Elapsed time between 2 dates, excluding weekend in SQL [duplicate] - sql

I would like to build an SQL query which calculates the difference between 2 dates, without counting the week-end days in the result.
Is there any way to format the dates to obtain this result ? For example for Oracle database :
select sysdate - creation_dttm from the_table


You should try with a function :
CREATE FUNCTION TOTAL_WEEKDAYS(date1 DATE, date2 DATE)
RETURNS INT
RETURN ABS(DATEDIFF(date2, date1)) + 1
- ABS(DATEDIFF(ADDDATE(date2, INTERVAL 1 - DAYOFWEEK(date2) DAY),
ADDDATE(date1, INTERVAL 1 - DAYOFWEEK(date1) DAY))) / 7 * 2
- (DAYOFWEEK(IF(date1 < date2, date1, date2)) = 1)
- (DAYOFWEEK(IF(date1 > date2, date1, date2)) = 7);
Test :
SELECT TOTAL_WEEKDAYS('2013-08-03', '2013-08-21') weekdays1,
TOTAL_WEEKDAYS('2013-08-21', '2013-08-03') weekdays2;
Result :
| WEEKDAYS1 | WEEKDAYS2 |
-------------------------
| 13 | 13 |


I have found another way to do calculate the difference, by using only SQL :
select sysdate - creation_dttm
- 2 * (to_char(sysdate, 'WW') - to_char(creation_dttm, 'WW'))
from the_table


I have found several of the answers on this thread to not do what they claim. After some experimentation, testing and adjusting, I have this to contribute.
declare #firstdate Date
declare #seconddate Date
set #firstDate = convert(date, '2016-03-07')
set #seconddate = convert(date, '2016-04-04')
select (datediff(dd, #firstDate, #secondDate)) -
(( DateDiff(wk, #firstDate, #secondDate) * 2) -
case when datepart(dw, #FirstDate) = 7 then 1 else 0 end -
case when datepart(dw, #secondDate) = 7 then -1 else 0 end)
Test harness included - you can just adjust the two dates and run your own tests. This assumes that the difference between two adjacent weekday dates is 1. If your country uses different days to signify weekend, then you will have to set the date-base accordingly so your "Saturday" is 7, and your "sunday" is 1.


From a previous post:
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SET #StartDate = '2008/10/01'
SET #EndDate = '2008/10/31'
SELECT
(DATEDIFF(dd, #StartDate, #EndDate) + 1)
-(DATEDIFF(wk, #StartDate, #EndDate) * 2)
-(CASE WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END)


Here is an example:
There are four variables, the first two are self-explanatory, just enter the date in YYYYMMDD format, the 3rd one is to set the number of normal work days in a given week, so if a site works 6 days a week, set it to 6, five days a week enter 5, etc. Finally, the
DATE_SEQ_NORML_FACTOR should be 1 when running against Oracle. This is to line up the Julian date to equal to 1 on Monday, 2 on Tuesday, etc when applying the MOD 7. Other DB will probably have different values between 0 and 6, so test it out before you use against other DBs.
Here are the limitations:
1. This formula assumes the first day of the week is MONDAY.
2. This formula assumes all days within the same week are CONTINUOUS.
3. This formula will work ONLY when the two dates involved in the calculation falls on a week day or work day, eg. the "Start Date" on a SATURDAY when the location works only MON-FRI will not work.
SELECT
&&START_DATE_YYYYMMDD "Start Date", --in YYYYMMDD format
&&END_DATE_YYYYMMDD "End Date", --in YYYYMMDD format
&&WK_WORK_DAY_CNT "Week Work Day Count", --Number of work day per week
&&DATE_SEQ_NORML_FACTOR "Normalization Factor", --set to 1 when run in Oracle
CASE
WHEN
FLOOR( TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR / 7 ) =
FLOOR( TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR / 7 )
THEN(
TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) -
TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + 1
)
ELSE(
(
&&WK_WORK_DAY_CNT - MOD( TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR , 7 ) + 1
) +
MOD( TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR , 7 ) +
(
(
(
TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) -
MOD( TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR , 7 )
) -
(
TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) +
(
7 -
(
MOD( TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR , 7 )
)
)
)
) / 7 * &&WK_WORK_DAY_CNT
)
) END "Week Day Count"
FROM DUAL


