PostgreSQL : cast string to date DD/MM/YYYY - string

Question

I'm trying to cast a CHARACTER VARYING column to a DATE but I need a date format like this : DD/MM/YYYY. I use the following SQL query :
ALTER TABLE test
ALTER COLUMN date TYPE DATE using to_date(date, 'DD/MM/YYYY');
The result is a date like this : YYYY-MM-DD.
How can I get the DD/MM/YYYYformat ?
Thanks a lot in advance !
Thomas

Answer 1

A DATE column does not have a format. You cannot specify a format for it.
You can use DateStyle to control how PostgreSQL emits dates, but it's global and a bit limited.
Instead, you should use to_char to format the date when you query it, or format it in the client application. Like:
SELECT to_char("date", 'DD/MM/YYYY') FROM mytable;
e.g.
regress=> SELECT to_char(DATE '2014-04-01', 'DD/MM/YYYY');
to_char
------------
01/04/2014
(1 row)

Answer 2

https://www.postgresql.org/docs/8.4/functions-formatting.html
SELECT to_char(date_field, 'DD/MM/YYYY')
FROM table

Answer 3

The documentation says
The output format of the date/time types can be set to one of the four
styles ISO 8601, SQL (Ingres), traditional POSTGRES (Unix date
format), or German. The default is the ISO format.
So this particular format can be controlled with postgres date time output, eg:
t=# select now();
now
-------------------------------
2017-11-29 09:15:25.348342+00
(1 row)
t=# set datestyle to DMY, SQL;
SET
t=# select now();
now
-------------------------------
29/11/2017 09:15:31.28477 UTC
(1 row)
t=# select now()::date;
now
------------
29/11/2017
(1 row)
Mind that as #Craig mentioned in his answer, changing datestyle will also (and in first turn) change the way postgres parses date.

Answer 4

In case you need to convert the returned date of a select statement to a specific format you may use the following:
select to_char(DATE (*date_you_want_to_select*)::date, 'DD/MM/YYYY') as "Formated Date"

Answer 5

Depends on which type you require as output, but here are 2 quick examples based on an intervals:
SELECT (now() - interval '15 DAY')::date AS order_date -> 2021-07-29
SELECT to_char(now() - interval '15 DAY', 'YYYY-MM-DD') -> 2021-07-29

Answer 6

Let's say your date column is order_date:
SELECT (
RIGHT(order_date, 4)
|| '-'
|| SUBSTRING(order_date, 4, 2)
|| '-'
|| LEFT(order_date, 2)
)::DATE
FROM test
OR
SELECT CAST(
RIGHT(order_date, 4)
|| '-'
|| SUBSTRING(order_date, 4, 2)
|| '-'
|| LEFT(order_date, 2)
AS DATE )
FROM test