Calculate working days - Monday to Fri in Tsql - tsql

Question

How can i calculate the last working five days which is monday to Friday. my current script gets the last monday's date, but i cannot get the last friday's date. Please help
declare #StartDate datetime
declare #EndDate datetime
--Calculate date range for report
select #EndDate = Cast(convert(char(10), getdate(), 101)+' 00:00:00' as datetime)
select #StartDate = DateAdd(d, -7, #EndDate)
select #EndDate = Cast(convert(char(10), getdate(), 101)+' 23:59:59' as datetime)
select #StartDate startdate
select #EndDate enddate

Answer 1

Datepart offers you an easy way to get the week day:
SET DATEFIRST 1 -- monday is first day of the week
SELECT DATEPART(weekday, '20110725')
-- result is 1
See T-SQL Date functions and SET DATEFIRST for more information.
Using the weekday, you can work out how many days ago last monday and friday are and use 'AddDate' (like you are doing now) to calculate those.
Note that you should really use DATEDIFF for this type of date range selection. If you select everything up to 23:59:59 there's always a chance that some records are left out. For example 23:59:59.001 is out of range but it's still on the same day. With DATEDIFF you can test whether it's on the same day, discarding the time part. No need to bother with casting to string, adding time and casting back.

Answer 2

The answer is more complex than people are assuming. What you need is to go 5 days back and find the first monday before that, and the first friday after that. You can eather use ##datefirst, a calculation or 'set firstdate 1' for that. I prefer not using the last one, because it can't be done in functions. As you can see i used the calculation, ##datefirst is just as good.
Assuming you want the last group of monday to friday that is in the past. This query will get that. If you trust your current monday, you can just add 5 days and subtract 1 minute (I wouldn't trust it, it only returns last monday if you run the query on a monday).
In my sql, I am not aiming for simplicity, I am aiming for effectivity.
DECLARE #getdate datetime = dateadd(day, cast(getdate() as int), 0)
-- the 'Declare' can also be written like this thanks to #Andriy M
--DECLARE #getdate = CAST(GETDATE() - 0.5 AS int)
SELECT #getdate - 5 - CAST(#getdate- 5 as int) % 7 monday,
dateadd(minute, -1, #getdate) - CAST(#getdate- 5 as int) % 7 friday
Result:
Monday Friday
2011-07-18 0:00:00 2011-07-22 23:59:00
*first solution was a day off #AndriyM pointed it out, it has been solved.
Answer to #Andriy M
For some reason it acted different than I expected. I can't explain it but try this
select cast(dateadd(day, cast(getdate() as int) - .5, 0) as datetime),
cast(dateadd(day, cast(getdate() as int), 0) as datetime),
cast(dateadd(day, cast(getdate() as int) + .5, 0) as datetime)
in the morning the last 2 fields has same value, in the evening the first 2 fields has the same value. I am as surprised as you are, I wish I could explain it. It was tested here
https://data.stackexchange.com/stackoverflow/query/new

Answer 3

These two questions:
Find last sunday
How to get last day of last week in sql?
might be of some help.
Although, if you already know how to find the last Monday, you can easily find the corresponding Friday by adding 4 days to the Monday date using the DATEADD() function. For example:
SELECT #EndDate = DATEADD(DAY, 4, #StartDate)

Answer 4

It's all relative to TODAY's date, right?
So find out what today is DATEPART(weekday, getdate())
And then turn that into an adjustment variable -- e.g.:
declare #adjustment int
set #adjustment = CASE DATEPART(weekday, getdate()) WHEN 'MONDAY' THEN 0 WHEN 'TUESDAY' THEN 1, ...END
So then the Monday you want would be today - 7 - #adjustment
...and the Friday you want would be Monday + 5
Declare #myMonday smalldatetime,
#myFriday smalldatetime
set #myMonday = getdate() - 7 - #adjustment
set #myFriday = #myMonday + 5

Answer 5

DECLARE #my int
DECLARE #myDeduct int
DECLARE #day INT
DECLARE #mydate DATETIME
SET #mydate = '2012-08-01'
SET #myDeduct = 0
SET DateFirst 1 -- Set it monday=1 (value)
--Saturday and Sunday on the first and last day of a month will Deduct 1
IF (DATEPART(weekday,(DATEADD(dd,-(DAY(#mydate)-1),#mydate))) > 5)
SET #myDeduct = #myDeduct + 1
IF (DATEPART(weekday,(DATEADD(dd,-(DAY(DATEADD(mm,1,#mydate))),DATEADD(mm,1,#mydate)))) > 5)
SET #myDeduct = #myDeduct + 1
SET #my = day(DATEADD(dd,-(DAY(DATEADD(mm,1,#mydate))),DATEADD(mm,1,#mydate)))
select (((#my/7) * 5 + (#my%7)) - #myDeduct) as Working_Day_per_month

