How to select sum of specific id in select query MySQL, Beego - mysql

I want to get a result like
result
-------------------------------------------------------
id | uuid | user_id |created_date | amount | name
-------------------------------------------------------
1 | ABC | 1 | 2019/5/1 | 5 | xa
2 | PQR | 2 | 2019/5/5 | 150 | xb
A query that I trying to use
SELECT(SELECT SUM(paid_amount) WHERE ID = t1.**HERE**) AS sub1,
(t1.amount - sub1) AS sub2
FROM invoice t1 CROSS JOIN
invoice_paid t2;
Table struct in my DB
table invoice_paid
------------------------------------
id | uuid | paid_date | paid_amount
------------------------------------
1 | ABC | 2019/5/1 | 15
2 | ABC | 2019/5/5 | 80
table invoice
-------------------------------------------------------
id | uuid | user_id |created_date | amount | name
-------------------------------------------------------
1 | ABC | 1 | 2019/5/1 | 100 | xa
2 | PQR | 2 | 2019/5/5 | 150 | xb
I can use sum only 1 condition like where id = 1 but how do I combine this query in select query with a join query.
I use beego(golang), MariaDB

You can use this query. It JOINs the invoice table to a derived table of SUMs of all the amounts paid per invoice from invoice_paid, subtracting that total from the invoice amount to get the outstanding amount:
SELECT i.id, i.uuid, i.user_id, i.created_date, i.amount - COALESCE(p.amount, 0) AS amount, i.name
FROM invoice i
LEFT JOIN (SELECT uuid, SUM(paid_amount) AS amount
FROM invoice_paid
GROUP BY uuid) p ON p.uuid = i.uuid
ORDER BY i.id
Output:
id uuid user_id created_date name amount
1 ABC 1 2019-05-01 00:00:00 xa 5
2 PQR 2 2019-05-05 00:00:00 xb 150
Demo on dbfiddle

Related

Get max value by priority when specific column is not null

I have database table like
transactions
-----------
id
code
date
amount
formalities
-----------
id
transaction_id
this is query to get max value of transactions
SELECT MAX(transaction_id) FROM `transactions` n LEFT JOIN `formalities` r ON r.transaction_id = n.id
But what I want to achieve here is to get the max value of id group by transactions code, but the value must check if the transactions have relation to formalities or not.
If yes, get max value where they relate.
If not just get the usual max value.
Is there a query to achieve something like this?
example:
transactions
--------------------
id | code | amount |
1 | ABC | 10000 |
2 | ABC | 20000 |
3 | KOO | 10000 |
4 | ABC | 20000 |
5 | KOO | 30000 |
6 | KOO | 10000 |
formalities
-----------
id | transaction_id |
1 | 3 |
2 | 5 |
The result I want is getting the following output from the transactions table
id
--
4
5 ( priority the one that have relation and get the max value)
Use a LEFT JOIN and get both - MAX(transactions.id) and MAX(formalities.transaction_id):
select t.code, max(f.transaction_id), max(t.id)
from transactions t
left join formalities f on f.transaction_id = t.id
group by t.code
The result would be
| code | max(f.transaction_id) | max(t.id) |
| ---- | --------------------- | --------- |
| ABC | NULL | 4 |
| KOO | 5 | 6 |
View on DB Fiddle
To "prioritize" transaction_id column from formalities table you can use COALESCE(), which will return the first non NULL value:
select t.code, coalesce(max(f.transaction_id), max(t.id)) as max_transaction_id
from transactions t
left join formalities f on f.transaction_id = t.id
group by t.code
Result:
| code | max_transaction_id |
| ---- | ------------------ |
| ABC | 4 |
| KOO | 5 |
View on DB Fiddle
I am not sure if I am getting the question right, but why don't you simply use -
SELECT MAX(transaction_id) FROM `transactions` n INNER JOIN `formalities` r ON r.transaction_id = n.id group by n.code

