I'm building a MySQL query but I can't seem to get it right.
I have four tables:
- customers
- orders
- sales_rates
- purchase_rates
There is a 1:n relation 'customernr' between customers and orders.
There is a 1:n relation 'ordernr' between orders and sales_rates.
There is a 1:n relation 'ordernr' between orders and purchase_rates.
What I would like to do is produce an output of all customers with their total purchase and sales amounts.
So far I have the following query.
SELECT c.customernr, c.customer_name, SUM(sr.sales_price) AS sales_price, SUM(pr.purchase_price) AS purchase_price
FROM orders o, customers c, sales_rates sr, purchase_rates pr
WHERE o.customernr = c.customernr
AND o.ordernr = sr.ordernr
AND o.ordernr = pr.ordernr
GROUP BY c.customer_name
The result of the sales_price and purchase_price is far too high. I seem to be getting double counts. What am I doing wrong? Is it possible to perform this in a single query?
Thank for your response!
It seems that the problem is that when you join the orders table to the tables with sales rates and purchase rates, you are getting the cartesian product of these two latter tables. I.e each row in these two tables are repeated once for each correponding row in the other table. The following query should solve this problem by summing the rates for each order before joining the sales rates and purchase rates to the other tables:
SELECT c.customernr, c.customer_name,
SUM(sr.sales_price) AS sales_price,
SUM(pr.purchase_price) AS purchase_price
FROM customers c
INNER JOIN orders o
ON o.customernr = c.customernr
LEFT JOIN (SELECT ordernr, SUM sales_price) AS sales_price
FROM sales_rates
GROUP BY ordernr) sr
ON sr.ordernr = o.ordernr
LEFT JOIN (SELECT ordernr, SUM(purchase_price) AS purchase_price
FROM purchase_rates
GROUP BY ordernr) pr
ON pr.ordernr = o.ordernr
GROUP BY c.customernr, c.customer_name;
It doesn't look like you are grouping by the customer. C.customerid or something like that.
Related
I have two MySQL-tables:
Persons (pid,name,companyID,companyName)
Orders (oid,companyID,details)
Now I want to count the number of order_id for each companyName as following:
Name Total
-------------------
CompanyName1 : 1200
CompanyName2 : 758
CompanyName3 : 11
I used this query but it's not working properly.
SELECT count(o.oid) as total,p.companyName
FROM orders as o, persons as p
WHERE o.companyID = p.companyID
GROUP BY p.companyName
Use join and group the result by p.companyID
SELECT p.companyName, count(o.oid) as total
FROM orders as o join persons as p
on o.companyID = p.companyID
GROUP BY p.companyID
If you are missing the companies without any orders you can use a left join.
SELECT p.companyName, count(*) as total
FROM persons p
LEFT JOIN orders o ON o.companyID = p.companyID
GROUP BY p.companyID, p.companyName
Please do not use the old, legacy join syntax any more - it is outdated since 1992.
Your data model looks messed up. That you have company ids and names in the person table but no corresponding companies table is highly suspicious.
In any case, presumably there can be multiple rows per company. You can condense the persons table and then join:
SELECT c.companyName, COUNT(*) as total
FROM orders o JOIN
(SELECT DISTINCT companyId, companyName
FROM persons p
) c
ON o.companyID = c.companyID
GROUP BY c.companyName;
However, you should fix the data model so you have a real bona fide companies table -- especially because you seem to care about that entity.
first sorry, i don't have fluid english.
I want select 3 rows in 3 different tables, two of them without Foreing Key/relation.
Need to select amount of customers in each store, and total amount of payments in these stores in one query.
Here are the tables:
Customers
Stores
Payments
I have tried these querys to get payments for each store and customers for each store, but don't know how can unify in one query:
Payments/store
SELECT count(a.payment_id) as alquileres, b.store_id
FROM customer b, payment a
WHERE a.customer_id = b.customer_id
GROUP BY b.store_id;
customers/store
SELECT count(customer_id), store_id
FROM customer
GROUP BY store_id;
But when I add count(customer_id) in unique query don't have same results.
You can use one query:
select c.store_id,
count(distinct c.customer_id) as num_customers,
count(p.payment_id) as num_payments
from customers c left join
payments p
on p.customer_id = c.customer_id
group by c.store_id;
I have the following three tables:
I have Following Query to Join Above 3 Tables
customer.customer_id,
customer.name,
SUM(sales.total),
sales.created_at,
SUM(sales_payments.amount)
FROM
sales INNER JOIN customer ON customer.customer_id = sales.customer_id
INNER JOIN sales_payments ON sales.customer_id = sales_payments.customer_id
WHERE sales.created_at ='2020-04-03'
GROUP By customer.name
Result for Above Query is given below
Sum of sales.total is double of the actual sum of sales.total column which has 2-row count, I need to have the actual SUM of that column, without doubling the SUM of those rows, Thank you, for your help in advance..
PROBLEM
The problem here is that there are consecutive inner joins and the number of rows getting fetched in the second inner join is not restricted. So, as we have not added a condition on sales_payment_id in the join between the sales and sales_payment tables, one row in sales table(for customer_id 2, in this case) would be mapped to 2 rows in the payment table. This causes the same values to be reconsidered.
In other words, the mapping for customer_id 2 between the 3 tables is 1:1:2 rather than 1:1:1.
SOLUTION
Solution 1 : As mentioned by Gordon, you could first aggregate the amount values of the sales_payments table and then aggregate the values in sales table.
