Extract day of week from date field in PostgreSQL assuming weeks start on Monday

select extract(dow from datefield)

extract a number from 0 to 6, where 0 is Sunday; is there a way to get the day of the week in SQL assuming that weeks start on Monday (so 0 will be Monday)?

Solution 1:

From the manual

isodow

    The day of the week as Monday (1) to Sunday (7)

So, you just need to subtract 1 from that result:

psql (9.6.1)
Type "help" for help.

postgres=> select extract(isodow from date '2016-12-12') - 1;
  ?column?
-----------
         0
(1 row)
postgres=>

Solution 2:

Use date_part Function dow()

Here 0=Sunday, 1=Monday, 2=Tuesday, ... 6=Saturday

   select extract(dow from date '2016-12-18'); /* sunday */

Output : 0

    select extract(isodow from date '2016-12-12'); /* Monday  */

Ouput : 1

Solution 3:

If you want the text version of the weekday then you can use the to_char(date, format) function supplying a date and the format that you want.

According to https://www.postgresql.org/docs/current/functions-formatting.html#FUNCTIONS-FORMATTING-DATETIME-TABLE we have the following format options we can use for date. I have shown some examples for output. According to the documentation the abbreviated day values are 3 characters long in English, other locales may vary.

select To_Char("Date", 'DAY'), * from "MyTable"; -- TUESDAY
select To_Char("Date", 'Day'), * from "MyTable"; -- Tuesday
select To_Char("Date", 'day'), * from "MyTable"; -- tuesday
select To_Char("Date", 'dy'), * from "MyTable";  -- tue
select To_Char("Date", 'Dy'), * from "MyTable";  -- Tue
select To_Char("Date", 'DY'), * from "MyTable";  -- TUE