Join Table + Group by + Sum + Finding Nth highest sum value - sql

Question

I have 2 tables -
Broker Table ->Column names-->ID,Name,City,branch
Policy Table ->Column names-->Policy_num,Broker_id,Premium
I need to find the broker name having 4th highest sum of the premiums.
Please suggest the SQL query for this.
NOTE:I am looking for a query that can run on all platforms.So please don't use
TOP,LIMIT or ROWNUMBER functions.
I am looking for a query similar to mentioned below,which is just finding 4th highest salary from a table.
SELECT Salary
FROM EmployeeSalary Emp1
WHERE 3 = (
SELECT COUNT( DISTINCT ( Emp2.Salary ) )
FROM EmployeeSalary Emp2
WHERE Emp2.Salary > Emp1.Salary
)

Answer 1

First, you don't need a join, because all the information you need is in the policy table. You can join in broker at the end, if you want.
Second, the question is ambiguous, because it doesn't specify how to handle ties.
But basically, you can follow the same structure as the query in your question:
SELECT p.BrokerId
FROM Policy p
GROUP BY p.BrokerId
HAVING 4 = (SELECT COUNT(*)
FROM (SELECT SUM(p2.Premium) as sumpremium
FROM Policy p2
GROUP BY p2.BrokerId
)
WHERE sumpremium >= SUM(p.premium)
);
But ironically, this is a very, very poor way to approach the problem. Why? There is no guarantee that the SUM() on the two aggregation queries will be exactly the same.
This difference can occur in two ways. If premium is a floating point number, then the order of addition can make a difference. Adding values near zero to a large value loses precision. So check this out:
select ( (v1 + v2) + v3), (v1 + (v2 + v3))
from (select cast(100000000.0001 as float) as v1, cast(-100000001.0001 as float) as v2, cast(0.000002 as float) as v3) x;
Even with fixed point numbers you can still have problems, because adding all the positive numbers first can result in an overflow.
It is really better to do arithmetic calculations once rather than doing them multiple times and then comparing the values.

Answer 2

Try this-
SELECT *
FROM
(
SELECT B.Broker_ID,A.Name,
SUM(B.Premium) S_P,
ROW_NUMBER() OVER (ORDER BY SUM(B.Premium) DESC) RN
FROM Broker A
INNER JOIN Policy B ON A.ID = B.Broker_ID
GROUP BY B.Broker_ID,A.Name
)A
WHERE A.RN = 4
Another option you can use TOP twice to get your desired output as below-
SELECT TOP 1 * FROM
(
SELECT TOP 4
B.Broker_ID,A.Name,
SUM(B.Premium) S
FROM Broker A
INNER JOIN Policy B ON A.ID = B.Broker_ID
GROUP BY B.Broker_ID,A.Name
ORDER BY SUM(your_id_column) DESC
)A
ORDER BY S
Another option considering MSSQL-
SELECT * FROM
(
SELECT B.Broker_ID,A.Name,
SUM(B.Premium) S_P,
COUNT(*) OVER (ORDER BY SUM(B.Premium) DESC ROWS UNBOUNDED PRECEDING) AS RN
FROM Broker A
INNER JOIN Policy B ON A.ID = B.Broker_ID
GROUP BY B.Broker_ID,A.Name
)A
WHERE A.RN = 4
Another option considering your sample query-
SELECT * FROM
(
SELECT B.Broker_ID, A.Name, SUM(B.Premium) S
FROM Broker A INNER JOIN Policy B ON A.ID = B.Broker_ID
GROUP BY B.Broker_ID,A.Name
)A
WHERE 3 = (
SELECT COUNT(*)
FROM
(
SELECT B.Broker_ID, A.Name, SUM(B.Premium) S
FROM Broker A INNER JOIN Policy B ON A.ID = B.Broker_ID
GROUP BY B.Broker_ID,A.Name
) B
WHERE B.S > A.S
)