Sum, Subtract and Join of multiple mysql table columns

I have four mysql tables client, transaction, other_loan and payment. I want to get the sum of load_amount and additional from table transaction + sum of amount from other_loan and subtract it to the sum of payment_amount in table payment. How can I achieve it?
Result I want:
> ID | Name | Amount
> 1 | Robin | 8718
> 2 | Reynaldo | 21
> 3 | Leomar | 0
My Tables:
transaction
> tid | id| date | load_amount | additional
> 1 | 1 | 2018-12-01 | 90 | 0
> 2 | 1 | 2018-12-07 | 90 | 0
> 3 | 2 | 2018-12-08 | 49 | 2
table: other_loan
> oid | id| amount | date
> 1 | 1 | 7928 | 2018-12-10
> 2 | 1 | 750 | 2018-12-10
table: payment
> pid |id | payment_amount | date
> 1 | 1 | 50 | 2015-12-10
> 2 | 1 | 90 | 2015-12-10
> 3 | 2 | 30 | 2015-12-10
table: client
> id | Name |
> 1 | Robin |
> 2 | Cinderella |
> 3 | Leomar |
Because you have multiple transactions, other loan amounts and payments per customer, you can't do a straight JOIN of the tables to each other as it will cause replication of rows, resulting in incorrect values. Instead, we SUM all the values within each table on a client basis before doing the JOIN. Additionally, since some clients don't have entries in each table, you must use LEFT JOINs and COALESCE on the results so that empty rows don't cause SUMs to become NULL. This query should give you the results you want:
SELECT c.id, c.name,
COALESCE(t.transactions, 0) + COALESCE(o.amounts, 0) - COALESCE(p.payments, 0) AS amount
FROM client c
LEFT JOIN (SELECT id, SUM(load_amount) + SUM(additional) AS transactions
FROM transaction
GROUP BY id) t on t.id = c.id
LEFT JOIN (SELECT id, SUM(amount) AS amounts
FROM other_loan
GROUP BY id) o ON o.id = c.id
LEFT JOIN (SELECT id, SUM(payment_amount) AS payments
FROM payment
GROUP BY id) p ON p.id = c.id
GROUP BY c.id
Output (for your sample data):
id name amount
1 Robin 8718
2 Cinderella 21
3 Leomar 0
Demo on SQLFiddle

How to use Mysql SUM with JOIN

I have the following tables:
purchase_tbl
id | productId | purchaseQuantity
---+-----------+-----------------
1 | 1 | 30
2 | 2 | 30
3 | 1 | 10
4 | 2 | 10
sale_tbl
id | productId | saleQuantity
---+-----------+-------------
1 | 1 | 10
2 | 2 | 10
3 | 1 | 10
4 | 2 | 10
5 | 1 | 10
6 | 2 | 10
I need to get the output as this one:
productId | totalPurchasedQuantity| totalSaleQuantity
----------+-----------------------+------------------
1 | 40 | 30
2 | 40 | 30
I'm using this query and how to get the desired result?
SELECT purchase_tbl.productId
, SUM(purchase_tbl.purchaseQuantity) AS totalPurchaseQuantity
, SUM(sale_tbl.saleQuantity) AS totalSaleQuantity
FROM purchase_tbl
JOIN sale_tbl
ON purchase_tbl.productId = sale_tbl.productId
GROUP BY purchase_tbl.productId
Current output
productId | totalPurchaseQuantity | totalSaleQuantity
----------+-----------------------+------------------
1 | 120 | 60
2 | 120 | 60
You better group then in separate query, as table have multiple records for each product, which getting cross product.
SELECT purchase.productId, totalPurchaseQuantity, totalSaleQuantity
FROM
(SELECT purchase_tbl.productId
, SUM(purchase_tbl.purchaseQuantity) AS totalPurchaseQuantity
FROM purchase_tbl
GROUP BY purchase_tbl.productId) purchase
INNER JOIN
(SELECT sale_tbl.productId
, SUM(sale_tbl.saleQuantity) AS totalSaleQuantity
FROM sale_tbl
GROUP BY sale_tbl.productId
) sale ON sale.productId= purchase.productId;
To obtain your expected result you have to do the aggregation on the individual table before joining them. Your query with be like:
SELECT A.productId, A.totalpurchaseQuantity, B.totalsaleQuantity
FROM
(SELECT productId, SUM(purchaseQuantity)
totalpurchaseQuantity FROM purchase_tbl
GROUP BY productId) A JOIN
(SELECT productId, SUM(saleQuantity)
totalsaleQuantity FROM sale_tbl
GROUP BY productId) B ON
A.productId=B.productId;