Solution 2 : Alternatively (IMHO a better approach), you could add a foreign key between sales and sales_payment tables. For example, the sales_payment_id column of sales_payment table can be introduced in the sales table as well. This would facilitate the join between these tables and reduce additional overheads while querying data.
The query would then look like:
`SELECT c.customer_id,
c.name,
SUM(s.total),
s.created_at,
SUM(sp.amount)
FROM customer c
INNER JOIN sales s
ON c.customer_id = s.customer_id
INNER JOIN sales_payments sp
ON c.customer_id = sp.customer_id
AND s.sales_payments_id = sp.sales_payments_id
WHERE s.created_at ='2020-04-03'
GROUP BY c.customer_id,
c.name,
s.created_at ;`
Hope that helps!
You have multiple rows for sales_payments and sales per customer. You need to pre-aggregate to get the right value:
SELECT c.customer_id, c.name, s.created_at, s.total, sp.amount
FROM customer c JOIN
(SELECT s.customer_id, s.created_at, SUM(s.total) as total
FROM sales s
WHERE s.created_at ='2020-04-03'
GROUP BY s.customer_id, s.created_at
) s
ON c.customer_id = s.customer_id JOIN
(SELECT sp.customer_id, SUM(sp.amount) as amount
FROM sales_payments sp
GROUP BY sp.customer_id
) sp
ON s.customer_id = sp.customer_id
I'm trying to create a query to find what is the total amount owed by each customer to the company. It is the GROUP BY customerNumber in the sub query that is creating the problem.
SELECT customerName,
customers.customerNumber,
SUM(quantityOrdered * priceEach) - ( SELECT SUM(amount) AS MoneyPayed FROM payments GROUP BY customerNumber ) AS AmountOwed
FROM payments
INNER JOIN customers ON payments.customerNumber = customers.customerNumber
INNER JOIN orders ON customers.customerNumber = orders.customerNumber
INNER JOIN orderdetails ON orders.orderNumber = orderdetails.orderNumber
GROUP BY customerNumber;
The tables I'm trying to link are payments and orderdetails.
When I get rid of the GROUP BY I get results in negatives as the total SUM of amount is subtracted from each row of SUM(quantityOrdered * priceEach).
How can I change this so that I can return multiple rows from payments to subtract from SUM(quantityOrdered * priceEach) from the order details table.
Link to DB as StackOverflow doesn't allow me to post images
Thanks for help, sorry if format is bad, this is my first post.
You will need a couple of subqueries to meet your requirement. Let us break it down.
First, you need the total value of orders from each customer. You're very close to the correct query for that. It should be
SELECT orders.customerNumber,
SUM(orderdetails.quantityOrdered * orderdetails.priceEach) owed
FROM orders
JOIN orderdetails ON orders.orderNumber = orderdetails.orderNumber
GROUP BY orders.customerNumber
This subquery's result set gives customerNumber and owed, the amount owed. Notice that orders::orderdetails is a one::many relationship, so we're sure we're counting each detail just once, so the SUMs will be correct.
Next we need the amount paid by each customer. This subquery is fairly simple.
SELECT customerNumber,
SUM(amount) paid
FROM payments
GROUP BY customerNumber
Now for the operation you're missing in your question: we need to join these two subqueries to your customers table.
SELECT customers.customerName, customers.customerNumber
owed.owed - paid.paid balance
FROM customers
LEFT JOIN (
SELECT orders.customerNumber,
SUM(orderdetails.quantityOrdered * orderdetails.priceEach) owed
FROM orders
JOIN orderdetails ON orders.orderNumber = orderdetails.orderNumber
GROUP BY orders.customerNumber
) paid ON customers.customerNumber = paid.customerNumber
LEFT JOIN (
SELECT customerNumber,
SUM(amount) paid
FROM payments
GROUP BY customerNumber
) owed ON customers.customerNumber = owed.customerNumber
See how this works? We join a table and two subqueries. Each subquery has either zero or one rows for each row in the table, so we need not use SUMs or GROUP BY in the outer query.
There's only one complication left: what if a customer has never paid anything? Then the value of paid.paid will be NULL after the LEFT JOIN operation. That will force the value of owed - paid to be NULL. So we need more smarts in the SELECT statement to yield correct sums.
SELECT customers.customerName, customers.customerNumber
COALESCE(owed.owed,0) - COALESCE(paid.paid,0) balance
...
COALESCE(a,b) is equivalent to if a is not null then a else b.
Pro tip In queries or subqueries with JOIN operations, always mention table.column instead of just column. The next person to work on your query will thank you.
I have these two tables:
invoices (contains id field)
contracts (contains fk to invoice + 'code' field)
Let's say I have one record in the invoices table and two records in the contracts table. Both records in the contracts table point to the same invoice record.
Desire: I'd like to count the amount of invoices.
What I've got so far:
select
c.code, count(*)
from
invoices i
join
contracts c
on
c.invoice_id = i.id
group by
c.code
Although the count shows two instead of the desired 1. I understand that this is because of the join on the contract table, but not sure how to fix this.
Try with COUNT(DISTINCT i.id); it should count the different invoice id's in the resultset.
select
c.code, count(distinct i.id)
from
invoices i
join
contracts c
on
c.invoice_id = i.id
group by
c.code
You want the number of invoices?
select count(*)
from invoices i
You want invoices with contracts?
select count( distinct c.invoice_id)
from
contracts
Your code finds number of contracts per each invoice