I want to find the difference between two dates which should be exact in months like if the date difference is greater than 182 days them on 183rd Day it should show as 7 Months.I tried below one,
SELECT ROUND(cast(DATEDIFF(DD,'2018-01-01 18:45:30.203',GETDATE()) as float)/30,0)
but it has 15 days difference.
I wouldn’t use 30. It’s fail on some months. For example Jan 1 and March 2 since February doesn’t have at least 30 days. But I think this is what you are after. If the current day isn’t the first of the month then add a month.
SELECT
Case
when datepart(day,getdate()) > 1
Then datediff(month,'2018-01-01 18:45:30.203',GETDATE()) + 1
Else datediff(month,'2018-01-01 18:45:30.203',GETDATE())
End
I think that calculating the difference in months as an integer is very similar to calculating the age of a person. We can take the DATEDIFF in months, add this number of months to the first date and compare that to the second date to decide whether we have to subtract 1 from the difference:
DECLARE #Date1 datetime = '2018-01-01 18:45:30.203';
DECLARE #Date2 datetime = GETDATE();
SELECT
CASE
WHEN DATEADD(month, DATEDIFF(month, #Date1, #Date2), #Date1) > #Date2
THEN DATEDIFF(month, #Date1, #Date2) - 1
ELSE DATEDIFF(month, #Date1, #Date2)
END
I have a query that converts week and year to date.
But it returns the exact date.
But what i want is, I need the date to be such a day, that is same as the first day of the year.
dateadd (week, PromisedWeek-1, dateadd (year, PromisedYear-1900, 0)) - 4 -
datepart(dw, dateadd (week, PromisedWeek-1, dateadd (year, PromisedYear-1900, 0)) - 4) + 1
Hypothetical example.
My current query does is:
If week is 4 and year 2017, returns 26-Sun-2017
My need:
If week is 4 and year 2017 and January 1st was a Wednesday, its should return 29-Wednesday-2017.
Hoping that you guys get what I am trying to explain.
I need the query to return such a date which has the same day as that of the current year's 1st day.
Unless I'm reading this wrong, you're making things too hard on yourself. Just do this:
declare #year int; set #year = 2017;
declare #week int; set #week = 4;
select dateadd(week, #week, dateadd(year, #year - 1900, 0))
Result:
2017-01-29 00:00:00.000
The inner dateadd() gives you the first day of the requested year (Jan 1), regardless of weekday. If it's a Wednesday, you'll get a Wednesday. This year, it's Sunday. The outer dateadd() then adds full weeks to that, so you end up with the same day of the week, just like you asked.
If that's not what you want, please explain how it needs to be more complicated.
This is pretty easy if you have a calendar table. A calendar table is a table that contains every date and its parts for a given range. I created mine for 2010 to 2020 with this code:
declare #start_dt as date = '1/1/2010';
declare #end_dt as date = '1/1/2020';
declare #dates as table (
date_id date primary key,
date_year smallint,
date_month tinyint,
date_day tinyint,
weekday_id tinyint,
weekday_nm varchar(10),
month_nm varchar(10),
day_of_year smallint,
quarter_id tinyint,
first_day_of_month date,
last_day_of_month date,
start_dts datetime,
end_dts datetime
)
while #start_dt < #end_dt
begin
insert into #dates(
date_id, date_year, date_month, date_day,
weekday_id, weekday_nm, month_nm, day_of_year, quarter_id,
first_day_of_month, last_day_of_month,
start_dts, end_dts
)
values(
#start_dt, year(#start_dt), month(#start_dt), day(#start_dt),
datepart(weekday, #start_dt), datename(weekday, #start_dt), datename(month, #start_dt), datepart(dayofyear, #start_dt), datepart(quarter, #start_dt),
dateadd(day,-(day(#start_dt)-1),#start_dt), dateadd(day,-(day(dateadd(month,1,#start_dt))),dateadd(month,1,#start_dt)),
cast(#start_dt as datetime), dateadd(second,-1,cast(dateadd(day, 1, #start_dt) as datetime))
)
set #start_dt = dateadd(day, 1, #start_dt)
end
select *
into Calendar
from #dates
Once you have a calendar table you can query for the correct weekday in the PromisedWeek
declare #PromisedYear as numeric
declare #PromisedWeek as int
set #PromisedYear = 2017
set #PromisedWeek = 4
select concat(date_day, '-', weekday_nm, '-', #PromisedYear)
from Calendar as c
where
weekday_id =
(select weekday_ID
from Calendar
where date_year = #PromisedYear
and day_of_year = 1
)
and datepart(week, date_id) = #PromisedWeek
and date_year = #PromisedYear
Turns out that January 1st 2017 was a Sunday and the 4th Sunday in 2017 was January 22nd, so the return is 22-Sunday-2017
I have to get/create date from the user input of week of month (week number in that month - 1st,2nd,3rd,4th and last) and day of week (sunday,monday..) in SQL server.
