Order by day_of_week in MySQL

Question

How can I order the mysql result by varchar column that contains day of week name?

Note that MONDAY should goes first, not SUNDAY.

Answer 1

Either redesign the column as suggested by Williham Totland, or do some string parsing to get a date representation.

If the column only contains the day of week, then you could do this:

ORDER BY FIELD(<fieldname>, 'MONDAY', 'TUESDAY', 'WEDNESDAY', 'THURSDAY', 'FRIDAY', 'SATURDAY', 'SUNDAY');

Answer 2

Why not this?

ORDER BY (
    CASE DAYOFWEEK(dateField)
    WHEN 1 THEN 7 ELSE DAYOFWEEK(dateField)
    END
)

I believe this orders Monday to Sunday...

Answer 3

I'm thinking that short of redesigning the column to use an enum instead, there's not a lot to be done for it, apart from sorting the results after you've gotten them out.

Edit: A dirty hack is of course to add another table with id:weekday pairs and using joins or select in selects to fake an enum.

Answer 4

This looks messy but still works and seems more generic:

select day, 
case day
  when 'monday' then 1
  when 'tuesday' then 2
  when 'wednesday' then 3
  when 'thursday' then 4
  when 'friday' then 5
  when 'saturday' then 6
  when 'sunday' then 7
end as day_nr from test order by day_nr;

Using if is even more generic and messier:

select id, day, 
if(day = 'monday',1,
  if(day = 'tuesday',2,
    if(day = 'wednesday',3,
      if(day = 'thursday',4,
        if(day = 'friday',5,
          if(day = 'saturday',6,7)
        )
      )
    )
  )
) as day_nr from test order by day_nr;

You can also hide the details of conversion from name to int in stored procedure.

Answer 5

... ORDER BY date_format(order_date, '%w') = 0, date_format(order_date, '%w') ;

Answer 6

I realise that this is an old thread, but as it comes to the top of google for certain search times I will use it to share my approach.

I wanted the same result as the original question, but in addition I wanted the ordering of the results starting from the current day of the week and then progressing through the rest of the days.

I created a separate table, in which the days were listed over a fortnight, so that no matter which day you started from you could run through a sequence of 7 days.

CREATE TABLE IF NOT EXISTS `Weekdays` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(50) NOT NULL DEFAULT '',
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=15 ;

INSERT INTO `Weekdays` (`id`, `name`) VALUES
(1, 'Monday'),
(2, 'Tuesday'),
(3, 'Wednesday'),
(4, 'Thursday'),
(5, 'Friday'),
(6, 'Saturday'),
(7, 'Sunday'),
(8, 'Monday'),
(9, 'Tuesday'),
(10, 'Wednesday'),
(11, 'Thursday'),
(12, 'Friday'),
(13, 'Saturday'),
(14, 'Sunday');

I then ran the query with a variable that determined the start point in sequence and used a join to get the order number for the days. For example to start the listing at Wednesday, I do the following:

SELECT @startnum := MIN(id) FROM Weekdays WHERE name='Wednesday';
SELECT * FROM Events INNER JOIN ( SELECT id as weekdaynum, name as dayname FROM Weekdays WHERE id>(@startnum-1) AND id<(@startnum+7) ) AS s2 ON s2.dayname=Events.day ORDER BY weekdaynum;

I hope this helps someone who stumbles onto this post.