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)?

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=>

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

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

with a as (select extract(isodow from date '2020-02-28') - 1 a ),
b as(select CASE 
         WHEN a.a=0 THEN 'Sunday'
		 WHEN a.a =1 THEN 'Monday'
		 WHEN a.a =2 THEN 'Tuesday'
		 WHEN a.a=3 THEN 'Wednesday'
		 WHEN a.a=4 THEN 'Thursday'
		 WHEN a.a=5 THEN 'Friday'
		 WHEN a.a=6 THEN 'Saturday'
     ELSE 'other'
   END from a ) 
 select * from b;