I have the table which I want select only top 2 sales in each class, the result is
Sophia A 40
Jennifer A 15
Greg B 50
Jeff B 20
Stella B 20
You could use RANK:
SELECT *
FROM (SELECT *, RANK() OVER(PARTITION BY class ORDER BY sales DESC) AS rnk
FROM tab) sub
WHERE rnk <=2;
window function dense_rank also fulfil expected result condition of sample data
select * from
(select * , dense_rank() over(partition by class order by scores DESC ) as rn
from tablea
) t where rn<=2
Related
I have the following table named customerOrders.
ID user order
1 1 2
2 1 3
3 1 1
4 2 1
5 1 5
6 2 4
7 3 1
8 6 2
9 2 2
10 2 3
I want to return to users with most orders. Currently, I have the following QUERY:
SELECT user, COUNT(user) AS UsersWithMostOrders
FROM customerOrders
GROUP BY user
ORDER BY UsersWithMostOrders DESC;
This returns me all the values grouped by total orders like.
user UsersWithMostOrders
1 4
2 4
3 1
6 1
I only want to return the users with most orders. In my case that would be user 1 and 2 since both of them have 4 orders. If I use TOP 1 or LIMIT, it will only return the first user. If I use TOP 2, it will only work in this scenario, it will return invalid data when top two users have different count of orders.
Required Result
user UsersWithMostOrders
1 4
2 4
You can use TOP 1 WITH TIES:
SELECT TOP 1 WITH TIES
[user], COUNT(*) AS UsersWithMostOrders
FROM customerOrders
GROUP BY [user]
ORDER BY UsersWithMostOrders DESC;
See the demo.
Results:
> user | UsersWithMostOrders
> ---: | ------------------:
> 1 | 4
> 2 | 4
Option 1
Should work with most versions of SQL.
select *
from (
select *,
rank() over(order by numOrders desc) as rrank
from (
select `user`, count(*) as numOrders
from customerOrders
group by `user`
) summed
) ranked
where rrank = 1
Play around with the code here
Option 2
If your version of SQL allows window functions (with), here is a much more readable solution which does the same thing
with summed as (
select `user`, count(*) as numOrders
from customerOrders
group by `user`
),
ranked as (
select *,
rank() over(order by numOrders desc) as rrank
from summed
)
select *
from ranked
where rrank = 1
Play around with the code here
You can use a CTE to attain this Req:
;WITH CTE AS(
SELECT [user], COUNT(user) AS UsersWithMostOrders
FROM #T
GROUP BY [user])
SELECT M.* from CTE M
INNER JOIN ( SELECT
MAX(UsersWithMostOrders) AS MaximumOrders FROM CTE) S ON
M.UsersWithMostOrders=S.MaximumOrders
Below Oracle Query can help:
WITH test_table AS
(
SELECT user, COUNT(order) AS total_order , DENSE_RANK() OVER (ORDER BY
total_order desc) AS rank_orders FROM customerOrders
GROUP BY user
)
select * from test_table where rank_orders = 1
Suppose I have a table of customer purchases ("my_table") like this:
--------------------------------------
customerid | date_of_purchase | price
-----------|------------------|-------
1 | 2019-09-20 | 20.23
2 | 2019-09-21 | 1.99
1 | 2019-09-21 | 123.34
...
I'd like to be able to find the nth highest spending customer in this table (say n = 5). So I tried this:
with cte as (
select customerid, sum(price) as total_pay,
row_number() over (partition by customerid order by total_pay desc) as rn
from my_table group by customerid order by total_pay desc)
select * from cte where rn = 5;
But this gives me nonsense results. For some reason rn doesn't seem to be unique (for example there are a bunch of customers with rn = 1). I don't understand why. Isn't rn supposed to be just a row number?
Remove the partition by in the definition of row_number():
with cte as (
select customerid, sum(price) as total_pay,
row_number() over (order by total_pay desc) as rn
from my_table
group by customerid
)
select *
from cte
where rn = 5;
You are already aggregating by customerid, so each customer has only one row. So the value of rn will always be 1.
My dataset name bollywood.csv:
This is my data. I need the actors who have the most lead roles in movies.
I Need the name of the lead actors and the number of films in which they have lead.
