Psql : Get Min, Max and Count records for each partner' invoice, and last payment - sql

Question

I have a table invoice like this :
id, partner_id, number, invoice_date
And a Payment Table like this:
id, payment_date, partner_id
I want to get min and max for both number & invoice_date, and count invoices, and last payment for each partner, something like this :
partner_id, min number, min date, max number, max date, count, last_pay
1, INV-2017-003, 02-01-2017, INV-2020-010, 01-01-2020, 142, 02-12-2019
5, INV-2019-124, 05-03-2019, INV-2020-005, 01-01-2020, 150, 01-01-2020
....

Answer 1

You can join those three tables including partners and grouping by parners' id column along with related aggregations :
select pr.id, min(invoice_date), max(invoice_date), count(*), max(payment_date) as last_pay
from partners pr
left join invoices i on i.partner_id = pr.id
left join payments p on p.partner_id = pr.id
group by pr.id
Update : You can use min() over (), max() over () and row_number() analytic functions to get the desired code depending on max and min dates :
select *
from
(
select pr.id,
min(invoice_date) over (partition by pr.id order by invoice_date) as min_invoice_date,
max(invoice_date) over (partition by pr.id order by invoice_date desc) as max_invoice_date,
max(code) over (partition by pr.id order by invoice_date desc) as max_code,
min(code) over (partition by pr.id order by invoice_date) as min_code,
count(*) over (partition by pr.id) as cnt,
max(payment_date) over (partition by pr.id) as last_pay,
row_number() over (partition by pr.id order by invoice_date desc) as rn
from partners pr
left join invoices i on i.partner_id = pr.id
left join payments p on p.partner_id = pr.id
) q
where rn = 1

Answer 2

Why isn't this simple aggregation?
select i.partner_id,
min(i.number) as min_number,
min(i.invoice_date) as min_invoice_date,
max(i.number) as min_number,
max(i.invoice_date) as min_invoice_date,
count(distinct i.invoice_id) as num_invoices,
max(p.payment_date) as max_payment_date
from invoices i left join
payments p
on p.invoice_id = i.invoice_id
group by i.partner_id;
If you want the number on the earliest invoice (and min() doesn't work), then you can do this with a "first" aggregation function. Unfortunately, Postgres doesn't directly support one. But it does through array functions:
select i.partner_id,
(array_agg(i.number order by i.invoice_date asc))[1] as min_number,
min(i.invoice_date) as min_invoice_date,
(array_agg(i.number order by i.invoice_date desc))[1] as min_number,
max(i.invoice_date) as min_invoice_date,
count(distinct i.invoice_id) as num_invoices,
max(p.payment_date) as max_payment_date
from invoices i left join
payments p
on p.partner_id = i.partner_id
group by i.partner_id;

Answer 3

This is similar to #BarbarosÖzhan, but calculates the min/max before the join (if you got multiple rows per partner for both invoices and payments the COUNT will be wrong otherwise). Additionally there's only a single PARTTION/ORDER which should result in a more efficient plan.
SELECT i.*, p.last_pay
FROM
( -- 1st row has all the min values = filtered using row_number
SELECT
partner_id
,number AS min_code
,invoice_date AS min_invoice_date
-- value from row with max date
,Last_Value(number)
Over (PARTITION BY partner_id
ORDER BY invoice_date
ROWS BETWEEN Unbounded Preceding AND Unbounded Following) AS max_code
,Last_Value(invoice_date)
Over (PARTITION BY partner_id
ORDER BY invoice_date
ROWS BETWEEN Unbounded Preceding AND Unbounded Following) AS max_invoice_date
,Count(*)
Over (PARTITION BY partner_id) AS Cnt
,Row_Number()
Over (PARTITION BY partner_id ORDER BY invoice_date) AS rn
FROM invoices
) AS i
LEFT JOIN
( -- max payment date per partner
SELECT partner_id, Max(payment_date) AS last_pay
FROM payments
GROUP BY partner_id
) AS p
ON p.partner_id = i.partner_id
WHERE i.rn = 1