How to select the sum() of a group of rows and the sum() of another group

I have created a SQLfiddle demo with sample data and desired result here :(http://sqlfiddle.com/#!9/dfe73a/7)
sample data
-- table company
+--------+---------+
| id | name |
+--------+---------+
| 1 | foo |
| 2 | bar |
+--------+---------+
-- table sales
+--------+---------------+-----------------+
| id | company_id | total_amount |
+--------+---------------+-----------------+
| 1 | 1 | 300.0 |
| 2 | 1 | 300.0 |
| 2 | 1 | 100.0 |
+--------+---------------+-----------------+
-- table moves
+--------+---------------+-----------------+
| id | company_id | balance_move |
+--------+---------------+-----------------+
| 1 | 1 | 700.0 |
| 2 | 1 | -300.0 |
| 2 | 1 | -300.0 |
+--------+---------------+-----------------+
I need to select every company along with the sum of it's total amount of sales and the sum of it's total balance moves
desired result
+----+----------------------+---------------------+
| id | total_amount_sum | balance_move_sum |
+----+----------------------+---------------------+
| 1 | 700 | 100 |
+----+----------------------+---------------------+
| 2 | (null) | (null) |
+----+----------------------+---------------------+
I tried this SQL query
SELECT
company.id,
sum(total_amount) total_amount_sum,
sum(balance_move) balance_move_sum
FROM company
LEFT JOIN sales ON company.id = sales.company_id
LEFT JOIN moves ON company.id = moves.company_id
GROUP BY company.id
But the sum() functions add all the redundant values came from the joins which result in 2100 (700*3) for total amount and 300 (100*3) for net balance
bad SQL statement result
+----+----------------------+---------------------+
| id | total_amount_sum | balance_move_sum |
+----+----------------------+---------------------+
| 1 | 2100 | 300 |
+----+----------------------+---------------------+
| 2 | (null) | (null) |
+----+----------------------+---------------------+
Is it possible to achieve the result I want ?
You're repeating rows by doing your joins.
Company: 1 row per company
After Sales join: 3 rows per company (1x3)
After Moves join: 9 rows per company (3x3)
You end up triplicating your SUM because of this.
One way to fix is to use derived tables like this, which calculate the SUM first, then join the resulting rows 1-to-1.
SELECT
company.id,
total_amount_sum,
balance_move_sum
FROM company
LEFT JOIN (SELECT SUM(total_amount) total_amount_sum, company_id
FROM sales
GROUP BY company_id
) sales ON company.id = sales.company_id
LEFT JOIN (SELECT SUM(balance_move) balance_move_sum, company_id
FROM moves
GROUP BY company_id
) moves ON company.id = moves.company_id
Using sub-queries to calculate the two sums separately will work.
SELECT
company.id,
(Select sum(total_amount) from sales where sales.company_id = company.id) total_amount_sum,
(Select sum(balance_move) from moves where moves.company_id = company.id) balance_move_sum
FROM company

Join table with MAX value from another

select * from order
-------------------
|orderID|productID|
-------------------
| 1 | 234 |
| 2 | 234 |
| 3 | 123 |
-------------------
select * from product_supplier
-------------------------------------------
|ID|supplierID|productID|cost_price|latest|
-------------------------------------------
|1 | 1 | 234 | 1.00 | 0 |
|2 | 1 | 234 | 0.50 | 1 |
|3 | 2 | 123 | 0.75 | 1 |
-------------------------------------------
desired result
------------------------------
|orderID|productID|cost_price|
------------------------------
| 1 | 234 | 1.00 |
| 2 | 234 | 1.00 |
| 3 | 123 | 0.75 |
------------------------------
I'm looking join the two tables above to get the orderID, productID and the largest cost_price for a given productID.
SELECT orderID, productID, cost_price
FROM order LEFT JOIN product_supplier
ON order.productID=product_supplier.productID AND MAX(cost_price);
gives ERROR 1111 (HY000): Invalid use of group function
How can i restrict the joined table to one row per order, joined with the largest corresponding cost_price value in product_supplier?
The canonical way of approaching this is to use a subquery to identify the products and their maximum prices from the product_supplier table, and then to join this subquery to order to get the result set you want.
SELECT t1.orderID,
t1.productID,
COALESCE(t2.cost_price, 0.0) AS cost_price -- missing products will appear
FROM order t1 -- with a zero price
LEFT JOIN
(
SELECT productID, MAX(cost_price) AS cost_price
FROM product_supplier
GROUP BY productID
) t2
ON t1.productID = t2.productID AND
t1.cost_price = t2.cost_price
when you need to use aggregate you must use group by and in compare condtion use Having
SELECT orderID, productID, MAX(cost_price)
FROM order LEFT JOIN product_supplier
ON order.productID=product_supplier.productID having cost_price=MAX(cost_price) group by productID ;