SQL server 2008, datediff returns wrong value - sql-server-2008

I have two columns of Data type time. I'm using datediff to find the difference in hours.
The problem is when I try to find the difference between 00:00:00 to 06:00:00, it returns -18.
How can I fix it?
note: I need to calculate more difference with this so I can't just divide it with -3
my function- (datediff(HOUR, startHour, endHour))*60
Thanks in advance


You are not comparing 00:00:00 to 06:00:00. You have some date component
This gives -18 as an example
DECLARE #starthour datetime = '00:00:00';
DECLARE #endhour datetime = '18991231 06:00:00';
SELECT #starthour, #endhour, DATEDIFF(hour, #starthour, #endhour);
SET #starthour = '20120507 00:00:00';
SET #endhour = '20120506 06:00:00';
SELECT #starthour, #endhour, DATEDIFF(hour, #starthour, #endhour);


Reverse the parameters:
my function- (datediff(HOUR, endHour, startHour))*60
Edit:
The function DATEDIFF works with dates and for some reason, it thinks you're subtracting 6AM - Midnight (next day), which is 18 hours.
The following code sample works fine for me in SQL 2008, so you need to check your data type to make sure you're using TIME and not DATETIME or SMALLDATETIME.
declare #t1 time
set #t1 = '00:00:00'
declare #t2 time
set #t2 = '06:00:00'
select datediff(hour, #t1, #t2)
-- returns 6


Syntax:
DATEDIFF (datepart, startdate, enddate )
Examples:
SELECT DATEDIFF(hour, '00:00:00', '06:00:00');
Result: 6
SELECT DATEDIFF(hour, '00:00:00', '06:00:00')*60;
Result: 360
Refer DATEDIFF (Transact-SQL)


I'm late to the game, but I faced a similar problem.
DATEDIFF(HOUR,'23:00:00','06:00:00')
Result: -17
Reversing the two times (as suggested by another answer above) does not represent the time frame I want to capture. I don't want to know how many hours are between 6am and 11pm. I want 11pm to 6am.
To resolve this, wrap the DATEDIFF() in a CASE statement and add 24 when the start time is greater than the end time:
DECLARE #startTime TIME(0) = '23:00:00';
DECLARE #endTime TIME(0) = '06:00:00';
SELECT CASE WHEN #startTime > #endTime THEN 24 + DATEDIFF(HOUR,#startTime,#endTime) ELSE DATEDIFF(HOUR,#startTime,#endTime) END
Result: 7

Related

Get difference in hours between two dates

How can I subtract two dates (entry_date from status_change_date) to get hours between them? I do this based on an IF statement where both dates are in YYYY,MM,DD 12:00:00 format
(IF Invoice_Num != next Inv_Num then entry_date - status_change_date else...
It returns an answer in days instead of hours. I also tried using minutes(entry_day) and hours(status_change_date) in Crystal after the code was done, but I still got day values.

declare #startDate datetime
declare #endDate datetime
set #startDate = '2016-12-05 08:05:00';
set #endDate = '2016-12-05 09:52:00';
select CONVERT(varchar(3),DATEDIFF(minute,#startDate, #endDate)/60) + ':' +
RIGHT('0' + CONVERT(varchar(2),DATEDIFF(minute,#startDate,#endDate)%60),2)
as TotalHours
Check this example, get the hours and minutes between to datetime vars

From document TIMESTAMPDIFF(unit,datetime_expr1,datetime_expr2),
SELECT TIMESTAMPDIFF(HOUR, '2016-12-05 12:00:00', '2016-12-06 12:00:00');
This will get you 24 hours.
SQLFiddle Demo

Date difference in Hours format from bigint in SQL Server

I have a column of bigint type containing date & time information (like 1353056515, 1353067040, 1360839600 etc.)
My requirement is to get time difference in HOURS format between column which I mentioned above and current datetime.
I tried to find the solution, but those were so confusing. I'm new to SQL Server.
Please help.

Please try this.
declare #mydate datetime
DECLARE #LocalTimeOffset BIGINT
,#AdjustedLocalDatetime BIGINT
SET #LocalTimeOffset = DATEDIFF(second,GETDATE(),GETUTCDATE())
SET #AdjustedLocalDatetime = 1416474000 - #LocalTimeOffset
SELECT DATEADD(second,#AdjustedLocalDatetime, CAST('1970-01-01 00:00:00' AS datetime))
-- It will give you date 2014-11-20 14:30:00.000
The data difference operation:
select DATEDIFF(hour,#mydate,GETDATE())
or
Create Function
create FUNCTION dbo.fn_ConvertToDateTime (#Datetime BIGINT)
RETURNS DATETIME
AS
BEGIN
DECLARE #mydate datetime
DECLARE #LocalTimeOffset BIGINT
,#AdjustedLocalDatetime BIGINT
SET #LocalTimeOffset = DATEDIFF(second,GETDATE(),GETUTCDATE())
SET #AdjustedLocalDatetime = #Datetime - #LocalTimeOffset
SELECT #mydate=DATEADD(second,#AdjustedLocalDatetime, CAST('1970-01-01 00:00:00' AS datetime))
return #mydate
END;
GO
select mydate= dbo.fn_ConvertToDateTime (1416474000)
select DATEDIFF(hour,#mydate,GETDATE())
Hope that helps.

