SQL- how to retrieve by similar dates - sql

Question

Okay, so I have a table with a user_id column and a submitted_dtm column.
I want to find instances where users submitted multiple records within 1 day of each other, and count how many times that has happened.
I've tried something like
select * from table_t t where
(select count(*) from table_t t2 where
t.user_id = t2.user_id and
t.pk!=t2.pk and
t.submitted_dtm between t2.submitted_dtm-.5 and t2.submitted_dtm+.5)>0;
The problem is that this query returns a result for each record in a date group. Instead, I just want a result per date group. Ideally, I'd just get the count in that group.
That is, if I have 6 records:
user_id submitted_dtm
--------------------------
1 12/04/2017 1:15
1 12/04/2017 5:50
2 11/25/2017 2:00
2 11/25/2017 3:25
2 11/25/2017 6:05
2 10/06/2017 4:00
I want 2 results, a count of 2 and a count of 3.
Is it possible to do this in sql?

Answer 1

Following up on Dessma's answer.
select user_id, trunc(submitted_dtm), count(1)
from table_t
group by user_id, trunc(submitted_dtm)
having count(1) > 1;
Sqlfiddle

Answer 2

In Oracle 12.1 and higher, you can solve such problems easily with the match_recognize clause. Link to documentation (with examples) below; my only note about the solution below is that I left the date in DATE data type, especially important if the output is used in further computations. If it isn't, you can wrap within TO_CHAR() with whatever format model is appropriate for your users.
https://docs.oracle.com/database/121/DWHSG/pattern.htm#DWHSG8956
with
inputs ( user_id, submitted_dtm ) as (
select 1, to_date('12/04/2017 1:15', 'mm/dd/yyyy hh24:mi') from dual union all
select 1, to_date('12/04/2017 5:50', 'mm/dd/yyyy hh24:mi') from dual union all
select 2, to_date('11/25/2017 2:00', 'mm/dd/yyyy hh24:mi') from dual union all
select 2, to_date('11/25/2017 3:25', 'mm/dd/yyyy hh24:mi') from dual union all
select 2, to_date('11/25/2017 6:05', 'mm/dd/yyyy hh24:mi') from dual union all
select 2, to_date('10/06/2017 4:00', 'mm/dd/yyyy hh24:mi') from dual
)
-- End of simulated inputs (for testing only, not part of the solution).
-- SQL query begins below this line. Use your actual table and column names.
select user_id, submitted_dtm, cnt
from inputs
match_recognize(
partition by user_id
order by submitted_dtm
measures trunc(a.submitted_dtm) as submitted_dtm,
count(*) as cnt
pattern ( a b+ )
define b as trunc(submitted_dtm) = trunc(a.submitted_dtm)
);
USER_ID SUBMITTED_DTM CNT
---------- ------------------- ----------
1 2017-12-04 00:00:00 2
2 2017-11-25 00:00:00 3

Answer 3

I don't have data to test it but I suspect something like this would do the trick :
SELECT user_id,To_char(t.submitted_dtm, 'dd/mm/yyyy'), COUNT(*)
FROM table_t t
INNER JOIN table_t t2
ON t.user_id = t2.user_id
AND t.pk != t2.pk
AND t.submitted_dtm BETWEEN t2.submitted_dtm - .5 AND
t2.submitted_dtm + .5
GROUP BY user_id,To_char(t.submitted_dtm, 'dd/mm/yyyy')
HAVING COUNT(*) > 1

Answer 4

This is a general idea of how to get the instances.
select user_id, t1.submitted_dtm t1submitted, t2.submitted_dtm t2submtted
from table_t t1 join table_t t2 using (user_id)
where t2.submitted_dtm > t1.submitted_dtm
and t2.submitted_dtm - t1.submitted_dtm <= 1;
The last line could be modified somehow depending on what you mean by within a day.
To count the instances, create a derived table from the above and select count(*) from it.