Start Date Of The Current Week

Feb 17, 2006

Can anyone explain the result of the sql statement

Select DATEADD(wk, DATEDIFF(wk, 6,getdate()), 6)


Can we use this to get the start date of the current week always???

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I am trying to always get the start of the week of the current quarter in my criteria

This is the statement for the current quarter
Dateadd(qq, Datediff(qq,0,GetDate()), 0)

This is the statement for the current week
DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 0)

How to calculate from the start of the week of the current quarter...

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seriously now

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I can't seem to upload images so here's the link: [URL] ....

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I have two tables:

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- [URL] ...
- [URL] ...

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CREATE TABLE weekdays
(
datevalue datetime NOT NULL
, numericvalue INT NOT NULL
);
INSERT INTO weekdays (datevalue, numericvalue) VALUES

[code]....

The output should look like this

weeknototalvalue
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362015-09-02 00:00:00.000
372015-09-07 00:00:00.000
372015-09-08 00:00:00.000
382015-09-12 00:00:00.000
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Is there a SQL function that will do this?


Thanks,
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Mar 19, 2005

I wrote the following function to find the start of week date for a given date and a given start day of week.

For example:
If the day passed is Saturday, 2005-03-19, and Sunday is the start of the week, it returns: 2005-03-13 00:00:00.000

If the day passed is Monday, 2005-03-14, and Sunday is the start of the week, it returns: 2005-03-13 00:00:00.000

If the day passed is Monday, 2005-03-14, and Monday is the start of the week, it returns: 2005-03-14 00:00:00.000


Does anyone have a simpler algorithim for start of week that they care to post?


Edit (2006/4/15):
Posted a companion function F_END_OF_WEEK, on this topic:
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=64760


There are other Start of Time Period Functions posted here:
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=64755


There are other End Date of Time Period Functions here:
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=64759





create function dbo.F_START_OF_WEEK
(
@DATEdatetime,
-- Sun = 1, Mon = 2, Tue = 3, Wed = 4
-- Thu = 5, Fri = 6, Sat = 7
-- Default to Sunday
@WEEK_START_DAYint= 1
)
/*
Find the fisrt date on or before @DATE that matches
day of week of @WEEK_START_DAY.
*/
returnsdatetime
as
begin
declare @START_OF_WEEK_DATEdatetime
declare @FIRST_BOWdatetime

-- Check for valid day of week
if @WEEK_START_DAY between 1 and 7
begin
-- Find first day on or after 1753/1/1 (-53690)
-- matching day of week of @WEEK_START_DAY
-- 1753/1/1 is earliest possible SQL Server date.
select @FIRST_BOW = convert(datetime,-53690+((@WEEK_START_DAY+5)%7))
-- Verify beginning of week not before 1753/1/1
if @DATE >= @FIRST_BOW
begin
select @START_OF_WEEK_DATE =
dateadd(dd,(datediff(dd,@FIRST_BOW,@DATE)/7)*7,@FIRST_BOW)
end
end

return @START_OF_WEEK_DATE

end
go

-- Sample function calls

select dbo.F_START_OF_WEEK(getdate(),default) -- Returns Date for Sunday

select dbo.F_START_OF_WEEK(getdate(),1)-- Returns Date for Sunday
select dbo.F_START_OF_WEEK(getdate(),2)-- Returns Date for Monday
select dbo.F_START_OF_WEEK(getdate(),3)-- Returns Date for Tuesday
select dbo.F_START_OF_WEEK(getdate(),4)-- Returns Date for Wednesday
select dbo.F_START_OF_WEEK(getdate(),5)-- Returns Date for Thursday
select dbo.F_START_OF_WEEK(getdate(),6)-- Returns Date for Friday
select dbo.F_START_OF_WEEK(getdate(),7)-- Returns Date for Saturday

select dbo.F_START_OF_WEEK(getdate(),0)-- Returns NULL
select dbo.F_START_OF_WEEK(getdate(),8)-- Returns NULL



Edited 2005/4/9:

I thought I would also post an alternate way of doing the Start of Week instead of using the F_START_OF_WEEK function. These queries demo doing the Start of Week inline in a query, and use a similar algorithm to find the start of week, but the start day of week is hard coded.

I posted two versions. The first version is simpler, but it has a minor flaw that returns a false result if the start of week would be before 1753/1/1. For the vast majority of applications this would not be a problem. In the second, the algorithm is modified slightly to cause it to overflow if you pick a date that would result in a start of week before 1753/1/1. Note that the F_START_OF_WEEK function returns a NULL in this situation.

The demo queries use the F_TABLE_NUMBER_RANGE that is posted in another thread in order to generate dates to demonstrated the results:
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=47685



-- First demo query for Start of Week
-- Returns bad result if the start of week would be before 1753/1/1
select
DATE,
Sun = dateadd(dd,(datediff(dd,-53684,a.DATE)/7)*7,-53684),
Mon = dateadd(dd,(datediff(dd,-53690,a.DATE)/7)*7,-53690),
Tue = dateadd(dd,(datediff(dd,-53689,a.DATE)/7)*7,-53689),
Wed = dateadd(dd,(datediff(dd,-53688,a.DATE)/7)*7,-53688),
Thu = dateadd(dd,(datediff(dd,-53687,a.DATE)/7)*7,-53687),
Fri = dateadd(dd,(datediff(dd,-53686,a.DATE)/7)*7,-53686),
Sat = dateadd(dd,(datediff(dd,-53685,a.DATE)/7)*7,-53685)
from
(
select
DATE = convert(datetime,number)
from
F_TABLE_NUMBER_RANGE(36524,40000)
) a


-- Second demo query for Start of Week
-- Modified to cause an error instead of returning a bad date
-- if the start of week would be before 1753/1/1
select
DATE,
Sun = dateadd(dd,((datediff(dd,-53684,a.DATE+7)/7)*7)-7,-53684),
Mon = dateadd(dd,((datediff(dd,-53690,a.DATE+7)/7)*7)-7,-53690),
Tue = dateadd(dd,((datediff(dd,-53689,a.DATE+7)/7)*7)-7,-53689),
Wed = dateadd(dd,((datediff(dd,-53688,a.DATE+7)/7)*7)-7,-53688),
Thu = dateadd(dd,((datediff(dd,-53687,a.DATE+7)/7)*7)-7,-53687),
Fri = dateadd(dd,((datediff(dd,-53686,a.DATE+7)/7)*7)-7,-53686),
Sat = dateadd(dd,((datediff(dd,-53685,a.DATE+7)/7)*7)-7,-53685)
from
(
select
DATE = convert(datetime,number)
from
F_TABLE_NUMBER_RANGE(36524,40000)
) a






CODO ERGO SUM

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GO
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
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205 | 8/1/2005| 2250 | 99
206 | 8/1/2005| 1950 | 88
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208 | 8/7/2005| 2255 | 77
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...
...
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