The number looks like a UNIX timestamp, which is the number of seconds since 1/1/1970 00:00:00 UTC. You can get the UNIX timestamp from a datetime with a simple DATEDIFF:
declare #date datetime='2014-11-20T10:00:00'
declare #epoch datetime='19700101'
select DATEDIFF(s,#epoch,#date)
To get the number of hours between two timestamps you simply need to divide the difference by 3600 (the number of seconds in an hour). You don't need to convert the timestamps to dates before calculating the difference :
declare #dateTS int=DATEDIFF(s,#epoch,#date)
declare #nowTS int=DATEDIFF(s,#epoch,GETUTCDATE())
select (#nowTS-#dateTS)/3600.0
Note that I used 3600.0 to get a decimal result. If 3600 is used the fractional part will be truncated and you'll get 0 for differences less than 1 hour. This is OK if you want to return whole hours.
In a query you could write something like this:
select (DATEDIFF(s,#epoch,GETUTCDATE())-[MY_TS_COLUMN])/3600.0 AS Hours
from [MY_TABLE]

Most efficient way of flooring todays date

This seems like overkill but is the only way I have been able to floor todays datetime to 00:00:00.000 at database level:
select CAST(FLOOR(CAST(CURRENT_TIMESTAMP AS float)) AS DATETIME)
I have tried using:
select FLOOR(getdate())
But get the following message:
Implicit conversion from data type datetime to float is not allowed. Use the CONVERT function to run this query.
Can anyone recommend another way of doing this?

Since you are using SQL Server 2008 you could make use of the date data type.
declare #Today date
set #Today = getdate()
select #Today
Or without the variable.
select cast(getdate() as date)
If you need to have the value as a datetime just cast it back to a datetime.
select cast(cast(getdate() as date) as datetime)

There are a lot of ways of doing this i have seen the floor one before. Here are a few more.
select cast(cast(CURRENT_TIMESTAMP as date) as datetime)
SELECT DATEADD(DAY, DATEDIFF(DAY, 0, CURRENT_TIMESTAMP), 0)
SELECT CAST(CAST(CURRENT_TIMESTAMP - 0.50000004 AS int) AS datetime)
I normaly do the Cast to date version.

How to get week number of the month from the date in sql server 2008

In SQL Statement in microsoft sql server, there is a built-in function to get week number but it is the week of the year.
Select DatePart(week, '2012/11/30') // **returns 48**
The returned value 48 is the week number of the year.
Instead of 48, I want to get 1, 2, 3 or 4 (week number of the month). I think the week number of the month can be achieved by modules with Month Number of this week. For e.g.
Select DATEPART(week, '2012/11/30')%MONTH('2012/11/30')
But I want to know is there other built-in functions to get WeekNumber of the month in MS SQL SERVER.

Here are 2 different ways, both are assuming the week starts on monday
If you want weeks to be whole, so they belong to the month in which they start:
So saturday 2012-09-01 and sunday 2012-09-02 is week 4 and monday 2012-09-03 is week 1 use this:
DECLARE #date date = '2012-09-01'
SELECT (day(datediff(d,0,#date)/7*7)-1)/7+1
If your weeks cut on monthchange so saturday 2012-09-01 and sunday 2012-09-02 is week 1 and monday 2012-09-03 is week 2 use this:
DECLARE #date date = '2012-09-01'
SELECT
datediff(ww,datediff(d,0,dateadd(m,datediff(m,7,#date),0)
)/7*7,dateadd(d,-1,#date))+1
I received an email from Gerald. He pointed out a flaw in the second method. This should be fixed now
I received an email from Ben Wilkins. He pointed out a flaw in the first method. This should be fixed now

DECLARE #DATE DATETIME
SET #DATE = '2013-08-04'
SELECT DATEPART(WEEK, #DATE) -
DATEPART(WEEK, DATEADD(MM, DATEDIFF(MM,0,#DATE), 0))+ 1 AS WEEK_OF_MONTH

No built-in function. It depends what you mean by week of month. You might mean whether it's in the first 7 days (week 1), the second 7 days (week 2), etc. In that case it would just be
(DATEPART(day,#Date)-1)/7 + 1
If you want to use the same week numbering as is used with DATEPART(week,), you could use the difference between the week numbers of the first of the month and the date in question (+1):
(DATEPART(week,#Date)- DATEPART(week,DATEADD(m, DATEDIFF(m, 0, #Date), 0))) + 1
Or, you might need something else, depending on what you mean by the week number.