We've incorporated the following logic into several reports with great success. Sorry, I no longer recall the source of this script to give them credit.
DECLARE #range INT;
SET #range = DATEDIFF(DAY, #FirstDate, #SecondDate);
SET #NumWorkDays = (
SELECT
#range / 7 * 5 + #range % 7 - (
SELECT COUNT(*)
FROM (
SELECT 1 AS d
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
) weekdays
WHERE d <= #range % 7
AND DATENAME(WEEKDAY, #SecondDate - d) IN ('Saturday', 'Sunday'))
);


I've updated #JOpuckman's function to take into account that different regions don't always have a weekend of Saturday and Sunday. Here's the code in case anyone else needs to apply this globally;
DECLARE #FirstDate DateTime
DECLARE #SecondDate DateTime
SET #FirstDate = '08-20-2012'
SET #SecondDate = '08-24-2012'
DECLARE #range INT;
DECLARE #WeekendDayNameStart VARCHAR(50)
DECLARE #WeekendDayNameEnd VARCHAR(50)
SET #WeekendDayNameStart = 'FRIDAY'
SET #WeekendDayNameEnd = (
SELECT CASE #WeekendDayNameStart
WHEN 'SUNDAY' THEN 'MONDAY'
WHEN 'MONDAY' THEN 'TUESDAY'
WHEN 'TUESDAY' THEN 'WEDNESDAY'
WHEN 'WEDNESDAY' THEN 'THURSDAY'
WHEN 'THURSDAY' THEN 'FRIDAY'
WHEN 'FRIDAY' THEN 'SATURDAY'
WHEN 'SATURDAY' THEN 'SUNDAY'
END
)
DECLARE #NumWorkDays INT
SET #range = DATEDIFF(DAY, #FirstDate, #SecondDate);
SET #NumWorkDays = (
SELECT
#range / 7 * 5 + #range % 7 - (
SELECT COUNT(*)
FROM (
SELECT 1 AS d
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
) weekdays
WHERE d <= #range % 7
AND DATENAME(WEEKDAY, #SecondDate - d) IN (#WeekendDayNameStart, #WeekendDayNameEnd))
);
-- Calculate whether the current date is a working day
DECLARE #CurDateExtra INT
SET #CurDateExtra =
(
CASE DATENAME(WEEKDAY, #SecondDate)
WHEN #WeekendDayNameStart THEN 0
WHEN #WeekendDayNameEnd THEN 0
ELSE 1
END
)
SET #NumWorkDays = #NumWorkDays + #CurDateExtra
SELECT #NumWorkDays


Calculates the difference in days between the two dates
Calculates the difference in week numbers and year numbers, subtracts the week numbers and then multiplies the result by 2 to calculate number of non-workdays between the two dates. If year numbers are different calculation of 52 * year difference + week number.
((sysdate - ced.created_dt) + ((((to_char(ced.created_dt,'IW') - ((to_char(sysdate,'YY') - to_char(ced.created_dt,'YY'))* 52))
- to_char(to_char(sysdate,'IW')))) * 2)) duration_in_weekdays


You can try this:
SELECT
Id,
DATEDIFF(d, datefrom, dateto) AS TotDays,
DATEDIFF(wk, datefrom, dateto) AS Wkds,
DATEDIFF(d, datefrom, dateto) - DATEDIFF(wk, datefrom, dateto) AS Days
FROM
YOURTABLE


If your company has a DATES table listing the date and relative work day, as most companies do, you could build a temp table that excludes duplicate relative work days that will also give you the ability to exclude holidays from your calculations as well as weekends.

Related

difference between two dates without weekends and holidays Sql query ORACLE

I have 2 tables: the 1st one contains the start date and the end date of a purchase order,
and the 2nd table contains year hollidays
-purchase order
-Holidays
I'm tryign to calculate the number of business days between 2 dates without the weekends and the holidays.
the output should be like this:
Start Date | End Date | Business Days
Could you please help me

You can remove the non-weekend holidays with a query like this:
select (t.end_date - t.start_date) - count(c.date)
from table1 t left join
calendar c
on c.date between t1.start_date and t1.end_date and
to_char(c.date, 'D') not in ('1', '7')
group by t.end_date, t.start_date;
Removing the weekend days is then more complication. Full weeks have two weekend days, so that is easy. So a good approximation is:
select (t.end_date - t.start_date) - (count(c.date) +
2 * floor((t.end_date - t.start_date) / 7))
from table1 t left join
calendar c
on c.date between t1.start_date and t1.end_date and
to_char(c.date, 'D') not in ('1', '7')
group by t.end_date, t.start_date;
This doesn't get the day of week, which is essentially if the end date is before the start date, then it is in the following week. However, this logic gets rather complicated the way that Oracle handles day of the week, so perhaps the above approximation is sufficient.

