How to calculate a third monday of every month using SQL statement? - mysql

Question

i need to trigger a notification. this notification has to be triggered every third monday of every month.

Answer 1

SELECT
(
DAYOFWEEK(NOW()) = 2
AND
DAYOFMONTH(NOW()) BETWEEN 15 AND 21
)
AS send_notice_today;

Answer 2

Try using dayofweek and dayofmonth functions. http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_dayofweek
Somehow you can check how many weeks are there from 1st of month to curdate() with dayofmonth (using an operation mod 7), and dayofweek should be 5 (thursday)

Answer 3

So perhaps MORE GENERALLY if you can get the "week of month" for the date using this:
(FLOOR((DAYOFMONTH(given) - 1) / 7)) AS 'week_of_month'
which I believe provides an accurate 0 based week-of-month index for a given date. then you can use the value to find any nth as in:
WHERE (week_of_month) = n AND weekday = {weekday}
you can also use the above to get the "last {weekday}" by:
WHERE (week_of_month >= 4) and weekday = {weekday}
note that the week_of_month can range from 0 to 5 (e.g., a 31 day month whose 1st falls on Saturday will have the 31st in the 6th week (5 as a 0 based index)
hope this helps ...
OK a bit more ... you might define the above as a function as in:
CREATE FUNCTION WEEKOFMONTH(given date) RETURNS int DETERMINISTIC RETURN (FLOOR((DAYOFMONTH(given) - 1) / 7))
and add another function:
CREATE FUNCTION WEEKNAME(given date) RETURNS text CHARACTER SET utf8 COLLATE utf8_unicode_ci DETERMINISTIC RETURN (CONCAT(ELT(WEEKOFMONTH(given)+1,'1st ','2nd ','3rd ','4th/Last ','5th/Last '),DAYNAME(given)))
then you can simply say things like
SELECT * FROM dataTable WHERE WEEKNAME(your_date_field) = "3rd Wednesday"
... I struggled with how the 4th/5th should be returned from WEEKDAY and settled on adding "/Last" for both under the theory that this is "good enough" should one want to test for either 4th, 5th or Last. Using this you can do:
SELECT * FROM dataTable WHERE WEEKNAME(your_date_field) LIKE "%Last%"

Answer 4

The date of the third monday of the current month would be given by the SQL statement:
SELECT date_add(date_sub(curdate(),INTERVAL dayofmonth(curdate())-1 DAY),
INTERVAL (7-weekday(date_sub(curdate(),INTERVAL dayofmonth(curdate())-1 DAY)))+14 DAY)

Answer 5

The approach that I'm taking is get the 1st Monday of the month, and depending on when in the month it is, add either 2 or 3 weeks to it (since when it falls out before/on Monday, you only need to walk 2 more weeks):
;with
filler as (select row_number() over (order by a) a from (select top 100 1 as a from syscolumns) a cross join (select top 100 1 as b from syscolumns) b),
dates as (select dateadd(month, a-1, '1/1/1900') date from filler where a <= 2000),
FirstMonday as (
select dateadd(day, case datepart(weekday,Date)
when 1 then 1
when 2 then 0
when 3 then 6
when 4 then 5
when 5 then 4
when 6 then 3
when 7 then 2
end, Date) as Date
,case when datepart(weekday,Date) = 1 then 3 else 2 end as Weeks
from dates
)
select dateadd(week, Weeks, Date) as ThirdMonday
from FirstMonday

Answer 6

This one calculates the first Monday of any month given the year-month-day
SET #firstday = '2015-04-01';
SELECT ADDDATE( #firstday , MOD((9-DAYOFWEEK(#firstday)),7)) as first_monday;
This one calculates the third Monday of any month given the year-month-day
SET #firstday = '2015-01-01';
SELECT ADDDATE( #firstday , MOD((23-DAYOFWEEK(#firstday)),21)) as third_monday;
This one calculates the third Friday of any month given the year-month-day
SET #firstday = '2015-09-01';
SELECT ADDDATE( #firstday , MOD((20-DAYOFWEEK(#firstday)),20)) as third_friday;
Thanks to #Brewal for the original question and #User2208436 for pointing us toward the answer.

Answer 7

Here's an answer that does not use DAYOFWEEK or DAYOFMONTH. It uses DATEADD and DATEPART only.
We'll need two helper functions:
CREATE FUNCTION dbo.day_of_week(#date Date)
RETURNS INT
AS BEGIN
-- (1 for Sunday, 2 for Monday, etc)
RETURN DATEPART(dw, DATEADD(year, year(#date)-1900, DATEADD(month, month(#date)-1, DATEADD(day, day(#date)-1, 0))))
END
GO
CREATE FUNCTION dbo.date_from_parts(#year INT, #month INT, #day INT)
RETURNS DATE
AS BEGIN
RETURN DATEADD(year, #year-1900, DATEADD(month, #month-1, DATEADD(day, #day-1, 0)))
END
GO
Then using the following example data:
DECLARE #day_of_week INT
SET #day_of_week = 2 -- Monday
DECLARE #year INT
DECLARE #month INT
SET #year = 2016
SET #month = 11
Let's first obtain the FIRST Monday of the month:
We will add an offset, (day_of_week - day of week of first day of the month) % 7, to the first day of the month.
(Also notice we need the construction ((x % n) + n) % n instead of just x % 7, to keep the answer within 0 and 6. For example, just SELECT -3 % 7 returns -3! See Mod negative numbers in SQL just like excel)
Now here's the final construction to obtain the first Monday of the month:
SELECT
DATEADD(
dd,
(((#day_of_week -
dbo.day_of_week(dbo.date_from_parts(#year, #month, 1))) % 7) + 7) % 7,
dbo.date_from_parts(#year, #month, 1)
)
To obtain the third Monday of the month, add 14 to the second term of this answer.

Answer 8

If #firstday is the first day of the month
select date_add(#firstday,
interval (if(weekday(#firstday)>0,21-weekday(#firstday),14)) day);
yields the 3rd monday of the month.
Tested with all months of 2018.
Seems simpler than previous solutions.
The key is the if function on weekday.
if weekday = 0 (first day is a monday), add 14 days
if weekday > 0, add 21 days and subtract the weekday

Answer 9

DECLARE #YEAR DATE='2019-01-01'
SELECT DATEADD( d, 23-(DATEPART(dw,#YEAR )%21),#YEAR )
This will help to get the third Monday of whatever month you want.