Examples:
4th Sunday of every month, Last Friday of every month, First Monday etc.
I was able to do it easily in .net but SQL server does seem limited in the date functions.
I am having to use lot of logic to get the date. To calculate the date using the above two parameters I had to use lot of datepart function.
Any suggestions on how to come up with the optimal SQL query for such a function?
I created a function other day for another OP GET Month, Quarter based on Work Week number
This function takes the current year as default it can be further modified to take Year as a parameter too.
an extension to that function can produce the results you are looking for ....
WITH X AS
(
SELECT TOP (CASE WHEN YEAR(GETDATE()) % 4 = 0 THEN 366 ELSE 365 END)-- to handle leap year
DATEADD(DAY
,ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1
, CAST(YEAR(GETDATE()) AS VARCHAR(4)) + '0101' )
DayNumber
From master..spt_values
),DatesData AS(
SELECT DayNumber [Date]
,DATEPART(WEEKDAY,DayNumber) DayOfTheWeek
,DATEDIFF(WEEK,
DATEADD(WEEK,
DATEDIFF(WEEK, 0, DATEADD(MONTH,
DATEDIFF(MONTH, 0, DayNumber), 0)), 0)
, DayNumber- 1) + 1 WeekOfTheMonth
FROM X )
SELECT * FROM DatesData
WHERE DayOfTheWeek = 6 -- A function would expect these two parameters
AND WeekOfTheMonth = 4 -- #DayOfTheWeek and #WeekOfTheMonth
Here is a general formula:
declare #month as datetime --set to the first day of the month you wish to use
declare #week as int --1st, 2nd, 3rd...
declare #day as int --Day of the week (1=sunday, 2=monday...)
--Second monday in August 2015
set #month = '8/1/2015'
set #week = 2
set #day = 2
select dateadd(
day,
((7+#day) - datepart(weekday, #month)) % 7 + 7 * (#week-1),
#month
)
You can also find the last, 2nd to last... etc with this reverse formula:
--Second to last monday in August 2015
set #month = '8/1/2015'
set #week = 2
set #day = 2
select
dateadd(
day,
-((7+datepart(weekday, dateadd(month,1,#month)-1)-#day)) % 7 - 7 * (#week-1),
dateadd(month,1,#month)-1
)
I have a table like this:
Month Value
2012-08-01 0.345
2012-09-01 0.543
2012-10-01 0.321
2012-11-01 0.234
2012-12-01 0.234
User inputs week range from '2012-09-29' to '2012-10-13'
Output should show results for all weeks in requested range and average values for each week with the following logic:
- if all weekdays are entirely in one month, just use monthly value for that month
- if weekdays are spread out over two months, calculate weekly value as average between those two months giving preference to the month that contains the most days of that week.
If someone can give me an idea how to do something like this in T-SQL that would be highly appreciated.
The last query is the example. The Calendar table is built on-request, but every database can do with a persisted Calendar table, on which you would filter the date range instead.
declare #tbl table (
Month datetime,
Value decimal(10,3));
insert #tbl select
'2012-08-01', 0.345 union all select
'2012-09-01', 0.543 union all select
'2012-10-01', 0.321 union all select
'2012-11-01', 0.234 union all select
'2012-12-01', 0.234;
declare #start datetime, #end datetime;
select #start = '2012-09-29', #end ='2012-10-13';
;with Calendar(TheDate,StartOfWeek,StartOfMonth) as(
select #start, #start+1-Datepart(dw,#start), #start-Day(#start)+1
union all
select TheDate+1, TheDate+1+1-Datepart(dw,TheDate+1),
TheDate+1-Day(TheDate+1)+1
from Calendar
where TheDate < #end
)
select case when #start > v.StartOfWeek
then #start else v.StartOfWeek end RangeStart,
case when #end < v.StartOfWeek+6
then #end else v.StartOfWeek+6 end RangeEnd,
cast(avg(m.value) as decimal(10,3)) [Average]
from Calendar v
join #tbl m on v.StartOfMonth = m.Month
group by v.StartOfWeek;
Output
RANGESTART RANGEEND Average
September, 29 2012 September, 29 2012 0.543
September, 30 2012 October, 06 2012 0.353
October, 07 2012 October, 13 2012 0.321
Your query would be something like this. The idea is find the first day of the next month DATEADD(mm, DATEDIFF(mm, 0, Month) + 1, 0. Calculate the number of days, and then you can get the monthly total, and calculate the average based on the difference between the first day of the current month and next month. (SQL syntax may need some clean up).