EDIT: I ignored the presence of the Oracle tag and jumped into scripting this for SQL Server. The concept doesn't change though.
To be super accurate, I would create a table whit the following format.
Year int, month int, DaysInMonth int, firstOccuranceOfSunday int
Create a Procedure to extract the weekends from a specific year and month on that table.
CREATE FUNCTION [dbo].[GetWeekendsForMonthYear]
(
#year int,
#month int
)
RETURNS #weekends TABLE
(
[Weekend] date
)
AS
BEGIN
declare #firstsunday int = 0
Declare #DaysInMonth int = 0
Select #DaysInMonth = DaysInMonth, #firstsunday = FirstSunday from Months
Where [Year] = #year and [month] = #month
Declare #FirstSaterday int = #firstsunday - 1
declare #CurrentDay int = 0
Declare #CurrentDayIsSunday bit = 0
if #FirstSaterday !< 1
Begin
insert into #Weekends values(DATEADD(year, #year -1900, DATEADD(month, #month -1, DATEADD(day, #Firstsaterday -1, 0))))
insert into #Weekends values(DATEADD(year, #year -1900, DATEADD(month, #month -1, DATEADD(day, #FirstSunday -1, 0))))
set #CurrentDayIsSunday = 1
set #CurrentDay = #firstsunday
END
else
begin
insert into #Weekends values(DATEADD(year, #year -1900, DATEADD(month, #month -1, DATEADD(day, #FirstSunday -1, 0))))
set #FirstSaterday = #firstsunday + 6
insert into #Weekends values(DATEADD(year, #year -1900, DATEADD(month, #month -1, DATEADD(day, #Firstsaterday -1, 0))))
set #CurrentDayIsSunday = 0
set #CurrentDay = #FirstSaterday
end
declare #done bit = 0
while #done = 0
Begin
if #CurrentDay <= #DaysInMonth
Begin
If #CurrentDayIsSunday = 1
begin
set #CurrentDay = #CurrentDay + 6
set #CurrentDayIsSunday = 0
if #CurrentDay <= #DaysInMonth
begin
insert into #Weekends Values(DATEADD(year, #year -1900, DATEADD(month, #month -1, DATEADD(day, #CurrentDay -1, 0))))
end
end
else
begin
set #CurrentDay = #CurrentDay + 1
set #CurrentDayIsSunday = 1
if #CurrentDay <= #DaysInMonth
begin
insert into #Weekends Values(DATEADD(year, #year -1900, DATEADD(month, #month -1, DATEADD(day, #CurrentDay -1, 0))))
end
end
end
ELSE
begin
Set #done = 1
end
end
RETURN
END
When called and provided with a year and month this will return a list of dates that represent weekends.
Now, using that function, create a procedure to call this function once for every applicable row in a specific date rang and return the values in a temptable.
Note, I'm posting this now so you can see what's going on but I am continuing to work on the code. I will post updates as they arise.
More to come: Get list of weekends(formatted) for a specific daterange, remove any dates from that list which can be found on your holidays table.
Unfortunately, I have to work tomorrow and am off to bed.

This query should produce exact number of business days for each range in purchase table:
with days as (
select rn, sd + level - 1 dt, sd, ed
from (select row_number() over (order by start_date) rn,
start_date sd, end_date ed from purchase_order)
connect by prior rn = rn and sd + level - 1 <= ed
and prior dbms_random.value is not null)
select sd start_date, ed end_date, count(1) business_days
from days d left join holidays h on holiday_date = d.dt
where dt - trunc(dt, 'iw') not in (5, 6) and h.holiday_date is null
group by rn, sd, ed
SQLFiddle demo
For each row in purchase_orders query generates dates from this range (this is done by subquery dates).
Main query checks if this is weekend day or holiday day and counts rest of dates.
Hierarchical query used to generate dates may cause slowdowns if there is big number of data in purchase_orders
or periods are long. In this case preferred way is to create calendar table, as already suggested in comments.