Answer 6

I'll throw my hat in the ring too. :-)
DECLARE #Date datetime = '01/11/2015'
DECLARE #StartDate datetime = DATEADD(d, (1 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date - 6) % 7)), #Date - 6) -- MONDAY
DECLARE #EndDate datetime = DATEADD(d, (5 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date - 6) % 7)), #Date - 6) -- FRIDAY
SELECT '#Date' as Variable ,CONVERT(date, #Date) as DateValue ,DATENAME(dw, #Date) as DayOfTheWeek
UNION SELECT '#StartDate' as Variable ,CONVERT(date, #StartDate) as DateValue ,DATENAME(dw, #StartDate) as DayOfTheWeek
UNION SELECT '#EndDate' as Variable ,CONVERT(date, #EndDate) as DateValue ,DATENAME(dw, #EndDate) as DayOfTheWeek
-- Variable DateValue DayOfTheWeek
-- ---------- ---------- ------------
-- #Date 2015-01-11 Sunday
-- #StartDate 2015-01-05 Monday
-- #EndDate 2015-01-09 Friday
BONUS: Here you can generate a quick table of the 5 weekdays using the same technique.
SELECT DATENAME(dw, DATEADD(d, TT.DaysToAdd, DATEADD(d, (1 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date - 6) % 7)), #Date - 6))) as DayOfTheWeek
, DATEADD(d, TT.DaysToAdd, DATEADD(d, (1 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date - 6) % 7)), #Date - 6)) as DateValue
FROM (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) - 1 as DaysToAdd FROM (VALUES(0),(0),(0),(0),(0)) a(n)) as TT
-- DayOfTheWeek DateValue
-- ------------------------------ -----------------------
-- Monday 2015-01-05 00:00:00.000
-- Tuesday 2015-01-06 00:00:00.000
-- Wednesday 2015-01-07 00:00:00.000
-- Thursday 2015-01-08 00:00:00.000
-- Friday 2015-01-09 00:00:00.000
Here's an explanation:
1) First we need to know what date to begin our evaluation from. For this example, we chose to use Sunday, January 11th, 2015.
DECLARE #Date2 datetime = '01/11/2015'
1b) Here's a nice to have bonus technique of getting the name for the day of the week, given a date value
SELECT #Date2 as DateValue, DATENAME(dw, #Date2) as DayOfTheWeek
2) Next we need to know deterministically (based on the US calendar) what the given day of the week is numerically between 1 and 7
NOTE: 1899.12.31 is the first Sunday before 1900.01.01, which is the MINIMUM value for the SmallDateTime data type.
NOTE: Yes, you could use DATEPART(dw, #Date) like this more simply, but it is not deterministic given that certain server environments could have different configurations
RESULTS: 1 = Sunday | 2 = Monday | 3 = Tuesday | 4 = Wednesday | 5 = Thursday | 6 = Friday | 7 = Saturday
SELECT ((DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date2) % 7) + 1) [DayOfWeek Deterministic (Based on US)]
3) Now, given any date, you should have a deterministic way of determining the Monday for that given week
SELECT DATEADD(d, (1 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date2) % 7)), #Date2) as [Monday Day of the Week - Deterministic (Based on US)]
4) Now, given any date, you should have a deterministic way of determining the Friday for that given week
SELECT DATEADD(d, (5 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date2) % 7)), #Date2) as [Monday Day of the Week - Deterministic (Based on US)]
5) The last date manipulation technique we need is to know how to get into a week that has the first full week of weekdays happening before it. For instance, if we are on a Sunday and we subtract 1 day from it, then we get to Saturday, which puts us in the previous week, which is the first full week of weekdays. Alternatively, if we also subtracted 1 day from Monday, it would only get us to Sunday, which is not the previous week, so subtracting 1 day is not enough. On the flip side, if we were on a Saturday and subtracted 7 days, it would take us past the previous full week of weekday, into the week before it, which is too far. Here's a run down of the analysis to figure out what the magic numbers is that you can subtract by that will work with any day of the week. As you can see below, the magic number is 6.
-- DAYS TO SUBTRACT
-- Day of the Week - 0 - 1 - 2 - 3 - 4 - 5 - 6 - 7
-- =============== ==== ==== ==== ==== ==== ==== ==== ====
-- Sunday Bad Good Good Good Good Good Good Good
-- Monday Bad Bad Good Good Good Good Good Good
-- Tuesday Bad Bad Bad Good Good Good Good Good
-- Wednesday Bad Bad Bad Bad Good Good Good Good
-- Thursday Bad Bad Bad Bad Bad Good Good Good
-- Friday Bad Bad Bad Bad Bad Bad Good Good
-- Saturday Good Good Good Good Good Good Good Bad
BONUS) If you want to have all the weekdays in little table, then you would want to also use a quick zero based "tally table". There are many ways to do this, so pick your flavor. Here are few.
SELECT * FROM (SELECT 0 as DaysToAdd UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4) as TT
SELECT * FROM (SELECT TOP 5 ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) - 1 as DaysToAdd FROM sys.all_columns a CROSS JOIN sys.all_columns b) as TT
SELECT * FROM (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) - 1 as DaysToAdd FROM (VALUES(0),(0),(0),(0),(0)) a(n)) as TT