SQL JOIN, GROUP BY on three tables to get totals

Question

I've inherited the following DB design. Tables are:

customers

customerid
customernumber

invoices

invoiceid
amount

invoicepayments

invoicepaymentid
invoiceid
paymentid

payments

paymentid
customerid
amount

My query needs to return invoiceid, the invoice amount (in the invoices table), and the amount due (invoice amount minus any payments that have been made towards the invoice) for a given customernumber. A customer may have multiple invoices.

The following query gives me duplicate records when multiple payments are made to an invoice:

SELECT i.invoiceid, i.amount, i.amount - p.amount AS amountdue
FROM invoices i
LEFT JOIN invoicepayments ip ON i.invoiceid = ip.invoiceid
LEFT JOIN payments p ON ip.paymentid = p.paymentid
LEFT JOIN customers c ON p.customerid = c.customerid
WHERE c.customernumber = '100'

How can I solve this?

Answer 1

I am not sure I got you but this might be what you are looking for:

SELECT i.invoiceid, sum(case when i.amount is not null then i.amount else 0 end), sum(case when i.amount is not null then i.amount else 0 end) - sum(case when p.amount is not null then p.amount else 0 end) AS amountdue
FROM invoices i
LEFT JOIN invoicepayments ip ON i.invoiceid = ip.invoiceid
LEFT JOIN payments p ON ip.paymentid = p.paymentid
LEFT JOIN customers c ON p.customerid = c.customerid
WHERE c.customernumber = '100'
GROUP BY i.invoiceid

This would get you the amounts sums in case there are multiple payment rows for each invoice

Answer 2

First of all, shouldn't there be a CustomerId in the Invoices table? As it is, You can't perform this query for Invoices that have no payments on them as yet. If there are no payments on an invoice, that invoice will not even show up in the ouput of the query, even though it's an outer join...

Also, When a customer makes a payment, how do you know what Invoice to attach it to ? If the only way is by the InvoiceId on the stub that arrives with the payment, then you are (perhaps inappropriately) associating Invoices with the customer that paid them, rather than with the customer that ordered them... . (Sometimes an invoice can be paid by someone other than the customer who ordered the services)

Answer 3

Thank you very much for the replies!

Saggi Malachi, that query unfortunately sums the invoice amount in cases where there is more than one payment. Say there are two payments to a $39 invoice of $18 and $12. So rather than ending up with a result that looks like:

1   39.00 9.00

You'll end up with:

1   78.00 48.00

Charles Bretana, in the course of trimming my query down to the simplest possible query I (stupidly) omitted an additional table, customerinvoices, which provides a link between customers and invoices. This can be used to see invoices for which payments haven't made.

After much struggling, I think that the following query returns what I need it to:

SELECT DISTINCT i.invoiceid, i.amount, ISNULL(i.amount - p.amount, i.amount) AS amountdue
FROM invoices i
LEFT JOIN invoicepayments ip ON i.invoiceid = ip.invoiceid
LEFT JOIN customerinvoices ci ON i.invoiceid = ci.invoiceid
LEFT JOIN (
  SELECT invoiceid, SUM(p.amount) amount
  FROM invoicepayments ip 
  LEFT JOIN payments p ON ip.paymentid = p.paymentid
  GROUP BY ip.invoiceid
) p
ON p.invoiceid = ip.invoiceid
LEFT JOIN payments p2 ON ip.paymentid = p2.paymentid
LEFT JOIN customers c ON ci.customerid = c.customerid
WHERE c.customernumber='100'

Would you guys concur?

Answer 4

I have a tip for those, who want to get various aggregated values from the same table.

Lets say I have table with users and table with points the users acquire. So the connection between them is 1:N (one user, many points records).

Now in the table 'points' I also store the information about for what did the user get the points (login, clicking a banner etc.). And I want to list all users ordered by SUM(points) AND then by SUM(points WHERE type = x). That is to say ordered by all the points user has and then by points the user got for a specific action (eg. login).

The SQL would be:

SELECT SUM(points.points) AS points_all, SUM(points.points * (points.type = 7)) AS points_login
FROM user
LEFT JOIN points ON user.id = points.user_id
GROUP BY user.id

The beauty of this is in the SUM(points.points * (points.type = 7)) where the inner parenthesis evaluates to either 0 or 1 thus multiplying the given points value by 0 or 1, depending on wheteher it equals to the the type of points we want.