Since you already have a table of holidays you can count the holidays between the starting and ending date and subtract that from the difference in days between your ending and starting date. For the weekends, you either need a table containing weekend days similar to your table of holidays, or you can generate them as below.
with sample_data(id, start_date, end_date) as (
select 1, date '2015-03-06', date '2015-03-7' from dual union all
select 2, date '2015-03-07', date '2015-03-8' from dual union all
select 3, date '2015-03-08', date '2015-03-9' from dual union all
select 4, date '2015-02-07', date '2015-06-26' from dual union all
select 5, date '2015-04-17', date '2015-08-16' from dual
)
, holidays(holiday) as (
select date '2015-01-01' from dual union all -- New Years
select date '2015-01-19' from dual union all -- MLK Day
select date '2015-02-16' from dual union all -- Presidents Day
select date '2015-05-25' from dual union all -- Memorial Day
select date '2015-04-03' from dual union all -- Independence Day (Observed)
select date '2015-09-07' from dual union all -- Labor Day
select date '2015-11-11' from dual union all -- Veterans Day
select date '2015-11-26' from dual union all -- Thanks Giving
select date '2015-11-27' from dual union all -- Black Friday
select date '2015-12-25' from dual -- Christmas
)
-- If your calendar table doesn't already hold weekends you can generate
-- the weekends with these next two subfactored queries (common table Expressions)
, firstweekend(weekend, end_date) as (
select next_day(min(start_date),'saturday'), max(end_date) from sample_data
union all
select next_day(min(start_date),'sunday'), max(end_date) from sample_data
)
, weekends(weekend, last_end_date) as (
select weekend, end_date from firstweekend
union all
select weekend + 7, last_end_date from weekends where weekend+7 <= last_end_date
)
-- if not already in the same table combine distinct weekend an holiday days
-- to prevent double counting (in case a holiday is also a weekend).
, days_off(day_off) as (
select weekend from weekends
union
select holiday from holidays
)
select id
, start_date
, end_date
, end_date - start_date + 1
- (select count(*) from days_off where day_off between start_date and end_date) business_days
from sample_data;
ID START_DATE END_DATE BUSINESS_DAYS
---------- ----------- ----------- -------------
1 06-MAR-2015 07-MAR-2015 1
2 07-MAR-2015 08-MAR-2015 0
3 08-MAR-2015 09-MAR-2015 1
4 07-FEB-2015 26-JUN-2015 98
5 17-APR-2015 16-AUG-2015 85

SQL Set Time to Midnight in function

I have This function
ALTER FUNCTION [General].[GetWeekEnding]
(
#Date DATETIME
)
RETURNS DATETIME
AS
BEGIN
-- Return the result of the function
RETURN (DATEADD(day, -1 - (DATEPART(dw, #Date) + ##DATEFIRST - 2) % 7, #Date) + 7)
END
I need to also set the time to 00:00:00.000 as well as finding the week ending of a provided date any thoughts?

I use a UDF for adding time components to dates e.g.
CREATE FUNCTION [dbo].[DateTimeAdd]
(
#datepart date,
#timepart time
)
RETURNS datetime2
AS
BEGIN
RETURN DATEADD(dd, DATEDIFF(dd, 0, #datepart), CAST(#timepart AS datetime2));
END
Then in your case you can use it like this:
SELECT dbo.DateTimeAdd(DATEADD (D, -1 * DatePart (DW, GetDate()) + 7, GetDate()), DATEADD(hh, 0, CAST(DATEADD(DAY, DATEDIFF(DAY, -1, GETDATE()), -1) AS TIME)))

