Extract records by amount summary from second table - mysql

Question

My query:
SELECT fd.*
FROM `fin_document` as fd
LEFT JOIN `fin_income` as fi ON fd.id=fi.document_id
WHERE fd.dt_payment < NOW()
HAVING SUM(fi.amount) < fd.total_amount
which is obviously not correct, has to retrieve all records from fin_document where dt_payment is earlier than NOW(). This part is ok. But I have to filter them by the payments made on this documents. One document can have more than 1 payment ( 2,3,4,5 ...). In fin_income are those payments. There is column document_id in fin_income table which is foreign key fin_income.document_id=fin_document.id. The problem ( at least for me ) is that I don't have a specific id criterion and the amount is made from all records from fin_income table. I also have to find records that still don't have payments on them ( they don't have rows in fin_income ).
fin_document:
+-------------------+---------------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------------+---------------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| dt_payment | date | YES | | NULL |
| total_amount | decimal(10,2) | NO | | 0.00 |
+-------------------+---------------------------+------+-----+---------+----------------+
fin_income:
+------------------+---------------+------+-----+-------------------+----------------+
| Field | Type | Null | Key | Default | Extra |
+------------------+---------------+------+-----+-------------------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| document_id | int(11) | YES | MUL | NULL | |
| amount | decimal(10,2) | YES | | 0.00 |
+------------------+---------------+------+-----+-------------------+----------------+

Answer 1

I'm not sure if I understand you correctly, but you can try this:
SELECT fd.*, SUM(IFNULL(fi.amount, 0)) as sum_amount, COUNT(fi.amount) as count_amount
FROM `fin_document` as fd
LEFT JOIN `fin_income` as fi ON fd.id=fi.document_id
WHERE fd.dt_payment < NOW()
GROUP BY fd.id
HAVING sum_amount < fd.total_amount # condition for searching by sum of payments
AND count_amount = {needed_count} # condition for searching by count of payments;
# documents without payments will have
# sum and count equal to 0
All aggregations are made in SELECT part, then all documents are grouped by id to avoid duplicates in result and make possible to use aggregation results (SUM, COUNT). And finally you can apply needed conditions (about date, paid sum or count of payments).
Note: pay attention that GROUP BY significantly increases execution time for a lot of data.

Answer 2

You may just need a correlated sub query to test income
drop table if exists fin_document,fin_income;
create table fin_document
(id int(11),
dt_payment date ,
total_amount decimal(10,2)
) ;
create table fin_income
( id int(11) ,
document_id int(11) ,
amount decimal(10,2)
);
insert into fin_document values
(1,'2019-05-31',1000),
(2,'2019-06-10',1000),
(3,'2019-07-10',1000);
insert into fin_income values
(1,1,5),(1,1,5);
SELECT fd.*,(select coalesce(sum(fi.amount),0) from fin_income fi where fd.id=fi.document_id) income
FROM `fin_document` as fd
WHERE fd.dt_payment < NOW() and
fd.total_amount > (select coalesce(sum(fi.amount),0) from fin_income fi where fd.id=fi.document_id);
+------+------------+--------------+--------+
| id | dt_payment | total_amount | income |
+------+------------+--------------+--------+
| 1 | 2019-05-31 | 1000.00 | 10.00 |
| 2 | 2019-06-10 | 1000.00 | 0.00 |
+------+------------+--------------+--------+
2 rows in set (0.00 sec)