Can anyone help me how to write this query without using count?
"Some directors directed more than one movie. For all such directors, return the titles of all movies directed by them, along with the director name. Sort by director name, then movie title.(without COUNT.)
mID | title | director
--------------------------------------
101 |Gone with the Wind |Victor Fleming
102 |Star Wars |George Lucas
103 |The Sound of Music |Robert Wise
104 |E.T. |Steven Spielberg
105 |Titanic |James Cameron
106 |Snow White |<null>
107 |Avatar |James Cameron
108 |Raiders of the Lost Ark |Steven Spielberg
You can compare for each director the min and max of the mid.
If they are different - there is more than 1.
Select mid, title, director
From tbl
where director in (Select director
From tbl
Group by director
Having max(mid) > min (mid))
order by director, title
You can use ROW_NUMBER() and a CTE or sub-query. Not using COUNT() is pretty silly though. It's an ideal case for such an aggregate.
with cte as(
select
director
,title
,row_number() over (partition by director order by title) as rn
from
yourTable)
select
director,
title
from cte
where director in (select director from cte where rn > 1)
order by
director,
title
How about this?
select t.*
from t join
(select director, sum(1) as cnt
from t
group by director
) d
on t.director = d.director
where cnt > 1
order by director, title;
Related
I have a table Movie with columns Movie and Viewer where each movie is viewed by any user any number of times, so the table can have multiple same entries. I want to find the Top N most viewed movies and then the Top K viewers for each of the Top N movies. How can I apply group by or partition by effectively in such scenario ? Or if there is any better approach to this, please share. Thanks!
Movie
User
Avengers
John
Batman
Chris
Batman
Ron
X-Men
Chris
X-Men
Ron
Matrix
John
Batman
Martin
Matrix
Chris
Batman
Chris
X-Men
Ron
So, in this table Batman is the most watched movie is Batman followed by X-Men so I want the result table to look like :
Movie
User
View count
Batman
Chris
2
Batman
Ron
1
Batman
Martin
1
X-Men
Ron
2
X-Men
Chris
1
Matrix
John
1
Matrix
Chris
1
Avengers
John
1
I understand that I can group by movie and then do order by count(*) desc but this doesn't give me the second column which is grouped by viewer and the count for each viewer also.
Consider below approach (assuming Top 3 movies with Top 2 users)
select movie, user, view_count
from (
select distinct *,
count(*) over(partition by movie) movie_views,
count(*) over(partition by movie, user) view_count
from your_table
)
qualify dense_rank() over(order by movie_views desc) <=3
and row_number() over(partition by movie order by view_count desc) <=2
-- order by movie_views desc, view_count desc
if applied to sample data in your question - output is
Given a dataset Roster_table as such:
Group ID
Group Name
Name
Phone
42
Red Dragon
Jon
123455678
32
Green Lizard
Liz
932143211
19
Blue Falcon
Ben
134554678
42
Red Dragon
Reed
432143211
42
Red Dragon
Brad
231314155
19
Blue Falcon
Chad
214124412
How do I get the following query output combining rows with the same Group ID from the dataset, and the new column Count in descending order:
Group ID
Group Name
Count
42
Red Dragon
3
19
Blue Falcon
2
32
Green Lizard
1
SELECT * FROM Roster_table
Please try this where alias tot_count is used in ORDER BY clause.
-- PostgreSQL(v11)
SELECT Group_ID
, MAX(Group_Name) Group_Name
, COUNT(1) tot_count
FROM Roster_table
GROUP BY Group_ID
ORDER BY tot_count DESC;
Please check from url https://dbfiddle.uk/?rdbms=postgres_11&fiddle=b66f9f0d40e804e89be12e3530fe00a0
Based on Rahul Biswas's answer:
Solution without using Max function
SELECT Group_ID, Group_Name, COUNT(*)
FROM Roster_table
GROUP BY Group_ID, Group_Name
ORDER BY COUNT(*) DESC
Credit goes to Eric S.
If I have the table "members" (shown below), how would I go about getting the record of the first occurrence of a membership_id (Oracle).
