How to get first day of every corresponding month in mysql?

Question

I want to get first day of every corresponding month of current year. For example, if user selects '2010-06-15', query demands to run from '2010-06-01' instead of '2010-06-15'.

Please help me how to calculate first day from selected date. Currently, I am trying to get desirable using following mysql select query:

Select
  DAYOFMONTH(hrm_attendanceregister.Date) >=
  DAYOFMONTH(
    DATE_SUB('2010-07-17', INTERVAL - DAYOFMONTH('2010-07-17') + 1 DAY
  )
FROM
  hrm_attendanceregister;

Thanks

Answer 1

Is this what you are looking for:

select CAST(DATE_FORMAT(NOW() ,'%Y-%m-01') as DATE);

Answer 2

You can use the LAST_DAY function provided by MySQL to retrieve the last day of any month, that's easy:

SELECT LAST_DAY('2010-06-15');

Will return:

2010-06-30

Unfortunately, MySQL does not provide any FIRST_DAY function to retrieve the first day of a month (not sure why). But given the last day, you can add a day and subtract a month to get the first day. Thus you can define a custom function:

DELIMITER ;;
CREATE FUNCTION FIRST_DAY(day DATE)
RETURNS DATE DETERMINISTIC
BEGIN
  RETURN ADDDATE(LAST_DAY(SUBDATE(day, INTERVAL 1 MONTH)), 1);
END;;
DELIMITER ;

That way:

SELECT FIRST_DAY('2010-06-15');

Will return:

2010-06-01

Answer 3

There is actually a straightforward solution since the first day of the month is simply today - (day_of_month_in_today - 1):

select now() - interval (day(now())-1) day

Contrast that with the other methods which are extremely roundabout and indirect.

---

Also, since we are not interested in the time component, curdate() is a better (and faster) function than now(). We can also take advantage of subdate()'s 2-arity overload since that is more performant than using interval. So a better solution is:

select subdate(curdate(), (day(curdate())-1))