My code is:
select lead, count(*) as nos from bollywood group by lead order by nos desc;
And the result is:
Amitabh 3
Akshay 3
John 3
Riteish 2
Shahrukh 2
Sunny 2
Emraan 2
Katrina 2
Nawazuddin 2
Tiger 2
Sharman 2
Manoj 2
Vidya 1
Tusshar 1
Tannishtha 1
Sushant 1
SunnyDeol 1
Sonam 1
Sonakshi 1
Siddarth 1
Shahid 1
Sandeep 1
Salman 1
If you want all actors with most lead roles (possibly multiple records):
select lead, count(*) as nos
from bollywood
group by lead
having count(*) =
(select max(cnt) from
(select lead, count(*) cnt
from bollywood
group by lead ) tblBolly )
Use rownum pseudocolumn with order by (oracle).
select from (
select lead, count(*) as nos
from bollywood
group by lead
order by lead desc
)
where rownum = 1
Use window functions:
select lead, cnt
from (select lead, count(*) as cnt,
rank() over (order by count(*) desc) as rnk
from bollywood
group by lead
) b
where rnk = 1;
I'm having problem with getting only TOP 2 values for each group (groups are in column).
Example :
ID Group Value
1 A 30
2 A 150
3 A 40
4 A 70
5 B 0
6 B 100
7 B 90
I expect my output to be
ID Group Value
1 A 150
2 A 70
3 B 100
4 B 90
Simply, for each group I want just 2 rows with the highest Value
Most databases support the ANSI standard row_number() function. You would use it as:
select group, value
from (select t.*,
row_number() over (partition by group order by value desc) as seqnum
from t
) t
where seqnum <= 2;
To set the id you can use row_number() in the outer query:
select row_number() over (order by group, value) as id,
group, value
from (select t.*,
row_number() over (partition by group order by value desc) as seqnum
from t
) t
where seqnum <= 2;
However, changing the id seems suspicious.
You can use CTE with rank function ROW_NUMBER() .
Here is query to get your result.
;WITH cte AS
( SELECT Group, value,
ROW_NUMBER() OVER (PARTITION BY Group ORDER BY value DESC) AS rn
FROM test
)
SELECT Group, value FROM cte
WHERE rn <= 2
ORDER BY value
I'm trying to get the Employee with the highest sales
Employee DeptNo Date Sales
Chris 2 2012/1/1 1000
Joe 1 2012/1/1 900
Arthur 3 2012/1/1 1100
Chris 2 2012/3/1 1200
Joe 1 2012/2/1 1500
Arthur 3 2010/2/1 1200
Joe 1 2010/3/1 900
Arthur 3 2010/3/1 1100
Arthur 3 2010/4/1 1200
Joe 1 2012/4/1 1500
Chris 2 2010/4/1 1800
I've tried using two subqueries, and then comparing them together to find the higher value
SELECT c1.Employee,
c1.TOTAL_SALES
FROM (SELECT Employee,
Sum(sales) AS TOTAL_SALES
FROM EmployeeSales
GROUP BY Employee) c1,
(SELECT Employee,
Sum(sales) AS TOTAL_SALES
FROM EmployeeSales
GROUP BY Employee) c2
WHERE ( c1.TOTAL_SALES > c2.TOTAL_SALES
AND c1.Employee > c2.Employee )
But the resulting query gives me two rows of
Employee TOTAL_SALES
joe 4800
joe 4800
What am I doing wrong?
I would use a CTE.
;With [CTE] as (
Select
[Employee]
,sum([Sales]) as [Total_Sales]
,Row_Number()
Over(order by sum([sales]) Desc) as [RN]
From [EmployeeSales]
Group by [Employee]
)
Select
[Employee]
,[Total_Sales]
From [CTE]
Where [RN] = 1
Example of working code SQL Fiddle:
http://sqlfiddle.com/#!3/bd772/2
To return all employees with the highest total sales, you can use SQL Server's proprietary TOP WITH TIES:
SELECT TOP (1) WITH TIES name, SUM(sales) as total_sales
FROM employees
GROUP BY name
ORDER BY SUM(sales) DESC
SELECT name, SUM(sales) as total_sales
FROM employees
GROUP BY name
ORDER by total_sales DESC
LIMIT 1;
A better solution is to group by an employee id so we are sure they are the same person. Since there can be two Chris's.
I would use a window partition
select * from
(
select
employee
, sum(sales) as sales
, row_number() over
(
order by sum(sales) desc
) as rank
from EmployeeSales
group by employee
) tmp
where tmp.rank = 1
And I agree with what someone said (Shawn) about having an employeeID and group by that for this, rather than the name.
(I removed the partition from the row_number() call as it is not needed for this)
you can use CTE for that
WITH CTE
AS ( select employee , sum(sales) as sales,
ROW_NUMBER() OVER (PARTITION BY employee ORDER BY sum(sales) desc) RN
FROM EmployeeSales)
SELECT employee ,
sales
FROM CTE
WHERE RN =1