Joining two aggregate queries from the same table - SQL Server - sql

Question

I have two queries, both are aggregated from the same table. I'm not sure if I have to join these two queries together or if it can be done with one select statement. The goal is to output a table that aggregates total charges and total refunds for each student.
Query #1:
select
s.learners_id, sum(charge.total_amount) charge_amount
from
fact_student_transactions_t charge
inner join
dim_students_t s on s.students_sk_id = charge.students_sk_id
left join
object_statuses_t os on os.object_statusid = charge.transaction_status_id
where
os.status_name = 'Success'
and charge.tran_type = 'CHARGE'
and charge.curr_in = 1
group by
s.learners_id
Query #2:
select
s.learners_id, sum(refund.total_amount) refund_amount
from
fact_student_transactions_t refund
inner join
dim_students_t s on s.students_sk_id = refund.students_sk_id
left join
object_statuses_t os on os.object_statusid = refund.transaction_status_id
where
os.status_name = 'Success'
and refund.tran_type = 'Refund'
and refund.trans_description not in ('Amount Successfully Transfered to Prepaid Balance.', 'Amount Successfully Transfered.')
and refund.payment_method != 'Transfer'
and refund.curr_in = 1
group by
s.learners_id

Answer 1

You can use conditional aggregation. Also, because of the nature of the where clause, the left join is unnecessary -- the unmatched records are filtered out anyway.
So:
select s.learners_id,
sum(case when st.tran_type = 'CHARGE' then st.total_amount else 0 end) as charge_amount,
sum(case when st.tran_type = 'Refund' and
st.trans_description not in ('Amount Successfully Transfered to Prepaid Balance.', 'Amount Successfully Transfered.')
and
st.payment_method <> 'Transfer'
then st.total_amount else 0
end) as refund_amount
from fact_student_transactions_t st join
dim_students_t s
on s.students_sk_id = t.students_sk_id join
object_statuses_t os
on os.object_statusid = t.transaction_status_id
where os.status_name = 'Success' and
st.curr_in = 1 and
st.tran_type in ('CHARGE', 'Refund')
group by s.learners_id