declare #startdate datetime
declare #enddate datetime
set #startdate = '2012-09-05'
set #enddate ='2012-10-13'
Select Monthtotal/DateDiff(d,Month,NextMonth)
FROM
(Select
Month, DATEADD(mm, DATEDIFF(mm, 0, Month) + 1, 0) NextMonth,
DateDiff(d,Month, DATEADD(mm, DATEDIFF(mm, 0, Month) + 1, 0) * Value as Monthtotal
FROM DatesTable
WHERE
#startdate >= Month and
#enddate <= DATEADD(mm, DATEDIFF(mm, 0, Month) + 1, 0))
I am trying to group records by week, storing the aggregated date as the first day of the week. However, the standard technique I use for rounding off dates does not appear to work correctly with weeks (though it does for days, months, years, quarters and any other timeframe I've applied it to).
Here is the SQL:
select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), 0);
This returns 2011-08-22 00:00:00.000, which is a Monday, not a Sunday. Selecting ##datefirst returns 7, which is the code for Sunday, so the server is setup correctly in as far as I know.
I can bypass this easily enough by changing the above code to:
select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), -1);
But the fact that I have to make such an exception makes me a little uneasy. Also, apologies if this is a duplicate question. I found some related questions but none that addressed this aspect specifically.
To answer why you're getting a Monday and not a Sunday:
You're adding a number of weeks to the date 0. What is date 0? 1900-01-01. What was the day on 1900-01-01? Monday. So in your code you're saying, how many weeks have passed since Monday, January 1, 1900? Let's call that [n]. Ok, now add [n] weeks to Monday, January 1, 1900. You should not be surprised that this ends up being a Monday. DATEADD has no idea that you want to add weeks but only until you get to a Sunday, it's just adding 7 days, then adding 7 more days, ... just like DATEDIFF only recognizes boundaries that have been crossed. For example, these both return 1, even though some folks complain that there should be some sensible logic built in to round up or down:
SELECT DATEDIFF(YEAR, '2010-01-01', '2011-12-31');
SELECT DATEDIFF(YEAR, '2010-12-31', '2011-01-01');
To answer how to get a Sunday:
If you want a Sunday, then pick a base date that's not a Monday but rather a Sunday. For example:
DECLARE #dt DATE = '1905-01-01';
SELECT [start_of_week] = DATEADD(WEEK, DATEDIFF(WEEK, #dt, CURRENT_TIMESTAMP), #dt);
This will not break if you change your DATEFIRST setting (or your code is running for a user with a different setting) - provided that you still want a Sunday regardless of the current setting. If you want those two answers to jive, then you should use a function that does depend on the DATEFIRST setting, e.g.
SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, CURRENT_TIMESTAMP), CURRENT_TIMESTAMP);
So if you change your DATEFIRST setting to Monday, Tuesday, what have you, the behavior will change. Depending on which behavior you want, you could use one of these functions:
CREATE FUNCTION dbo.StartOfWeek1 -- always a Sunday
(
#d DATE
)
RETURNS DATE
AS
BEGIN
RETURN (SELECT DATEADD(WEEK, DATEDIFF(WEEK, '19050101', #d), '19050101'));
END
GO
...or...
