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
Related
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
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
I don't know how to explain the scenario using words. So am writing the examples:
I have a table named tblType:
type_id | type_name
---------------------
1 | abb
2 | cda
3 | edg
4 | hij
5 | klm
And I have another table named tblRequest:
req_id | type_id | user_id | duration
-------------------------------------------
1 | 4 | 1002 | 20
2 | 1 | 1002 | 60
3 | 5 | 1008 | 60
....
So what am trying to do is, fetch the SUM() of duration for each type, for a particular user.
This is what I tried:
SELECT
SUM(r.`duration`) AS `duration`,
t.`type_id`,
t.`type_name`
FROM `tblRequest` AS r
LEFT JOIN `tblType` AS t ON r.`type_id` = t.`type_id`
WHERE r.`user_id` = '1002'
GROUP BY r.`type_id`
It might return something like this:
type_id | type_name | duration
-------------------------------
1 | abb | 60
4 | hij | 20
It works. But the issue is, I want to get 0 as value for other types that doesn't have a row in tblRequest. I mean I want the output to be like this:
type_id | type_name | duration
-------------------------------
1 | abb | 60
2 | cda | 0
3 | edg | 0
4 | hij | 20
5 | klm | 0
I mean it should get the rows of all types, but 0 as value for those type that doesn't have a row in tblRequest
You could perform the aggregation on tblRequest and only then join it, using a left join to handle missing rows and coalesce to convert the nulls to 0s:
SELECT t.type_id, type_name, COALESCE(sum_duration, 0) AS duration
FROM tblType t
LEFT JOIN (SELECT type_id, SUM(duration) AS sum_duration
FROM tblRequest
WHERE user_id = '1002'
GROUP BY type_id) r ON t.type_id = r.type_id
Select a.type_id, isnull(sum(b.duration), 0)
From tblType a Left Outer Join tblRequest b
ON a.type_id = b.type_id and b.user_id = 1002
Group by a.type_id
Help please, I have a table like this:
| ID | userId | amount | type |
-------------------------------------
| 1 | 10 | 10 | expense |
| 2 | 10 | 22 | income |
| 3 | 3 | 25 | expense |
| 4 | 3 | 40 | expense |
| 5 | 3 | 63 | income |
I'm looking for a way to use one query and retrive the balance of each user.
The hard part comes when the amounts has to be added on expenses and substracted on incomes.
This would be the result table:
| userId | balance |
--------------------
| 10 | 12 |
| 3 | -2 |
You need to get each totals of income and expense using subquery then later on join them so you can subtract expense from income
SELECT a.UserID,
(b.totalIncome - a.totalExpense) `balance`
FROM
(
SELECT userID, SUM(amount) totalExpense
FROM myTable
WHERE type = 'expense'
GROUP BY userID
) a INNER JOIN
(
SELECT userID, SUM(amount) totalIncome
FROM myTable
WHERE type = 'income'
GROUP BY userID
) b on a.userID = b.userid
SQLFiddle Demo
This is easiest to do with a single group by:
select user_id,
sum(case when type = 'income' then amount else - amount end) as balance
from t
group by user_id
You could have 2 sub-queries, each grouped by id: one sums the incomes, the other the expenses. Then you could join these together, so that each row had an id, the sum of the expenses and the sum of the income(s), from which you can easily compute the balance.
I have scheduled payments, so I have these tables:
customer
+id
paymentSchedule
+id
+customer_id
+amount //total price
+dueDate //date to be paid
payments
+id
+date
+customer_id
+paymentSchedule_id
+amount //amount paid, it can be a partial payment
How do I write a query to get Today's due amount by customer.
I mean I need to join the tables (thats my main problem) and then substract the
sum of the payments.mount minus the sum of the scheduledPaymens.amount
but.. how?
Thanks in advance
This is probably not 100%, but should be pretty solid to help you tweak:
SELECT customer_id, (due.amount - paid.amount) as amountDue
FROM
(SELECT customer_id, SUM(amount) as amount
FROM paymentSchedule
WHERE dateDate <= getDate()
and customer_id = #custid) as due
LEFT JOIN
(SELECT customer_id, SUM(amount) as amount
FROM payments
WHERE customer_id = #custid) as paid ON paid.customer_id = due.customer_id
Ok, this is how I understood the problem. I simplified the tables because they where just complicating things, and adding dates is just straight forward.
PaymentSchedule
+----+-------------+-----------------+
| id | customer_id | original_amount |
+----+-------------+-----------------+
| 1 | Tom | 100 |
| 2 | Tom | 200 |
| 3 | Tom | 300 |
| 4 | Moe | 400 |
+----+-------------+-----------------+
Payments
+----+--------------------+-------------+
| id | paymentSchedule_id | paid_amount |
+----+--------------------+-------------+
| 1 | 1 | 70 |
| 2 | 2 | 150 |
| 3 | 2 | 50 |
| 4 | 4 | 300 |
| 5 | 4 | 25 |
+----+--------------------+-------------+
Result of query
+-------------+-------------------+-----------------+-----------+----------------+
| CUSTOMER_ID | PAYMENTSCHEDULEID | ORIGINAL_AMOUNT | TOTALPAID | PENDINGPAYMENT |
+-------------+-------------------+-----------------+-----------+----------------+
| Tom | 1 | 100 | 70 | 30 |
| Tom | 2 | 200 | 200 | 0 |
| Tom | 3 | 300 | 0 | 300 |
| Moe | 4 | 400 | 325 | 75 |
+-------------+-------------------+-----------------+-----------+----------------+
Query with double select
select *, s.original_amount - s.TotalPaid as PendingPayment from (
select
ps.customer_id, ps.id as PaymentScheduleId, ps.original_amount,
coalesce(sum(p.paid_amount), 0) as TotalPaid
from paymentSchedule ps
left join payments p on p.paymentSchedule_id = ps.id
group by ps.customer_id, PaymentScheduleId, ps.original_amount
) as S
Query with single select
select
ps.customer_id, ps.id as PaymentScheduleId, ps.original_amount,
coalesce(sum(p.paid_amount), 0) as TotalPaid,
ps.original_amount - coalesce(sum(p.paid_amount), 0) as PendingPayment
from paymentSchedule ps
left join payments p on p.paymentSchedule_id = ps.id
group by ps.customer_id, PaymentScheduleId, ps.original_amount
The result of both queries is the same. I just wonder which one runs faster. You can try both and tell us :)
Let me know if this this is the result you expected
Something like this should be a good starting point for you to tweak.
SELECT c.*
FROM customer c
INNER JOIN paymentSchedule ps
ON c.id = ps.customer_id
LEFT JOIN payments p
ON ps.id = p.paymentSchedule_id
WHERE ps.dueDate = 'This depends on how you store dueDate'
AND ps.amount - p.amount > 0