Number of e.g. Mondays left in month

Question

How do you most easily calculate how many e.g. Mondays are left in a month using MySQL (counting today)?

Bonus points for a solution that solves it for all days of the week in one query.

Desired output (run on Tuesday August 17th 2010):

dayOfWeek   left
1           2      -- Sunday
2           2      -- Monday
3           3      -- Tuesday (yep, including today)
4           2      -- Wednesday
5           2      -- Thursday
6           2      -- Friday
7           2      -- Saturday

Answer 1

i found something

according to this article "find next monday"

http://www.gizmola.com/blog/archives/99-Finding-Next-Monday-using-MySQL-Dates.html

    SELECT DATE_ADD(CURDATE(), INTERVAL (9 - IF(DAYOFWEEK(CURDATE())=1, 8,
 DAYOFWEEK(CURDATE()))) DAY) AS NEXTMONDAY;

what we need to do is calculate the days between end month and next Monday, and divide in 7 .

update (include current day) :

so the result is like :

for Monday

       SELECT  CEIL( ((DATEDIFF(LAST_DAY(NOW()),DATE_ADD(CURDATE(), 
INTERVAL (9 - IF(DAYOFWEEK(CURDATE())=1, 8, DAYOFWEEK(CURDATE()))) DAY)))+1)/7) 
    + IF(DAYOFWEEK(CURDATE())=2,1,0)

for Tuesday :

    SELECT  CEIL( ((DATEDIFF(LAST_DAY(NOW()),DATE_ADD(CURDATE(), 
INTERVAL (10 - IF(DAYOFWEEK(CURDATE())=1, 8, DAYOFWEEK(CURDATE()))) DAY)))+1)/7)
     + IF(DAYOFWEEK(CURDATE())=3,1,0)

Answer 2

Create a date table that contains one row for each day that you care about (say Jan 1 2000 - Dec 31 2099):

create table dates (the_date date primary key);

delimiter $$

create procedure populate_dates (p_start_date date, p_end_date date)
begin
declare v_date date;
set v_date = p_start_date;
while v_date <= p_end_date
do
  insert ignore into dates (the_date) values (v_date);
  set v_Date = date_add(v_date, interval 1 day);
end while;
end $$

delimiter ;

call populate_dates('2000-01-01','2099-12-31');

Then you can run a query like this to get your desired output:

set @date = curdate();

select dayofweek(the_date) as dayOfWeek, count(*) as numLeft 
from dates 
where the_date >= @date
and the_date <  str_to_date(period_add(date_format(@date,'%Y%m'),1),'%Y%m') 
group by dayofweek(the_date);

That will exclude days of the week that have 0 occurrences left in the month. If you want to see those you can create another table with the days of the week (1-7):

create table days_of_week (
  id tinyint unsigned not null primary key, 
  name char(10) not null
);

insert into days_of_week (id,name) values (1,'Sunday'),(2,'Monday'),
  (3,'Tuesday'),(4,'Wednesday'),(5,'Thursday'),(6,'Friday'),(7,'Saturday');

And query that table with a left join to the dates table:

select w.id, count(d.the_Date) as numLeft 
from days_of_week w 
left outer join dates d on w.id = dayofweek(d.the_date) 
  and d.the_date >= @date 
  and d.the_date <  str_to_date(period_add(date_format(@date,'%Y%m'),1),'%Y%m') 
group by w.id;

Answer 3

Have a look at my responses to;

http://stackoverflow.com/questions/3249877/mysql-using-the-dates-in-a-between-condition-for-the-results

and

http://stackoverflow.com/questions/3093924/select-all-months-within-given-date-span-including-the-ones-with-0-values

for a way I think would work nicely, similar to @Walker's above, but without having to do the dayofweek() function within the query, and possibly more flexible too. One of the responses has a link to a SQL dump of my table which can be imported if it helps!