Since SQL Server 2005, the best way to remove time is to cast to the date data type:
select cast(#Date as date)
I would suggest that you have your function return a date instead of a datetime.
To get to the week ending, you need a lookup table to map days of the week to integers. You can do this with datepart. However, that is subject to system settings. Here is a function:
Create FUNCTION [GetWeekEnding] (
#Date DATETIME,
#WeekEndingDOW varchar(10)
)
RETURNS DATE
AS
BEGIN
-- Return the result of the function
declare #newdate datetime;
with lookup as (
select 'Sunday' as dow, 0 as daynum union all
select 'Monday' as dow, 1 as daynum union all
select 'Tuesday' as dow, 2 as daynum union all
select 'Wednesday' as dow, 3 as daynum union all
select 'Thursday' as dow, 4 as daynum union all
select 'Friday' as dow, 5 as daynum union all
select 'Saturday' as dow, 6 as daynum
)
select #newdate = #Date - (select daynum from lookup where datename(dw, #date) = dow) + (select daynum from lookup where #WeekEndingDOW = dow);
select #newdate = (case when #newdate < #date then #newdate + 7 else #newdate end)
RETURN cast(#newdate as date)
END;
Note that this function does use datetime internally. For some reason that I cannot fathom, you can say "#date + 1" to mean "add one day to the date value" when #date is a datetime. However, this does not work when #date is a date. (This is also true of the dateadd function.)

How to get all the weekend dates of the current year in SQL?

I tried but could not get the right solution. I want an SQL query that lists all the weekend dates of the current year.
I tried this SQL query:
WITH hier(num, lvl) AS (
SELECT 0, 1
UNION ALL
SELECT 100, 1
UNION ALL
SELECT num + 1, lvl + 1
FROM hier
WHERE lvl < 100
)
SELECT lvl [Week],
convert(date,DATEADD(dw, -DATEPART(dw, DATEADD(wk,DATEDIFF(wk,0,'12/31/'+convert(nvarchar,YEAR(getdate()))), 0)+6 ),
DATEADD(wk, DATEDIFF(wk,0,'12/31/'+convert(nvarchar,YEAR(getdate()))), 0)+6 ) - num * 7,101) [End Date]
FROM hier a
where num < 52
ORDER BY [End Date] asc
Its output is like this:
Week End date
52 2012-01-14
51 2012-01-21
50 2012-01-28
49 2012-02-04
I want the dates to start from the beginning – so, the above is missing one weekend, which is 2012-07-01. Also, I want the week numbers to show as 1, 2, 3... instead of 52, 51....

Check out this blog post.
Your question is explained in detail.
DECLARE #Year AS INT,
#FirstDateOfYear DATETIME,
#LastDateOfYear DATETIME
-- You can change #year to any year you desire
SELECT #year = 2010
SELECT #FirstDateOfYear = DATEADD(yyyy, #Year - 1900, 0)
SELECT #LastDateOfYear = DATEADD(yyyy, #Year - 1900 + 1, 0)
-- Creating Query to Prepare Year Data
;WITH cte AS (
SELECT 1 AS DayID,
#FirstDateOfYear AS FromDate,
DATENAME(dw, #FirstDateOfYear) AS Dayname
UNION ALL
SELECT cte.DayID + 1 AS DayID,
DATEADD(d, 1 ,cte.FromDate),
DATENAME(dw, DATEADD(d, 1 ,cte.FromDate)) AS Dayname
FROM cte
WHERE DATEADD(d,1,cte.FromDate) < #LastDateOfYear
)
SELECT FromDate AS Date, Dayname
FROM CTE
WHERE DayName IN ('Saturday','Sunday') -- For Weekend
/*
WHERE DayName LIKE 'Sunday'
WHERE DayName NOT IN ('Saturday','Sunday') -- For Weekday
WHERE DayName LIKE 'Monday' -- For Monday
WHERE DayName LIKE 'Sunday' -- For Sunday
*/
OPTION (MaxRecursion 370)

Will this help
DECLARE #startDate DATETIME, #endDate DATETIME
SELECT #startDate = '2012-01-01', #endDate = '2012-12-31'
;WITH Calender AS (
SELECT #startDate AS dt
UNION ALL
SELECT dt + 1 FROM Calender
WHERE dt + 1 <= #endDate
)
SELECT
dt
,NameMonth = DATENAME(Month, dt)
,NameDay = DATENAME (Weekday,dt)
,WeekofYr = DATEPART(WEEK, dt) FROM Calender
WHERE DATENAME (Weekday,dt) IN ('Sunday')
Option(MaxRecursion 0)
Result(Partial)
dt NameMonth NameDay WeekofYr
2012-01-01 00:00:00.000 January Sunday 1
2012-01-08 00:00:00.000 January Sunday 2
...............................................
...............................................
2012-12-30 00:00:00.000 December Sunday 53

