Reputation: 10805
Im looking similar thing in PostgreSQL I can do in Excel using FLOOR function.
For example:
=FLOOR(199999;100000)
Will give me 100 000
I tried in pgsql:
SELECT round(199999 ,-5)
But this rounds number UP -> 200000.
Upvotes: 0
Views: 2497
Reputation: 14861
Good function, but it can be taken a step further and create a SQL function.
create or replace
function floor_precision(_value int, _precision int)
returns integer
language sql
immutable strict
as $$
select round(_value - (pow(10, _precision) / 2)::int, _precision * -1)::int;
$$;
Upvotes: 0
Reputation: 23676
round()
always rounds 0-4 down and 5-9 up, so you cannot achieve a floor round with this.
This would be be the way to do it.
SELECT
floor(199999 / 100000) * 100000
which equals
SELECT
floor(199999 / pow(10,5)) * pow(10,5)
With that you could write your own function:
CREATE OR REPLACE FUNCTION floor_precision(_value int, _precision int)
RETURNS integer AS $$
DECLARE
rounded integer;
BEGIN
SELECT floor(_value / pow(10, _precision)) * pow(10, _precision)
INTO rounded;
RETURN rounded;
END;
$$ LANGUAGE plpgsql;
SELECT floor_precision(199999, 5)
Alternatively you could to this iof you like to use the round()
function with its second parameter:
SELECT
round(199999 - 50000, -5)
which equals
SELECT
round(199999 - (pow(10,5) / 2)::int, -5)
Naturally, you can create an own function here as well:
CREATE OR REPLACE FUNCTION floor_precision(_value int, _precision int)
RETURNS integer AS $$
DECLARE
rounded integer;
BEGIN
SELECT round(_value - (pow(10, _precision) / 2)::int, _precision * -1)
INTO rounded;
RETURN rounded;
END;
$$ LANGUAGE plpgsql;
SELECT floor_precision(199999, 5)
Second variant seems to be much faster according to the fiddle's execution plan.
Upvotes: 4