Group by two fields, and having count() on first field

Question

I have a table that stored users play list, a video can be viewed by multiple users for multiple times. A records goes like this:

videoid, userid, time
123,     abc   , 2013-09-11

It means user(abc) has watched video(123) on 2013-09-11

Now I want to find distinct users watched video list (no duplication), and only show the users that have watched more than two videos.

SELECT videoid, userid 
FROM table_play_list
WHERE SOME CONDICTION
GROUP BY userid, videoid

The sql only select distinct users watchlist, I also want to filter users that have watched more than two different videos.

I know I have to google and read the documentation first, some said 'HAVING' could solve this, unfortunately, I could not make it.

Answer 1

If I understand correctly, you are looking for users who watched more than two different videos. You can do this by using count(distinct) with a partition by clause:

select userid, videoid
from (SELECT userid, videoid, count(distinct videoid) over (partition by userid) as cnt
      FROM table_play_list
      WHERE <ANY CONDITION>
     ) t
where cnt > 2;

Answer 2

Try like this,

SELECT userid, count(*)
FROM   table_play_list
--WHERE SOME CONDITION
GROUP BY user_id
having count(*) >2;

Try this if you need to get the count based on userid and videoid(users who watch the same video more than two times).

SELECT userid, videoid, count(*)
FROM   table_play_list
--WHERE SOME CONDITION
GROUP BY user_id, video_id
having count(*) >2;

Answer 3

This is probably best handled with analytics (window functions). Without analytics you will probably need a self-join.

SQL> WITH table_play_list AS (
  2     SELECT 123 videoid, 'a' userid FROM dual UNION ALL
  3     SELECT 125 videoid, 'a' userid FROM dual UNION ALL
  4     SELECT 123 videoid, 'b' userid FROM dual UNION ALL
  5     SELECT 123 videoid, 'b' userid FROM dual UNION ALL
  6     SELECT 123 videoid, 'c' userid FROM dual
  7  )
  8  SELECT videoid, userid,
  9         COUNT(*) over(PARTITION BY userid) nb_video
 10    FROM table_play_list;
 
   VIDEOID USERID   NB_VIDEO
---------- ------ ----------
       123 a               2
       125 a               2
       123 b               2
       123 b               2
       123 c               1

This lists all user/video and the total number of videos watched by each user. As you can see user b has watched the same video twice, I don't know if it's possible in your system.

You can filter with a subquery:

SQL> WITH table_play_list AS (
  2     SELECT 123 videoid, 'a' userid FROM dual UNION ALL
  3     SELECT 125 videoid, 'a' userid FROM dual UNION ALL
  4     SELECT 123 videoid, 'b' userid FROM dual UNION ALL
  5     SELECT 123 videoid, 'b' userid FROM dual UNION ALL
  6     SELECT 123 videoid, 'c' userid FROM dual
  7  )
  8  SELECT *
  9    FROM (SELECT videoid, userid,
 10                 COUNT(*) over(PARTITION BY userid) nb_video
 11            FROM table_play_list)
 12   WHERE nb_video > 1;
 
   VIDEOID USERID   NB_VIDEO
---------- ------ ----------
       123 a               2
       125 a               2
       123 b               2
       123 b               2

Answer 4

The below will give users who have watched more than two different videos.

 SELECT userid, count(distinct video_id)
 FROM   table_play_list
 WHERE SOME CONDICTION
 GROUP BY user_id
 having count(distinct video_id) >2;

Answer 5

• If you use Oracle PL/SQL you can use like this:

SELECT column1, column2  
FROM 
    (
    SELECT column1, column2, COUNT(column1)
    OVER (PARTITION BY column1) AS cnt 
    FROM test 
    GROUP BY column1, column2
    ORDER BY column1 
    ) 
WHERE cnt > 2

• If you use standard SQL you can use like this:

SELECT column1, column2 
FROM test 
WHERE column1 IN
    (
        SELECT column1 
        FROM
         (
             SELECT column1, column2 
             FROM test 
             GROUP BY column1, column2 
             ORDER BY column1
         )  
         GROUP BY column1 
         HAVING COUNT(column1) > 2
     ) 
GROUP BY column1, column2 
ORDER BY column1