SQL Inner Join Group By - mysql

Question

Table - Customers
id FirstName LastName
23 James Smith
24 Tom Raven
25 Bob King
Table - Orders
id CustomerID
30 23
31 24
32 23
33 25
Table - Products
OrderID Product1 Product2 Product3
30 1 0 0
31 2 0 0
32 0 1 1
33 2 1 0
I want to count the total number of Products for each Customer so answers would be:
CustomerID FirstName LastName Total
23 James Smith 3
24 Tom Raven 2
25 Bob King 3
So far I've got for the SQL query:
SELECT Customers.id, Orders.id,
FROM Customers
INNER JOIN Orders ON Customers.id = Orders.CustomerID
INNER JOIN Products ON Orders.id = Products.OrderID
Not sure how to count up the products though.
Any help would be much appreciated.

Answer 1

select c.id as CustomerID
,(sum(p.Product1) + sum(p.Product2) + sum(p.Product3)) as ProductCount
from Customers c
inner join Orders o on c.id = o.CustomerID
inner join Products p on o.id = p.OrderID
group by c.id

Answer 2

You could use
SELECT c.id as CustomerID, c.firstname, c.lastname,
sum(p.Product1 + p.Product2 + p.Product3) as total
FROM Customers c
INNER JOIN Orders o
ON c.id=o.CustomerID
INNER JOIN Products p
ON o.id=p.OrderID
group by c.id,c.firstname, c.lastname;
And as #Thorsten Kettner's comment, you should consider normalizing your table design.

Answer 3

Add a GROUP BY clause! Normally it includes the selected columns that are not arguments to a set function.
SELECT O.CustomerID, C.FirstName, C.LastName, count(*) as Total
FROM Customers C
INNER JOIN Orders O ON C.id = O.CustomerID
INNER JOIN Products P ON O.id = P.OrderID
GROUP BY O.CustomerID, C.FirstName, C.LastName
PS. Now using table aliases, to save some typing.

Answer 4

SELECT Customers.id, Customers.firstname, Customers.lastname,count(*) as total FROM Customers
INNER JOIN Orders
ON Customers.id=Orders.CustomerID
INNER JOIN Products
ON Orders.id=Products.OrderID
group by Customers.id,Customers.firstname, Customers.lastname

Answer 5

A more conventional approach might be as follows:
DROP TABLE IF EXISTS customers;
CREATE TABLE customers
(customer_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,firstname VARCHAR(20) NOT NULL
,lastname VARCHAR(20) NOT NULL
);
INSERT INTO customers VALUES
(23,'James','Smith'),
(24,'Tom','Raven'),
(25,'Bob','King');
DROP TABLE IF EXISTS orders;
CREATE TABLE orders
(order_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,customer_id INT NOT NULL
);
INSERT INTO orders VALUES
(30 ,23),
(31 ,24),
(32 ,23),
(33 ,25);
DROP TABLE IF EXISTS order_detail;
CREATE TABLE order_detail
(order_id INT NOT NULL
,product_id INT NOT NULL
,quantity INT NOT NULL
,PRIMARY KEY(order_id,product_id)
);
INSERT INTO order_detail VALUES
(30 ,1 ,1),
(31 ,1 ,2),
(33 ,1 ,2),
(32 ,2 ,1),
(33 ,2 ,1),
(32 ,3 ,1);
SELECT c.*
, SUM(od.quantity) total
FROM customers c
JOIN orders o
ON o.customer_id = c.customer_id
JOIN order_detail od
ON od.order_id = o.order_id
GROUP
BY c.customer_id;
+-------------+-----------+----------+-------+
| customer_id | firstname | lastname | total |
+-------------+-----------+----------+-------+
| 23 | James | Smith | 3 |
| 24 | Tom | Raven | 2 |
| 25 | Bob | King | 3 |
+-------------+-----------+----------+-------+

Answer 6

SELECT DISTINCT CustomerID, FirstName, LastName, Total
FROM ( SELECT id AS CustomerID, FirstName, LastName FROM Customers ) AS c
NATURAL JOIN
( SELECT id AS OrderID, CustomerID FROM Orders ) AS o
NATURAL JOIN
( SELECT OrderID, SUM( Product1 + Product2 + Product3 ) AS Total
FROM Products
GROUP
BY OrderID ) AS p;