Get number of weekdays (Sundays, Mondays, Tuesdays) between two dates SQL - sql

Question

So, I'd like to, for a Start Date and End Date, determine how many particular days of the week occur between these two dates.
So how many mondays, tuesdays, etc
I know I can do this with a loop between the Start and End Date and check each day, but the difference could possibly be a great number of days. I'd prefer something that didn't require a loop. Any ideas? (Must be supported in SQL Server 2005+)

Answer 1

Given what I think you're trying to get, this should do it:
SET DATEFIRST 1
DECLARE
#start_date DATETIME,
#end_date DATETIME
SET #start_date = '2011-07-11'
SET #end_date = '2011-07-22'
;WITH Days_Of_The_Week AS (
SELECT 1 AS day_number, 'Monday' AS day_name UNION ALL
SELECT 2 AS day_number, 'Tuesday' AS day_name UNION ALL
SELECT 3 AS day_number, 'Wednesday' AS day_name UNION ALL
SELECT 4 AS day_number, 'Thursday' AS day_name UNION ALL
SELECT 5 AS day_number, 'Friday' AS day_name UNION ALL
SELECT 6 AS day_number, 'Saturday' AS day_name UNION ALL
SELECT 7 AS day_number, 'Sunday' AS day_name
)
SELECT
day_name,
1 + DATEDIFF(wk, #start_date, #end_date) -
CASE WHEN DATEPART(weekday, #start_date) > day_number THEN 1 ELSE 0 END -
CASE WHEN DATEPART(weekday, #end_date) < day_number THEN 1 ELSE 0 END
FROM
Days_Of_The_Week

Answer 2

I'm not sure what the OP is after, this will give a count per weeks day:
SET DATEFIRST 1
DECLARE #StartDate datetime
,#EndDate datetime
SELECT #StartDate='7/13/2011'
,#EndDate='7/28/2011'
;with AllDates AS
(
SELECT #StartDate AS DateOf, datename(weekday,#StartDate) AS WeekDayName, datepart(weekday,#StartDate) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, datename(weekday,DateOf+1), datepart(weekday,DateOf+1)
FROM AllDates
WHERE DateOf<#EndDate
)
SELECT COUNT(*) CountOf,WeekDayName FROM AllDates GROUP BY WeekDayName,WeekDayNumber ORDER BY WeekDayNumber
OUTPUT:
CountOf WeekDayName
----------- ------------------------------
2 Monday
2 Tuesday
3 Wednesday
3 Thursday
2 Friday
2 Saturday
2 Sunday
(7 row(s) affected)
this will give a count of Monday to Friday days:
SET DATEFIRST 1
DECLARE #StartDate datetime
,#EndDate datetime
SELECT #StartDate='7/13/2011'
,#EndDate='7/28/2011'
;with AllDates AS
(
SELECT #StartDate AS DateOf, datepart(weekday,#StartDate) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, datepart(weekday,DateOf+1)
FROM AllDates
WHERE DateOf<#EndDate
)
SELECT COUNT(*) AS WeekDayCount FROM AllDates WHERE WeekDayNumber<=5
OUTPUT:
WeekDayCount
------------
12
(1 row(s) affected)
If you have a holiday table, you can join it in and remove those as well. Here is a slightly different version that may preform better:
SET DATEFIRST 1
DECLARE #StartDate datetime
,#EndDate datetime
SELECT #StartDate='7/13/2011'
,#EndDate='7/28/2011'
;with AllDates AS
(
SELECT #StartDate AS DateOf, datepart(weekday,getdate()) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, (WeekDayNumber+1) % 7
FROM AllDates
WHERE DateOf<#EndDate
)
SELECT COUNT(*) AS WeekDayCount FROM AllDates WHERE WeekDayNumber>0 AND WeekDayNumber<6
--I don't like using "BETWEEN", ">", ">=", "<", and "<=" are more explicit in defining end points
produces same output as original query.