Just look at the date and see what range it falls in.
Range 1-7 is the 1st week, Range 8-14 is the 2nd week, etc.
SELECT
CASE WHEN DATEPART(day,yourdate) < 8 THEN '1'
ELSE CASE WHEN DATEPART(day,yourdate) < 15 then '2'
ELSE CASE WHEN DATEPART(day,yourdate) < 22 then '3'
ELSE CASE WHEN DATEPART(day,yourdate) < 29 then '4'
ELSE '5'
END
END
END
END

Similar to the second solution, less code:
declare #date datetime = '2014-03-31'
SELECT DATEDIFF(week,0,#date) - (DATEDIFF(week,0,DATEADD(dd, -DAY(#date)+1, #date))-1)

Check this out... its working fine.
declare #date as datetime = '2014-03-10'
select DATEPART(week,#date) - DATEPART(week,cast(cast(year(#date) as varchar(4))+'-' + cast(month(#date) as varchar(2)) + '-01' as datetime))+1

WeekMonth = CASE WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 22 THEN '5'
WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 15 THEN '4'
WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 8 THEN '3'
WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 1 THEN '2'
ELSE '1'
END

There is no inbuilt function to get you the week number. I dont think dividing will help you anyway as the number of weeks in a month is not constant.
http://msdn.microsoft.com/en-us/library/bb675168.aspx
I guess you can divide the number(48) by 4 and take the modules of the same and project that as the week number of that month, by adding one to the result.

Here's a suggestion for getting the first and last days of the week for a month:
-- Build a temp table with all the dates of the month
drop table #tmp_datesforMonth
go
declare #begDate datetime
declare #endDate datetime
set #begDate = '6/1/13'
set #endDate = '6/30/13';
WITH N(n) AS
( SELECT 0
UNION ALL
SELECT n+1
FROM N
WHERE n <= datepart(dd,#enddate)
)
SELECT DATEADD(dd,n,#BegDate) as dDate
into #tmp_datesforMonth
FROM N
WHERE MONTH(DATEADD(dd,n,#BegDate)) = MONTH(#BegDate)
--- pull results showing the weeks' dates and the week # for the month (not the week # for the current month)
select MIN(dDate) as BegOfWeek
, MAX(dDate) as EndOfWeek
, datediff(week, dateadd(week, datediff(week, 0, dateadd(month, datediff(month, 0, dDate), 0)), 0), dDate) as WeekNumForMonth
from #tmp_datesforMonth
group by datediff(week, dateadd(week, datediff(week, 0, dateadd(month, datediff(month, 0, dDate), 0)), 0), dDate)
order by 3, 1

A dirty but easy one liner using Dense_Rank function. Performance WILL suffer, but effective none the less.
DENSE_RANK()over(Partition by Month(yourdate),Year(yourdate) Order by Datepart(week,yourdate) asc) as Week

Here is the query that brings the week number on whatever the startday and endday of the week it may be.
SET DATEFIRST 2
DECLARE #FROMDATE DATE='12-JAN-2015'
-- Get the first day of month
DECLARE #ALLDATE DATE=DATEADD(month, DATEDIFF(month, 0, #FROMDATE), 0)
DECLARE #FIRSTDATE DATE
;WITH CTE as
(
-- Get all dates in that month
SELECT 1 RNO,CAST(#ALLDATE AS DATE) as DATES
UNION ALL
SELECT RNO+1, DATEADD(DAY,1,DATES )
FROM CTE
WHERE DATES < DATEADD(MONTH,1,#ALLDATE)
)
-- Retrieves the first day of week, ie, if first day of week is Tuesday, it selects first Tuesday
SELECT TOP 1 #FIRSTDATE = DATES
FROM CTE
WHERE DATEPART(W,DATES)=1
SELECT (DATEDIFF(DAY,#FIRSTDATE,#FROMDATE)/7)+1 WEEKNO
For more information I have answered for the below question. Can check that.
How do I find week number of a date according to DATEFIRST

floor((day(#DateValue)-1)/7)+1

Here you go....
Im using the code below..
DATEPART(WK,#DATE_INSERT) - DATEPART(WK,DATEADD(DAY,1,DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#DATE_INSERT),0)))) + 1