Expected results
123 John Doe A P
313 Michael Casey A A
113 Luke Skywalker A P
Table - members
membership_id first_name last_name status type
123 John Doe A P
313 Michael Casey A A
113 Luke Skywalker A P
123 Bob Dole A A
313 Lucas Smith A A
SELECT membership_id,
first_name,
last_name,
status,
type
FROM( SELECT membership_id,
first_name,
last_name,
status,
type,
rank() over (partition by membership_id
order by type desc) rnk
FROM members )
WHERE rnk = 1
will work for your sample data set. If you can have ties-- that is, multiple rows with the same membership_id and the same maximum type-- this query will return all those rows. If you only want to return one of the rows where there is a tie, you would either need to add additional criteria to the order by to ensure that all ties are broken or you would need to use the row_number function rather than rank which will arbitrarily break ties.
Select A.*
FROM Members AS A inner join
(Select membership_id, first(first_name) AS FN, first(last_name) AS LN
From Members
Group by membership_id) AS B
ON A.membership_id=B.membership_id and A.first_name=B.FN and A.last_name=B.LN
Hope that helps!
select *
from members
where rowid in (
select min(rowid)
from members
group by membership_id
)
Oracle 11g:
I want results to list by highest count, then ch_id. When I use group by to get the count then I loose the granularity of the detail. Is there an analytic function I could use?
SALES
ch_id desc customer
=========================
ANAR Anari BOB
SWIS Swiss JOE
SWIS Swiss AMY
BRUN Brunost SAM
BRUN Brunost ANN
BRUN Brunost ROB
Desired Results
count ch_id customer
===========================================
3 BRUN ANN
3 BRUN ROB
3 BRUN SAM
2 SWIS AMY
2 SWIS JOE
1 ANAR BOB
Use the analytic count(*):
select * from
(
select count(*) over (partition by ch_id) cnt,
ch_id, customer
from sales
)
order by cnt desc
select total, ch_id, customer
from sales s
inner join (select count(*) total, ch_id from sales group by ch_id) b
on b.ch_id = s.chi_id
order by total, ch_id
ok - the other post that happened at the same time, using partition, is the better solution for Oracle. But this one works regardless of DB.
How would I be able to get N results for several groups in
an oracle query.
For example, given the following table:
|--------+------------+------------|
| emp_id | name | occupation |
|--------+------------+------------|
| 1 | John Smith | Accountant |
| 2 | Jane Doe | Engineer |
| 3 | Jack Black | Funnyman |
|--------+------------+------------|
There are many more rows with more occupations. I would like to get
three employees (lets say) from each occupation.
Is there a way to do this without using a subquery?
I don't have an oracle instance handy right now so I have not tested this:
select *
from (select emp_id, name, occupation,
rank() over ( partition by occupation order by emp_id) rank
from employee)
where rank <= 3
Here is a link on how rank works: http://www.psoug.org/reference/rank.html
This produces what you want, and it uses no vendor-specific SQL features like TOP N or RANK().
SELECT MAX(e.name) AS name, MAX(e.occupation) AS occupation
FROM emp e
LEFT OUTER JOIN emp e2
ON (e.occupation = e2.occupation AND e.emp_id <= e2.emp_id)
GROUP BY e.emp_id
HAVING COUNT(*) <= 3
ORDER BY occupation;
In this example it gives the three employees with the lowest emp_id values per occupation. You can change the attribute used in the inequality comparison, to make it give the top employees by name, or whatever.
Add RowNum to rank :
select * from
(select emp_id, name, occupation,rank() over ( partition by occupation order by emp_id,RowNum) rank
from employee)
where rank <= 3
tested this in SQL Server (and it uses subquery)
select emp_id, name, occupation
from employees t1
where emp_id IN (select top 3 emp_id from employees t2 where t2.occupation = t1.occupation)
just do an ORDER by in the subquery to suit your needs
I'm not sure this is very efficient, but maybe a starting place?
select *
from people p1
join people p2
on p1.occupation = p2.occupation
join people p3
on p1.occupation = p3.occupation
and p2.occupation = p3.occupation
where p1.emp_id != p2.emp_id
and p1.emp_id != p3.emp_id
This should give you rows that contain 3 distinct employees all in the same occupation. Unfortunately, it will give you ALL combinations of those.
Can anyone pare this down please?