CREATE FUNCTION dbo.StartOfWeek2 -- always the DATEFIRST weekday
(
#d DATE
)
RETURNS DATE
AS
BEGIN
RETURN (SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, #d), #d));
END
GO
Now, you have plenty of alternatives, but which one performs best? I'd be surprised if there would be any major differences but I collected all the answers provided so far and ran them through two sets of tests - one cheap and one expensive. I measured client statistics because I don't see I/O or memory playing a part in the performance here (though those may come into play depending on how the function is used). In my tests the results are:
"Cheap" assignment query:
Function - client processing time / wait time on server replies / total exec time
Gandarez - 330/2029/2359 - 0:23.6
me datefirst - 329/2123/2452 - 0:24.5
me Sunday - 357/2158/2515 - 0:25.2
trailmax - 364/2160/2524 - 0:25.2
Curt - 424/2202/2626 - 0:26.3
"Expensive" assignment query:
Function - client processing time / wait time on server replies / total exec time
Curt - 1003/134158/135054 - 2:15
Gandarez - 957/142919/143876 - 2:24
me Sunday - 932/166817/165885 - 2:47
me datefirst - 939/171698/172637 - 2:53
trailmax - 958/173174/174132 - 2:54
I can relay the details of my tests if desired - stopping here as this is already getting quite long-winded. I was a bit surprised to see Curt's come out as the fastest at the high end, given the number of calculations and inline code. Maybe I'll run some more thorough tests and blog about it... if you guys don't have any objections to me publishing your functions elsewhere.
For these that need to get:
Monday = 1 and Sunday = 7:
SELECT 1 + ((5 + DATEPART(dw, GETDATE()) + ##DATEFIRST) % 7);
Sunday = 1 and Saturday = 7:
SELECT 1 + ((6 + DATEPART(dw, GETDATE()) + ##DATEFIRST) % 7);
Above there was a similar example, but thanks to double "%7" it would be much slower.
For those who need the answer at work and creating function is forbidden by your DBA, the following solution will work:
select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-1), YourDate) as DATE) as WeekStart
From.....
This gives the start of that week. Here I assume that Sundays are the start of weeks. If you think that Monday is the start, you should use:
select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-2), YourDate) as DATE) as WeekStart
From.....
This works wonderfully for me:
CREATE FUNCTION [dbo].[StartOfWeek]
(
#INPUTDATE DATETIME
)
RETURNS DATETIME
AS
BEGIN
-- THIS does not work in function.
-- SET DATEFIRST 1 -- set monday to be the first day of week.
DECLARE #DOW INT -- to store day of week
SET #INPUTDATE = CONVERT(VARCHAR(10), #INPUTDATE, 111)
SET #DOW = DATEPART(DW, #INPUTDATE)
-- Magic convertion of monday to 1, tuesday to 2, etc.
-- irrespect what SQL server thinks about start of the week.
-- But here we have sunday marked as 0, but we fix this later.
SET #DOW = (#DOW + ##DATEFIRST - 1) %7
IF #DOW = 0 SET #DOW = 7 -- fix for sunday
RETURN DATEADD(DD, 1 - #DOW,#INPUTDATE)
END
Maybe you need this:
SELECT DATEADD(DD, 1 - DATEPART(DW, GETDATE()), GETDATE())
Or
DECLARE #MYDATE DATETIME
SET #MYDATE = '2011-08-23'
SELECT DATEADD(DD, 1 - DATEPART(DW, #MYDATE), #MYDATE)
Function
CREATE FUNCTION [dbo].[GetFirstDayOfWeek]
( #pInputDate DATETIME )
RETURNS DATETIME
BEGIN
SET #pInputDate = CONVERT(VARCHAR(10), #pInputDate, 111)
RETURN DATEADD(DD, 1 - DATEPART(DW, #pInputDate),
#pInputDate)
END
GO
Googled this script:
create function dbo.F_START_OF_WEEK
(
#DATE datetime,
-- Sun = 1, Mon = 2, Tue = 3, Wed = 4
-- Thu = 5, Fri = 6, Sat = 7
-- Default to Sunday
#WEEK_START_DAY int = 1
)
/*
Find the fisrt date on or before #DATE that matches
day of week of #WEEK_START_DAY.
*/
returns datetime
as
begin
declare #START_OF_WEEK_DATE datetime
declare #FIRST_BOW datetime
-- 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
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=47307
CREATE FUNCTION dbo.fnFirstWorkingDayOfTheWeek
(
#currentDate date
)
RETURNS INT
AS
BEGIN
-- get DATEFIRST setting
DECLARE #ds int = ##DATEFIRST
-- get week day number under current DATEFIRST setting
DECLARE #dow int = DATEPART(dw,#currentDate)
DECLARE #wd int = 1+(((#dow+#ds) % 7)+5) % 7 -- this is always return Mon as 1,Tue as 2 ... Sun as 7
RETURN DATEADD(dd,1-#wd,#currentDate)
END
For the basic (the current week's Sunday)
select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) as date)
If previous week:
select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) -7 as date)
Internally, we built a function that does it but if you need quick and dirty, this will do it.