Answer 3

This assume standard settings but cans be adapted
DECLARE #StartDate datetime, #EndDate datetime
SELECT #StartDate='20110601', #EndDate='20110630'
;WITH AllDates AS
(
SELECT #StartDate AS DateOf, datepart(weekday, #StartDate) AS WeekDayNumber
UNION ALL
SELECT DateOf+1, datepart(weekday, DateOf+1)
FROM AllDates
WHERE DateOf < #EndDate
)
SELECT SUM(CASE WHEN WeekDayNumber BETWEEN 2 AND 6 THEN 1 ELSE 0 END) AS WeekDayCount
FROM AllDates
OPTION (MAXRECURSION 0)

Answer 4

This should be valid for SQL Server, and should be internationlization safe (note: I do not have a server to test this against).
SELECT datediff(day, #start, #end) - datediff(week, #start, #end) * 2
- CASE WHEN datepart(weekday, #start)
IN (datepart(weekday, '1970-01-03'),
datepart(weekday, '1970-01-04'))
THEN 1
ELSE 0 END,
- CASE WHEN datepart(weekday, #end)
IN (datepart(weekday, '1970-01-03'),
datepart(weekday, '1970-01-04'))
THEN 1
ELSE 0 END
Give that a whirl.
Given the clarification, this should get the number of each of the days.
Uses no recursion, and should be completely international-safe. You will have to adjust start/end date parameters for inclusion/exclusion as necessary (The DB2 version I was using to check this excluded the start date, but included the end date, for example).
WITH dayOfWeek (name, dayNumber) as (VALUES(dayname(weekday, '1970-01-01'), daypart(weekday, '1970-01-01')),
(dayname(weekday, '1970-01-02'), daypart(weekday, '1970-01-02')),
(dayname(weekday, '1970-01-03'), daypart(weekday, '1970-01-03')),
(dayname(weekday, '1970-01-04'), daypart(weekday, '1970-01-04')),
(dayname(weekday, '1970-01-05'), daypart(weekday, '1970-01-05')),
(dayname(weekday, '1970-01-06'), daypart(weekday, '1970-01-06')),
(dayname(weekday, '1970-01-07'), daypart(weekday, '1970-01-07')))
SELECT name, dayNumber, datediff(weeks, #start, #end)
+ CASE WHEN datepart(weekday, #end) >= dayNumber THEN 1 ELSE 0 END
- CASE WHEN datepart(weekday, #start) >= dayNumber THEN 1 ELSE 0 END
FROM dayOfWeek
Does that help any?

Answer 5

you can use DATEDIFF and DATEPART functions + some basic math to get the desired result without looping.

Answer 6

I would have simply added as a comment to the marked answer, but do not have enough "reputation". Instead of hardcoding the day_number which is dependent on datefirst, you could set it by discovering the weekday for each day of the week:
;WITH Days_Of_The_Week AS (
SELECT DATEPART(dw, '2007-01-01') AS day_number, 'Monday' AS day_name UNION ALL -- 2007-01-01 is a known Monday
SELECT DATEPART(dw, '2007-01-02') AS day_number, 'Tuesday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-03') AS day_number, 'Wednesday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-04') AS day_number, 'Thursday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-05') AS day_number, 'Friday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-06') AS day_number, 'Saturday' AS day_name UNION ALL
SELECT DATEPART(dw, '2007-01-07') AS day_number, 'Sunday' AS day_name
)

Answer 7

#start_date date = '2017-08-11',
#end_date date = '2017-08-27',
#weekday int = 7,
#count int output
As
Begin
Declare #i int = 0
set #count = 0
while(#i <= (select Datediff(Day, #start_date, #end_date)))
begin
if(Dateadd(Day, #i, #start_date) > #end_date)
break
if(Datepart(weekday, Dateadd(Day, #i, #start_date)) = #weekday)
set #count += 1
set #i += 1
end
select #count