How to handle duplicates created by LEFT JOIN - sql

Question

LEFT TABLE:
+------+---------+--------+
| Name | Surname | Salary |
+------+---------+--------+
| Foo | Bar | 100 |
| Foo | Kar | 300 |
| Fo | Ba | 35 |
+------+---------+--------+
RIGHT TABLE:
+------+-------+
| Name | Bonus |
+------+-------+
| Foo | 10 |
| Foo | 20 |
| Foo | 50 |
| Fo | 10 |
| Fo | 100 |
| F | 1000 |
+------+-------+
DESIRED OUTPUT:
+------+---------+--------+-------+
| Name | Surname | Salary | Bonus |
+------+---------+--------+-------+
| Foo | Bar | 100 | 80 |
| Foo | Kar | 300 | 0 |
| Fo | Ba | 35 | 110 |
+------+---------+--------+-------+
The closest I get is this:
SELECT
a.Name,
Surname,
sum(Salary),
sum(Bonus)
FROM (SELECT
Name,
Surname,
sum(Salary) as Salary
FROM input
GROUP BY 1,2) a LEFT JOIN (SELECT Name,
SUM(Bonus) as Bonus
FROM input2
GROUP BY 1) b
ON a.Name = b.Name
GROUP BY 1,2;
Which gives:
+------+---------+-------------+------------+
| Name | Surname | sum(Salary) | sum(Bonus) |
+------+---------+-------------+------------+
| Fo | Ba | 35 | 110 |
| Foo | Bar | 100 | 80 |
| Foo | Kar | 300 | 80 |
+------+---------+-------------+------------+
I can't figure out how to get rid of Bonus duplication. Ideal solution for me would be as specified in the 'DESIRED OUTPUT', which is adding Bonus to only one Name and for other records with the same Name adding 0.

Answer 1

You can use row_number():
select l.*, (case when l.seqnum = 1 then r.bonus else 0 end) as bonus
from (select l.*, row_number() over (partition by name order by salary) as seqnum
from "left" l
) l left join
(select r.name, sum(bonus) as bonus
from "right" r
group by r.name
) r
on r.name = l.name

Answer 2

Try a Row_number over the Name category partioned by Name. This will give you different numbers for your duplicates. You can then search for the case when this number is 1 and return the result you want. Else return 0. The code can look something like this.
SELECT
a.Name,
Surname,
sum(Salary),
Case when Duplicate_Order = 1
then bonus
else 0
end as 'Bonus'
FROM (SELECT
Name,
Surname,
sum(Salary) as Salary
,ROW_NUMBER() over (partition by Name order by name) as [Duplicate_Order]
FROM input
GROUP BY 1,2) a
LEFT JOIN (SELECT Name,
SUM(Bonus) as Bonus
FROM input2
GROUP BY 1) b
ON a.Name = b.Name
GROUP BY 1,2;
Hope that helps!

Answer 3

You can use Correlated Subquery with sum() aggregation to compute the bonus column, and then apply lag() window analytic function to get the zeros for successively identical valued column values for the name column :
select Name, Surname, Salary,
bonus - lag(bonus::int,1,0) over (partition by name order by salary) as bonus
from
(
select i1.*,
( select sum(Bonus)
from input2 i2
where i1.Name = i2.Name
group by i2.Name ) as bonus
from input i1
) ii
order by name desc, surname;
Demo