Try Below Code:
declare #dt datetime='2018-03-15 05:16:00.000'
IF (Select (DatePart(DAY,#dt)%7))>0
Select (DatePart(DAY,#dt)/7) +1
ELSE
Select (DatePart(DAY,#dt)/7)

There is an inbuilt option to get the week number of the year
**select datepart(week,getdate())**

You can simply get week number by getting minimum week number of month and deduct it from week number. Suppose you have a table with dates
select
emp_id, dt , datepart(wk,dt) - (select min(datepart(wk,dt))
from
workdates ) + 1 from workdates

Solution:
declare #dt datetime='2018-03-31 05:16:00.000'
IF (Select (DatePart(DAY,#dt)%7))>0
Select (DatePart(DAY,#dt)/7) +1
ELSE
Select (DatePart(DAY,#dt)/7)

declare #end_date datetime = '2019-02-28';
select datepart(week, #end_date) - datepart(week, convert(datetime, substring(convert(nvarchar, convert(datetime, #end_date), 127), 1, 8) + '01')) + 1 [Week of Month];

Here is the tried and tested solution for this query in any situation - like if 1st of the month is on Friday , then also this will work -
select (DATEPART(wk,#date_given)-DATEPART(wk,dateadd(d,1-day(#date_given),#date_given)))+1
above are some solutions which will fail if the month's first date is on Friday , then 4th will be 2nd week of the month

Logic here works as well 4.3 weeks in every month. Take that from the DATEPART(WEEK) on every month but January. Just another way of looking at things. This would also account for months where there is a 5th week
DECLARE #date VARCHAR(10)
SET #date = '7/27/2019'
SELECT CEILING(DATEPART(WEEK,#date)-((DATEPART(MONTH,#date)-1)*4.3333)) 'Week of Month'

Below will only work if you have every week of the month represented in the select list. Else the rank function will not work, but it is a good solution.
SELECT DENSE_RANK() OVER (PARTITION BY MONTH(DATEFIELD)
ORDER BY DATEPART(WEEK,DATEFIELD) ASC) AS WeekofMont

try this one
declare #date datetime = '20210928'
select convert(int,(((cast(datepart(day,#date) as decimal(4,2))/7)-(1.00/7.00))+1.00))

select datepart(week,#date)-datepart(week,dateadd(day,1,eomonth(dateadd(m,-1,#date))))+1
or
select datepart(week,#date)-datepart(week,dateadd(d,-datepart(d,#date)+1,#date))+1
steps:
1,get the first day of month
2,week of year of the date - week of year of the first day of the month
3,+1
2023-1-1 is sunday
SET DATEFIRST 1;
DECLARE #date date = '2023-1-02';
select datepart(week,#date)-datepart(week,dateadd(day,1,eomonth(dateadd(m,-1,#date))))+1
return 2
SET DATEFIRST 7;
DECLARE #date date = '2023-1-02';
select datepart(week,#date)-datepart(week,dateadd(day,1,eomonth(dateadd(m,-1,#date))))+1
return 1

Code is below:
set datefirst 7
declare #dt datetime='29/04/2016 00:00:00'
select (day(#dt)+datepart(WEEKDAY,dateadd(d,-day(#dt),#dt+1)))/7

select #DateCreated, DATEDIFF(WEEK, #DateCreated, GETDATE())

Compare datetime in stored procedure

I have a sql 2008 database and I am creating a stored procedure that shall check if a datetime is more than 3 hours old but I don't know how to do it.
Do you have some way to do it?
the datetime is a field in the table.
BR

Rather than applying DATEDIFF to the column value, which will negate an index, I suggest using a comparison of the column to an expression (which can use an index).
If you want this as a filter:
SELECT columns
FROM dbo.table
WHERE DateTimeColumn < DATEADD(HOUR, -3, CURRENT_TIMESTAMP);
(If you want only the rows that are newer than 3 hours old, change < to > or >=.)
If you want to return all rows with a column showing whether it is more than 3 hours old:
SELECT columns, [3HoursOld] = CASE
WHEN DateTimeColumn < DATEADD(HOUR, -3, CURRENT_TIMESTAMP)
THEN 'Yes, older than 3 hours.'
ELSE 'No, not older than 3 hours.'
END
FROM dbo.table;

Take at look at the DATEDIFF function.
DATEDIFF ( datepart , startdate , enddate )
You would then use with datepart set to hh and the enddate set to the current time. To get the current database time you could use GETDATE(). Compare the result with 3 since it will return the number of hours passed.

#date is the date you want to compare
declare #date datetime
set #date= '2012-02-15 14:20:42.797'
SELECT DateDiff(hh, DATEADD(hh,-3,#date), GETDATE()) --if it's > 3
you better create a Boolean function that does the trick that you can use where ever you like