you can try this
DECLARE #FirstDateOfYear DATETIME
SET #FirstDateOfYear = ’2010-01-01′
SELECT DISTINCT DATEADD(d, number, #FirstDateOfYear),
CASE DATEPART(dw, DATEADD(d, number, #FirstDateOfYear))
WHEN 7 THEN ‘Saturday’
WHEN 1 THEN ‘Sunday’
ELSE ‘Work Day’
END
FROM master..spt_values
WHERE number BETWEEN 0 AND 364
AND (DATEPART(dw, DATEADD(d, number, #FirstDateOfYear)) = 1 OR DATEPART(dw, DATEADD(d, number, #FirstDateOfYear)) = 7)
ORDER BY DATEADD(d, number, #FirstDateOfYear)

Try to find the first Saturday by doing this:
Start on 2012-01-01
If it's not a Saturday, add a day
Goto 2
Then, into a temporary table, add that date and the following date (Sunday).
After that, loop the following:
Add 7 and 8 days to the last Saturday you found (you get the following Saturday and Sunday)
Check whether they are still in 2012
If they are, store them in temp table and goto 1
There may be more elegant ways, but that's my quick & dirty solution. As you didn't post any code of what you've tried, I'll leave the implementation up to you.

this also works
declare #dat datetime, #add int
set #dat = '20120101'
set #add = datepart(w,#dat)
set #add = 5 - #add -- friday
set #dat = dateadd(d,#add,#dat)
while #dat <= '20121231'
begin
print #dat
set #dat = dateadd(d,7,#dat)
end

;with AllDaysOfYear (Day) as (
select DATEADD(year,DATEDIFF(year,0,CURRENT_TIMESTAMP),0) --Jan 1st
union all
select DATEADD(day,1,Day) from AllDaysOfYear
where DATEPART(year,DATEADD(day,1,Day)) = DATEPART(year,CURRENT_TIMESTAMP)
)
select
ROW_NUMBER() OVER (ORDER BY Day) as WeekNo,
Day
from
AllDaysOfYear
where
DATEPART(weekday,Day) = DATEPART(weekday,'20120714')
option (maxrecursion 0)
First, generate a set of all of the days in the current year (AllDaysInYear). Then, select those whose weekday is a saturday. The value I've used ('20120714') isn't terribly important - it just has to be any saturday, from any year. I'm just using it to avoid needing to have particular DATEFIRST or language settings.

This query shows how to get the first day of this year and the first day of the next year in the first part. The first day of the next year is calculated once so as not to keep getting and comparing the year parts.
;WITH cte(TheDate,NextYear) AS
(
SELECT CAST(CONVERT(CHAR(4),GETDATE(),112)+'0101' AS DATETIME),
CAST(YEAR(GETDATE())*10000+10101 AS CHAR(8))
UNION ALL
SELECT DateAdd(d,1,TheDate),NextYear
FROM cte
WHERE DateAdd(d,1,TheDate)<NextYear
)
SELECT Week = DatePart(wk,TheDate),
TheDate
FROM cte
WHERE DateName(dw,TheDate) in ('Saturday')
ORDER BY TheDate
OPTION (MAXRECURSION 366)

with t as
(
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
)
select * from
(
select
current_timestamp
-datepart(dy,current_timestamp)
+row_number() over (order by t.b) d
from t, t t1, t t2
) tmp
where datepart(yyyy,d)=datepart(yyyy,current_timestamp)
and
DATENAME(dw,d)='sunday'

DECLARE #Year AS INT
SELECT #Year = 2020
;WITH weekends AS (
SELECT DATEFROMPARTS(#Year, 1, 1) AS dt
UNION ALL
SELECT DATEADD(DAY, 1, dt)
FROM weekends
WHERE dt < DATEFROMPARTS(#Year, 12, 31)
)
SELECT dt, DATENAME(MONTH, dt), DATENAME(DW, dt)
FROM weekends
WHERE DATEPART(DW, dt) IN (1, 7)
OPTION(MaxRecursion 366)