Since Julian date 0 is a Monday just add the number of weeks to Sunday
which is the day before -1 Eg. select dateadd(wk,datediff(wk,0,getdate()),-1)
I found some of the other answers long-winded or didn't actually work if you wanted Monday as the start of the week.
Sunday
SELECT DATEADD(week, DATEDIFF(week, -1, GETDATE()), -1) AS Sunday;
Monday
SELECT DATEADD(week, DATEDIFF(week, 0, GETDATE() - 1), 0) AS Monday;
Set DateFirst 1;
Select
Datepart(wk, TimeByDay) [Week]
,Dateadd(d,
CASE
WHEN Datepart(dw, TimeByDay) = 1 then 0
WHEN Datepart(dw, TimeByDay) = 2 then -1
WHEN Datepart(dw, TimeByDay) = 3 then -2
WHEN Datepart(dw, TimeByDay) = 4 then -3
WHEN Datepart(dw, TimeByDay) = 5 then -4
WHEN Datepart(dw, TimeByDay) = 6 then -5
WHEN Datepart(dw, TimeByDay) = 7 then -6
END
, TimeByDay) as StartOfWeek
from TimeByDay_Tbl
This is my logic. Set the first of the week to be Monday then calculate what is the day of the week a give day is, then using DateAdd and Case I calculate what the date would have been on the previous Monday of that week.
This is a useful function for me
/* MeRrais 211126
select [dbo].[SinceWeeks](0,NULL)
select [dbo].[SinceWeeks](5,'2021-08-31')
*/
alter Function [dbo].[SinceWeeks](#Weeks int, #From datetime=NULL)
Returns date
AS
Begin
if #From is null
set #From=getdate()
return cast(dateadd(day, -(#Weeks*7+datepart(dw,#From)-1), #From) as date)
END
I don't have any issues with any of the answers given here, however I do think mine is a lot simpler to implement, and understand. I have not run any performance tests on it, but it should be neglegable.
So I derived my answer from the fact that dates are stored in SQL server as integers, (I am talking about the date component only). If you don't believe me, try this SELECT CONVERT(INT, GETDATE()), and vice versa.
Now knowing this, you can do some cool math equations. You might be able to come up with a better one, but here is mine.
/*
TAKEN FROM http://msdn.microsoft.com/en-us/library/ms181598.aspx
First day of the week is
1 -- Monday
2 -- Tuesday
3 -- Wednesday
4 -- Thursday
5 -- Friday
6 -- Saturday
7 (default, U.S. English) -- Sunday
*/
--Offset is required to compensate for the fact that my ##DATEFIRST setting is 7, the default.
DECLARE #offSet int, #testDate datetime
SELECT #offSet = 1, #testDate = GETDATE()
SELECT CONVERT(DATETIME, CONVERT(INT, #testDate) - (DATEPART(WEEKDAY, #testDate) - #offSet))
I had a similar problem. Given a date, I wanted to get the date of the Monday of that week.
I used the following logic: Find the day number in the week in the range of 0-6, then subtract that from the originay date.
I used: DATEADD(day,-(DATEPART(weekday,)+5)%7,)
Since DATEPRRT(weekday,) returns 1 = Sundaye ... 7=Saturday,
DATEPART(weekday,)+5)%7 returns 0=Monday ... 6=Sunday.
Subtracting this number of days from the original date gives the previous Monday. The same technique could be used for any starting day of the week.
I found this simple and usefull. Works even if first day of week is Sunday or Monday.
DECLARE #BaseDate AS Date
SET #BaseDate = GETDATE()
DECLARE #FisrtDOW AS Date
SELECT #FirstDOW = DATEADD(d,DATEPART(WEEKDAY,#BaseDate) *-1 + 1, #BaseDate)
Maybe I'm over simplifying here, and that may be the case, but this seems to work for me. Haven't ran into any problems with it yet...
CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7 - 7) as 'FirstDayOfWeek'
CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7) as 'LastDayOfWeek'