CODEWITHSUNDEEP
sqlpostgresqlgroup-by
riders994
riders994

Reputation: 1316

Max Row In Group By

I've tried reading through a number of the solutions I've found, but there seem to be slight differences in the problems where I'm not sure how to adapt them to fit my own.

I have a table of transactions like so:

+------+-------+------------------------------+
| id   | rev_$ | product  | local_currency    |
+------+-------+------------------------------+
| 1    | 15    | shoe     | USD               |
| 2    | 10    | shirt    | USD               |
| 1    | 20    | shoe     | CAD               |
| 2    | 30    | shoe     | GBP               |
| 1    |  8    | shirt    | USD               |
| 2    | 15    | shirt    | USD               |
| 1    | 10    | shoe     | CAD               |
| 2    | 10    | shoe     | USD               |
+------+-------+------------------------------+

I want to aggregate the table so that

So that table should look like this, after aggregating:

+------+-------+------------------------------+
| id   | rev_$ | product  | local_currency    |
+------+-------+------------------------------+
| 1    | 45    | shoe     | CAD               |
| 1    |  8    | shirt    | USD               |
| 2    | 25    | shirt    | USD               |
| 2    | 40    | shoe     | GBP               |
+------+-------+------------------------------+

Similar questions: (1), (2)

Upvotes: 4

Views: 61

Answers (2)

user330315
user330315

Reputation:

You can use group by to calculate the total revenue, collect all currencies in an array and pick the one from the highest value:

select id, 
       sum(rev_$), 
       product, 
       (array_agg(local_currency order by rev_$ desc))[1] as local_currency
from orders
group by id, product
order by id, product;

array_agg(local_currency order by rev_$ desc) will create an array for all currencies that are part of the group defined by group by ordered by rev$ descending. So the first element ([1]) is the one corresponding to the "highest single value transaction"

Online example: https://rextester.com/VOK41538

Another option is to write an aggregate function that does this without the array:

create or replace function first_agg (p_one anyelement, p_other anyelement )
  returns anyelement 
  language sql 
  immutable strict
as
$$
  select p_one;
$$;

create aggregate first_element 
(
  sfunc    = first_agg,
  basetype = anyelement,
  stype    = anyelement
);

Then you can use it like this:

select id, 
       sum(rev_$), 
       product, 
       first_element(local_currency order by rev_$ desc) as local_currency
from orders
group by id, product
order by id, product;

Online example: https://rextester.com/YGRR9338

Upvotes: 5

ScaisEdge
ScaisEdge

Reputation: 133400

could using some subquery 

select  m2.id, sum(m2.rev_$), t2.local_currency
my_table m2
from  (
  select distinct local_currency, id  
  from my_table m1 
  inner join  (
    select  id, max(rev_$) max_rev
    from  my_table 
    group by id 
  ) t1  on t1.id = m1.id and t1.max_rev = m1.rev_$ 
) t2 ON m2.id= t2.id

Upvotes: 1

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