Calculate difference between 2 dates in SQL, excluding weekend days

I would like to build an SQL query which calculates the difference between 2 dates, without counting the week-end days in the result.
Is there any way to format the dates to obtain this result ? For example for Oracle database :
select sysdate - creation_dttm from the_table

You should try with a function :
CREATE FUNCTION TOTAL_WEEKDAYS(date1 DATE, date2 DATE)
RETURNS INT
RETURN ABS(DATEDIFF(date2, date1)) + 1
- ABS(DATEDIFF(ADDDATE(date2, INTERVAL 1 - DAYOFWEEK(date2) DAY),
ADDDATE(date1, INTERVAL 1 - DAYOFWEEK(date1) DAY))) / 7 * 2
- (DAYOFWEEK(IF(date1 < date2, date1, date2)) = 1)
- (DAYOFWEEK(IF(date1 > date2, date1, date2)) = 7);
Test :
SELECT TOTAL_WEEKDAYS('2013-08-03', '2013-08-21') weekdays1,
TOTAL_WEEKDAYS('2013-08-21', '2013-08-03') weekdays2;
Result :
| WEEKDAYS1 | WEEKDAYS2 |
-------------------------
| 13 | 13 |

I have found another way to do calculate the difference, by using only SQL :
select sysdate - creation_dttm
- 2 * (to_char(sysdate, 'WW') - to_char(creation_dttm, 'WW'))
from the_table

I have found several of the answers on this thread to not do what they claim. After some experimentation, testing and adjusting, I have this to contribute.
declare #firstdate Date
declare #seconddate Date
set #firstDate = convert(date, '2016-03-07')
set #seconddate = convert(date, '2016-04-04')
select (datediff(dd, #firstDate, #secondDate)) -
(( DateDiff(wk, #firstDate, #secondDate) * 2) -
case when datepart(dw, #FirstDate) = 7 then 1 else 0 end -
case when datepart(dw, #secondDate) = 7 then -1 else 0 end)
Test harness included - you can just adjust the two dates and run your own tests. This assumes that the difference between two adjacent weekday dates is 1. If your country uses different days to signify weekend, then you will have to set the date-base accordingly so your "Saturday" is 7, and your "sunday" is 1.

From a previous post:
DECLARE #StartDate DATETIME
DECLARE #EndDate DATETIME
SET #StartDate = '2008/10/01'
SET #EndDate = '2008/10/31'
SELECT
(DATEDIFF(dd, #StartDate, #EndDate) + 1)
-(DATEDIFF(wk, #StartDate, #EndDate) * 2)
-(CASE WHEN DATENAME(dw, #StartDate) = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw, #EndDate) = 'Saturday' THEN 1 ELSE 0 END)

Here is an example:
There are four variables, the first two are self-explanatory, just enter the date in YYYYMMDD format, the 3rd one is to set the number of normal work days in a given week, so if a site works 6 days a week, set it to 6, five days a week enter 5, etc. Finally, the
DATE_SEQ_NORML_FACTOR should be 1 when running against Oracle. This is to line up the Julian date to equal to 1 on Monday, 2 on Tuesday, etc when applying the MOD 7. Other DB will probably have different values between 0 and 6, so test it out before you use against other DBs.
Here are the limitations:
1. This formula assumes the first day of the week is MONDAY.
2. This formula assumes all days within the same week are CONTINUOUS.
3. This formula will work ONLY when the two dates involved in the calculation falls on a week day or work day, eg. the "Start Date" on a SATURDAY when the location works only MON-FRI will not work.
SELECT
&&START_DATE_YYYYMMDD "Start Date", --in YYYYMMDD format
&&END_DATE_YYYYMMDD "End Date", --in YYYYMMDD format
&&WK_WORK_DAY_CNT "Week Work Day Count", --Number of work day per week
&&DATE_SEQ_NORML_FACTOR "Normalization Factor", --set to 1 when run in Oracle
CASE
WHEN
FLOOR( TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR / 7 ) =
FLOOR( TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR / 7 )
THEN(
TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) -
TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + 1
)
ELSE(
(
&&WK_WORK_DAY_CNT - MOD( TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR , 7 ) + 1
) +
MOD( TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR , 7 ) +
(
(
(
TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) -
MOD( TO_CHAR( TO_DATE( &&END_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR , 7 )
) -
(
TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) +
(
7 -
(
MOD( TO_CHAR( TO_DATE( &&START_DATE_YYYYMMDD , 'YYYYMMDD') , 'J' ) + &&DATE_SEQ_NORML_FACTOR , 7 )
)
)
)
) / 7 * &&WK_WORK_DAY_CNT
)
) END "Week Day Count"
FROM DUAL

