How to avoid DRY in PostgreSQL with a nested query - sql

Question

This query works fine in this way:
SELECT DISTINCT inv.move_id,
inv.client_id,
inv.reference,
inv.date_inv,
inv.amount,
inv.state,
inv.client_type,
inv.agent_id,
(
SELECT SUM(nc.amount) AS total
FROM invoice nc
WHERE nc.journal = 15
AND nc.ref_move = inv.move_id
AND nc.state = inv.state
) AS credit, (
SELECT SUM(ip.amount)
FROM inv_payment ip
WHERE inv.reference = ip.reference
) AS paid,
(inv.amount) - (
coalesce((
SELECT SUM(nc.amount) AS total
FROM invoice nc
WHERE nc.journal = 15
AND nc.ref_move = inv.move_id
AND nc.state = inv.state
), 0) + coalesce((
SELECT SUM(ip.amount)
FROM inv_payment ip
WHERE inv.reference = ip.reference
), 0)
) AS outstanding
FROM invoice inv
INNER JOIN inv_move im ON im.id = inv.move_id
INNER JOIN client cl ON cl.id = inv.client_id
WHERE inv.agent_id = '%'
AND inv.date_inv
BETWEEN '%' AND '%'
AND inv.state = 'open'
ORDER BY inv.agent_id
I'm trying to DRY doing this:
SELECT DISTINCT inv.move_id,
inv.client_id,
inv.reference,
inv.date_inv,
inv.amount,
inv.state,
inv.client_type,
inv.agent_id,
(
SELECT SUM(nc.amount) AS total
FROM invoice nc
WHERE nc.journal = 15
AND nc.ref_move = inv.move_id
AND nc.state = inv.state
) AS credit, (
SELECT SUM(ip.amount)
FROM inv_payment ip
WHERE inv.reference = ip.reference
) AS paid,
(inv.amount) - (
coalesce((credit), 0) + coalesce((paid), 0)
) AS outstanding
FROM invoice inv
INNER JOIN inv_move im ON im.id = inv.move_id
INNER JOIN client cl ON cl.id = inv.client_id
WHERE inv.agent_id = '%'
AND inv.date_inv
BETWEEN '%' AND '%'
AND inv.state = 'open'
ORDER BY inv.agent_id
But I'm getting an error in this way.
Because I want to use the alias column for operation values in other column.
How could I do it?
I was trying to use a WITH PostgreSQL statement but I couldn't find a better solution in that way.

Answer 1

You CANT use a field calculated on the same level.
SELECT 1 + 2 as three, three * 2 as six
^^^^^
undefined
So either you create a subquery.
SELECT three*2 as six
FROM ( SELECT 1 + 2 as three
FROM table
) T
or for your example use a JOIN
using CTE:
WITH cte as (
SELECT ref_move, state, SUM(nc.amount) AS total
FROM invoice nc
WHERE nc.journal = 15
GROUP BY ref_move, state
)
SELECT *
FROM invoice i
JOIN cte c
ON i.ref_move= c.ref_move
AND i.state = c.state

Answer 2

Use subquery in FROM:
SELECT *, amount-credit-paid AS outstanding FROM (
SELECT DISTINCT inv.move_id,
inv.client_id,
inv.reference,
inv.date_inv,
inv.amount,
inv.state,
inv.client_type,
inv.agent_id,
(
SELECT SUM(nc.amount) AS total
FROM invoice nc
WHERE nc.journal = 15
AND nc.ref_move = inv.move_id
AND nc.state = inv.state
) AS credit, (
SELECT SUM(ip.amount)
FROM inv_payment ip
WHERE inv.reference = ip.reference
) AS paid
FROM invoice inv ...
) AS _;