inner join three tables results in multiplied values - mysql

Question

I'm trying to (let's say) gather a report on customers.
In that report I want to include sum of orders and ticket number for each client.
Tables:
Customer(id, name)
Order(id, customer_id, amount)
support_ticket(id, customer_id)
query:
select
c.id as 'Customer',
count(distinct t.id) as "Ticket count",
count(distinct o.id) as "Order count",
sum(o.amount) as 'Order Amount'
from customer as c
inner join `order` as o on c.id = o.customer_id
inner join support_ticket as t on c.id = t.customer_id
group by c.id
Since I join with customer.id on the two tables, I get all the rows "duplicated", since I get all possible combinations, so if the client as multiple tickets, the sum(o.amount) will we multiplied because of "duplicated rows"
sqlFiddle (mysql): http://sqlfiddle.com/#!9/ba39ba/13
sqlFiddle (pg): http://sqlfiddle.com/#!17/bc32e/7
It seems like a simple case but I've been looking at it too much I think, I can't find the proper way to do that report.
What am I doing wrong?

Answer 1

your best bet is to re-write the Aggregation off the Order table as as Derived Table;
EG
select
c.id as 'Customer',
count(distinct t.id) as "Ticket count",
o.amount as 'Order Amount' ,
o.[Order count]
from customer as c
inner join
(SELECT
o.customer_id,
sum(amount) as amount ,
count(distinct o.id) as "Order count"
from [order]
group by o.customer_id)
as o on c.id = o.customer_id
inner join support_ticket as t on c.id = t.customer_id
group by
c.id ,
o.amount ,
o.[Order count]
Note that the Derived Table Columns then are added to the group by clause at the bottom.
Cheers!

Answer 2

Just calculate order values in a sub-query and join it.
SELECT
c.id as 'Customer'
,count(DISTINCT st.id) as 'Ticket Count'
,o.`Order Count`
,o.amount as `Order Amount`
FROM customer c
INNER JOIN support_ticket st
on c.id = st.customer_id
INNER JOIN (
SELECT
customer_id
,SUM(amount) as 'amount'
,count(distinct id) as 'Order Count'
FROM `order`
group by customer_id
) o
on c.id = o.customer_id
GROUP BY c.id;

Answer 3

select c.id as 'Customer'
,t2.count_ticket as "Ticket count"
,t1.count_order as "Order count"
,t1.amount as 'Order Amount'
from customer as c
inner join (select customer_id
,count(id) as count_order
,sum(amount) as amount
from Order group by customer_id) t1
on c.id = t1.customer_id
inner join (select customer_id
,count(id) as count_ticket
from support_ticket group by customer_id) t2
on c.id = t2.customer_id

Answer 4

In cases like yours, when I think the solution of my problem should be fairly simple but I cant wrap my head around it, I tend to use a WITH clause.
Not because its better, but because it helps me to understand my code better by splitting up complexity. First I create a relatively simple temp. Solving the first part of my problem.
WITH temp AS (
SELECT
c.id AS "customer",
COUNT(DISTINCT o.id) AS "order_count",
SUM(o.amount) AS "order_amount"
FROM customer AS c
INNER JOIN "order" AS o on c.id = o.customer_id
GROUP BY c.id
)
Then I simply select the first half of my solution from temp, adding this way all intermediate results, and solve the second part of my initial sql.
SELECT
temp.customer,
COUNT(DISTINCT t.id) as "ticket_count",
temp.order_count,
temp.order_amount
FROM temp
INNER JOIN support_ticket as t on temp.customer = t.customer_id
GROUP BY temp.customer, temp.order_count, temp.order_amount
The principle is the same like in all previous answers, but SELECTS are separated and I can check them fast, and continue on if I'm happy with parts of the solution.