I did - filter the rows first, in a subquery. However, if you need to do this for all user id's in the same query, you don't need a subquery; you select from the table itself, and you need to add partition by user_id right at the top of the match_recognize clause, before order by date_.

Answer 2

You can use lag() in a subquery to retrieve the "previous" place, and then filter on rows where the previous place is different that the current place:
select place, user_id, date
from (
select t.*, lag(place) over(partition by user_id order by date) lag_place
from mytable t
) t
where lag_place is null or place <> lag_place
This gives you the expected output for all users. If you want only for user 4, then you can filter in the subquery (and there is no need to partition by user):
select place, user_id, date
from (
select t.*, lag(place) over(order by date) lag_place
from mytable t
where user_id = 4
) t
where lag_place is null or place <> lag_place