We've incorporated the following logic into several reports with great success. Sorry, I no longer recall the source of this script to give them credit.
DECLARE #range INT;
SET #range = DATEDIFF(DAY, #FirstDate, #SecondDate);
SET #NumWorkDays = (
SELECT
#range / 7 * 5 + #range % 7 - (
SELECT COUNT(*)
FROM (
SELECT 1 AS d
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
) weekdays
WHERE d <= #range % 7
AND DATENAME(WEEKDAY, #SecondDate - d) IN ('Saturday', 'Sunday'))
);

I've updated #JOpuckman's function to take into account that different regions don't always have a weekend of Saturday and Sunday. Here's the code in case anyone else needs to apply this globally;
DECLARE #FirstDate DateTime
DECLARE #SecondDate DateTime
SET #FirstDate = '08-20-2012'
SET #SecondDate = '08-24-2012'
DECLARE #range INT;
DECLARE #WeekendDayNameStart VARCHAR(50)
DECLARE #WeekendDayNameEnd VARCHAR(50)
SET #WeekendDayNameStart = 'FRIDAY'
SET #WeekendDayNameEnd = (
SELECT CASE #WeekendDayNameStart
WHEN 'SUNDAY' THEN 'MONDAY'
WHEN 'MONDAY' THEN 'TUESDAY'
WHEN 'TUESDAY' THEN 'WEDNESDAY'
WHEN 'WEDNESDAY' THEN 'THURSDAY'
WHEN 'THURSDAY' THEN 'FRIDAY'
WHEN 'FRIDAY' THEN 'SATURDAY'
WHEN 'SATURDAY' THEN 'SUNDAY'
END
)
DECLARE #NumWorkDays INT
SET #range = DATEDIFF(DAY, #FirstDate, #SecondDate);
SET #NumWorkDays = (
SELECT
#range / 7 * 5 + #range % 7 - (
SELECT COUNT(*)
FROM (
SELECT 1 AS d
UNION ALL SELECT 2
UNION ALL SELECT 3
UNION ALL SELECT 4
UNION ALL SELECT 5
UNION ALL SELECT 6
UNION ALL SELECT 7
) weekdays
WHERE d <= #range % 7
AND DATENAME(WEEKDAY, #SecondDate - d) IN (#WeekendDayNameStart, #WeekendDayNameEnd))
);
-- Calculate whether the current date is a working day
DECLARE #CurDateExtra INT
SET #CurDateExtra =
(
CASE DATENAME(WEEKDAY, #SecondDate)
WHEN #WeekendDayNameStart THEN 0
WHEN #WeekendDayNameEnd THEN 0
ELSE 1
END
)
SET #NumWorkDays = #NumWorkDays + #CurDateExtra
SELECT #NumWorkDays

Calculates the difference in days between the two dates
Calculates the difference in week numbers and year numbers, subtracts the week numbers and then multiplies the result by 2 to calculate number of non-workdays between the two dates. If year numbers are different calculation of 52 * year difference + week number.
((sysdate - ced.created_dt) + ((((to_char(ced.created_dt,'IW') - ((to_char(sysdate,'YY') - to_char(ced.created_dt,'YY'))* 52))
- to_char(to_char(sysdate,'IW')))) * 2)) duration_in_weekdays

You can try this:
SELECT
Id,
DATEDIFF(d, datefrom, dateto) AS TotDays,
DATEDIFF(wk, datefrom, dateto) AS Wkds,
DATEDIFF(d, datefrom, dateto) - DATEDIFF(wk, datefrom, dateto) AS Days
FROM
YOURTABLE

If your company has a DATES table listing the date and relative work day, as most companies do, you could build a temp table that excludes duplicate relative work days that will also give you the ability to exclude holidays from your calculations as well as weekends.

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

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+)

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

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.

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)

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?

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

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
)

#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