How can I SUM distinct records in a Postgres database where there are duplicate records? - postgresql

Question

Imagine a table that looks like this:
The SQL to get this data was just SELECT *
The first column is "row_id" the second is "id" - which is the order ID and the third is "total" - which is the revenue.
I'm not sure why there are duplicate rows in the database, but when I do a SUM(total), it's including the second entry in the database, even though the order ID is the same, which is causing my numbers to be larger than if I select distinct(id), total - export to excel and then sum the values manually.
So my question is - how can I SUM on just the distinct order IDs so that I get the same revenue as if I exported to excel every distinct order ID row?
Thanks in advance!

Answer 1

Easy - just divide by the count:
select id, sum(total) / count(id)
from orders
group by id
See live demo.
Also handles any level of duplication, eg triplicates etc.

Answer 2

You can try something like this (with your example):
Table
create table test (
row_id int,
id int,
total decimal(15,2)
);
insert into test values
(6395, 1509, 112), (22986, 1509, 112),
(1393, 3284, 40.37), (24360, 3284, 40.37);
Query
with distinct_records as (
select distinct id, total from test
)
select a.id, b.actual_total, array_agg(a.row_id) as row_ids
from test a
inner join (select id, sum(total) as actual_total from distinct_records group by id) b
on a.id = b.id
group by a.id, b.actual_total
Result
| id | actual_total | row_ids |
|------|--------------|------------|
| 1509 | 112 | 6395,22986 |
| 3284 | 40.37 | 1393,24360 |
Explanation
We do not know what the reasons is for orders and totals to appear more than one time with different row_id. So using a common table expression (CTE) using the with ... phrase, we get the distinct id and total.
Under the CTE, we use this distinct data to do totaling. We join ID in the original table with the aggregation over distinct values. Then we comma-separate row_ids so that the information looks cleaner.
SQLFiddle example
http://sqlfiddle.com/#!15/72639/3

Answer 3

Create custom aggregate:
CREATE OR REPLACE FUNCTION sum_func (
double precision, pg_catalog.anyelement, double precision
)
RETURNS double precision AS
$body$
SELECT case when $3 is not null then COALESCE($1, 0) + $3 else $1 end
$body$
LANGUAGE 'sql';
CREATE AGGREGATE dist_sum (
pg_catalog."any",
double precision)
(
SFUNC = sum_func,
STYPE = float8
);
And then calc distinct sum like:
select dist_sum(distinct id, total)
from orders
SQLFiddle

Answer 4

You can use DISTINCT in your aggregate functions:
SELECT id, SUM(DISTINCT total) FROM orders GROUP BY id
Documentation here: https://www.postgresql.org/docs/9.6/static/sql-expressions.html#SYNTAX-AGGREGATES

Answer 5

If we can trust that the total for 1 order is actually 1 row. We could eliminate the duplicates in a sub-query by selecting the the MAX of the PK id column. An example:
CREATE TABLE test2 (id int, order_id int, total int);
insert into test2 values (1,1,50);
insert into test2 values (2,1,50);
insert into test2 values (5,1,50);
insert into test2 values (3,2,100);
insert into test2 values (4,2,100);
select order_id, sum(total)
from test2 t
join (
select max(id) as id
from test2
group by order_id) as sq
on t.id = sq.id
group by order_id
sql fiddle

Answer 6

In difficult cases:
select
id,
(
SELECT SUM(value::int4)
FROM jsonb_each_text(jsonb_object_agg(row_id, total))
) as total
from orders
group by id

Answer 7

I would suggest just use a sub-Query:
SELECT "a"."id", SUM("a"."total")
FROM (SELECT DISTINCT ON ("id") * FROM "Database"."Schema"."Table") AS "a"
GROUP BY "a"."id"
The Above will give you the total of each id
Use below if you want the full total of each duplicate removed:
SELECT SUM("a"."total")
FROM (SELECT DISTINCT ON ("id") * FROM "Database"."Schema"."Table") AS "a"

Answer 8

Using subselect (http://sqlfiddle.com/#!7/cef1c/51):
select sum(total) from (
select distinct id, total
from orders
)
Using CTE (http://sqlfiddle.com/#!7/cef1c/53):
with distinct_records as (
select distinct id, total from orders
)
select sum(total) from distinct_records;