Local variable sum previous columns - mysql

Question

I am trying to use #variable1 + #variable2 into a query but is actually given 0 as result.
MariaDB
Server Version: 10.2.21
set #start_at = '2019-01-01';
set #end_at = '2019-01-16';
set #receivable = 0;
set #invoiced = 0;
SELECT DISTINCT Customer.custnr 'Customer Number',
Address.name 'Name',
#receivable := sum(case
WHEN [condition1 <= #start_at]
AND Transactions.`key` not in [subquery]
THEN Transactions.amount
ELSE 0 END) 'Account Receivable',
#invoiced := sum(case
WHEN [condition1 between #start_at and #end_at]
AND [condition2]
AND [condition3]
AND Transactions.`key` not in [subquery]
THEN Transactions.amount
ELSE 0 END) 'Invoiced',
#receivable + #invoiced 'Total'
FROM LocalCust
INNER JOIN Customer
on Customer.`key` = LocalCust.customerkey
INNER JOIN Address
on Address.`key` = Customer.addresskey
INNER JOIN Location
on Location.`key` = LocalCust.localkey
INNER JOIN Transactions
on Transactions.localcustkey = LocalCust.`Key`
GROUP BY Transactions.localcustkey;
Result:

Answer 1

Use a subquery and don't use variables at all:
SELECT x.*, (Account_Receivable + Invoice) as Total
FROM (SELECT c.custnr as Customer_Number, a.name,
sum(case when condition1 <= #start_at and
t.`key` not in [subquery]
then t.amount
else 0
end) as Account_Receivable,
sum(case when condition1 between #start_at and #end_at and
[condition2] and
[condition3] and
t.`key` not in [subquery]
then t.amount
else 0
end) as Invoiced
FROM LocalCust lc JOIN
Customer c
on c.`key` = lc.customerkey JOIN
Address a
on a.`key` = c.addresskey JOIN
Location l
on l.`key` = lc.localkey join
Transactions t
on t.localcustkey = lc.`Key`
GROUP BY c.custnr, a.name
) x;
Notes:
Table aliases make the query easier to write and to read.
SELECT DISTINCT is almost never needed with GROUP BY.
The GROUP BY keys should match the unaggregated columns in the SELECT.
Choose column aliases that do not need to be escaped. That is, no spaces.

Answer 2

You can not just put #receivable, #invoiced and #receivable + #invoiced in the same select statement. (They will not store the value in order. They will be executed at the same time.)
First, You need to store the values in #receivable, #invoiced then use a subquery to calculate the total:
set #start_at = '2019-01-01';
set #end_at = '2019-01-16';
set #receivable = 0;
set #invoiced = 0;
SELECT *, A.[Account Receivable] + A.[Invoiced] AS TOTAL FROM (
SELECT DISTINCT Customer.custnr 'Customer Number',
Address.name 'Name',
#receivable := sum(case
WHEN [condition1 <= #start_at]
AND Transactions.`key` not in [subquery]
THEN Transactions.amount
ELSE 0 END) 'Account Receivable',
#invoiced := sum(case
WHEN [condition1 between #start_at and #end_at]
AND [condition2]
AND [condition3]
AND Transactions.`key` not in [subquery]
THEN Transactions.amount
ELSE 0 END) 'Invoiced'
FROM LocalCust
INNER JOIN Customer
on Customer.`key` = LocalCust.customerkey
INNER JOIN Address
on Address.`key` = Customer.addresskey
INNER JOIN Location
on Location.`key` = LocalCust.localkey
INNER JOIN Transactions
on Transactions.localcustkey = LocalCust.`Key`
GROUP BY Transactions.localcustkey) A;