I am looking for function that give me the right week-number.
The first day of the year should always be on week number 1.
The first saturday of a year belongs to the first week.
In another words: here is what i am looking for, but not in PostgreSQL, but in sql-server.
I realy want to do that using a function that i will call later.
Thank you in advance for your help.
In SQL Server there's no need for a custom function; just use the built in DATEPART function with week as the first argument:
-- first day
select datepart(week, '2013-01-01');
Output: 1
-- first Saturday
select datepart(week, '2013-01-05');
Output: 1
-- first Sunday
select datepart(week, '2013-01-06');
Output: 2
-- today
select datepart(week, '2013-04-01');
Output: 14
Here's a live demo: http://www.sqlfiddle.com/#!3/d41d8/11802
select datepart(week, getdate()) as CurrentWeekofYear
Related
Using SQL Server: I am trying to find all records prior to the first day of the current month and set this as a parameter in an SSRS report (so I can't use a static value).
So, I need all records prior to the first day of the each current month going forward in column CREATEDDATETIME ('yyyy-mm-dd').
I have seen a lot of threads on how to find records for a specific month and various other searches but none specifically related to the above. Interested to see if the EOMONTH function will be of use here.
Thanks for the help and advice.
Here is an expression to use EOMONTH() function with optional parameter -1.
Explanations:
DateAdd: add 1 day to expression
getdate is current date
EOMONTH is end day of a given month; however, if you put an optional integer -1, this would mean last month
Thus: first day of current month is add one day to end of day last month
SELECT DATEADD(DAY,1,EOMONTH(getdate(),-1));
Result: 2018-04-01
SO in your query:
select *
from table
where CREATEDDATETIME < DATEADD(DAY,1,EOMONTH(getdate(),-1));
I would use datefromparts():
select t.*
from t
where CREATEDDATETIME < datefromparts(year(getdate()), month(getdate()), 1);
There's already two other answers that work, but for completeness here is a third common technique:
SELECT *
FROM [table]
WHERE CREATEDDATETIME < dateadd(month, datediff(month,0, current_timestamp), 0)
You might also get answers suggesting you build the date using strings. Don't do that. It's both the least efficient and most error prone option you could use.
all three answers are great, how ever you may find that it will select all the data prior to the 1st day of the current month until the 1st Createdate. This could cause the report to take forever to run, Maybe building in a limitation to the code I would use something like this to build a report that gives details for last month only.
Select [columns]
from [source]
where [Createdate] between
/*First day of last Month*/
DATEADD(mm, DATEDIFF(mm, 0, Getdate())-1, 0 and
/*First day of this Month*/
dateadd(mm,datediff(mm,0,Getdate()),0)
I'm currently studying SQL Server using the book Ben-Gan, Itzik. T-SQL Fundamentals. Below is a query used to select order placed at end of the month. (I know that function EOMONTH() can also be used)
SELECT orderid, orderdate, custid, empid
FROM Sales.Orders
WHERE orderdate = DATEADD( month, DATEDIFF( month, '18991231', orderdate), '18991231');
The author's explanation is:
This expression first calculates the difference in terms of whole
months between an anchor last day of some month (December 31, 1899, in
this case) and the specified date. Call this difference diff. By
adding diff months to the anchor date, you get the last day of the
target month.
However, I'm still a bit confused as to how it actually works. Would someone kindly explain it?
That seems like a rather arcane way to do this. What the code is doing is calculating the number of months since the last day of some month. Then, it adds this number of months to that date. Because of the rules of dateadd(), the month remains the last date.
However, I prefer a simpler method:
where day(dateadd(day, 1, orderdate)) = 1
I find this much clearer.
select DATEDIFF(MONTH, '20160131', '20160201')
give us 1 month and
SELECT DATEADD(month, 1, '20160131')
give us 2016-02-29 00:00:00.000
that's ok
I tried out the query myself and seem to have got the hang of it. here is what i wrote just in case anyone else is interested
SELECT DATEADD(month, DATEDIFF(MONTH, '20160131', '20160201'), '20160131');
result:
2016-02-29 00:00:00.000
so my interpretation is that adding one or more "month" to a particular date in which the last date of the month is 31 will always return a date in which the date is the last day of the month. if this sentence makes any sense...
I'm pretty new to SQL so I hope this isn't a dumb question, tried to google but couldn't find anything.
I'm summing sales of departments per week in SQL and am using TD_SYSFNLIB.WEEKNUMBER_OF_YEAR (trans_dt) to get the week number.
I think everything is working except I'd like to change the format of the weeks to the start date of the week, e.g. week 1 = 1/4/15
Also, i'm not sure how to handle the very first of the year week 0 since I think that should be grouped up with week 52 of last year.
The following date math trick should get you Beginning of Week as an actual date without having to join to the SYS_CALENDAR view or using a function:
SELECT CURRENT_DATE - ((CURRENT_DATE - DATE '0001-01-07) MOD 7) AS BOW;
Starting with TD14 there's NEXT_DAY which returns the following weekday, if you subtract 7 days you get the previous day:
next_day(trans_dt - 7, 'sunday')
I am new to T-SQL and needed urgent assistance here.
I am trying to get the week number from a given date.
I understand that there is a build in function for it but the value return is not exactly what I wanted.
For e.g., by using select datepart(wk, '2013-01-07'), it would return me '2'.. but the actually fact is it should return '1' instead of '2'.
Any ideas how to correct this issue?
You can use dy datepart specifier to get dayOfYear number and divide it by 7:
select (datepart(dy, '2013-01-05') - 1) / 7 + 1;
Working DEMO.
Try this
SELECT DATEPART(WEEK,GETDATE())
This depends on hop you define the first week. Does it always start on the same weekday? or does it always start on the first of January? If you want it to always start on the same weekday, then use Set datefirst to tell T-SQL what weekdaty you want to define as the start of the week. If you want it to always start on Jan 1, then just use day of year instead of week, subtract 1, integer divide by 7 and add 1.
declare #dat DateTime = getdate()
Select Select (datepart(dy, #dat)-1) / 7 + 1
Although going from memory, I believe the ISO standard for the first week of the year is the week in the year that the first Thursday of the year is in. This would possibly explain why the built in function gives a result different to that you require.
I want to calculate Last Week Number of Month in SQL. I am having Week Number and Year.
Eg. If I pass WeekNo=51 , Year=2008 , than function should return LastWeekofMonth= 52.
I want to calculate Week number using below standards.
According to ISO 8601:1988 that is used in Sweden the first week of the year is the first week that has at least four days within the new year.
So if your week starts on a Monday the first Thursday any year is within the first week. You can DateAdd or DateDiff from that.
Please Help me..........
Thanks in advance.
SELECT WEEK(LAST_DAY(STR_TO_DATE('2008-51-Mon', '%x-%v-%a')));
Should do the trick for getting the last week number of month with MySQL :
I first convert to a date, then I get the last day of the month (here: 2008-12-31), then I compute the week of the last day of the month (52).
It should be easy to turn it into a function.
Hope this helps.
This is fairly straightforward if you use a calendar table. The month you need is given by this query.
select iso_year, month_of_year
from calendar c
where iso_year = 2008 and iso_week = 51
group by iso_year, month_of_year
--
iso_year month_of_year
2008 12
So you can use that result in a join on the calendar table, like this.
select max(c.iso_week) as last_week_of_month
from calendar c
inner join
(select iso_year, month_of_year
from calendar c
where iso_year = 2008 and iso_week = 51
group by iso_year, month_of_year) m
on m.iso_year = c.iso_year and m.month_of_year = c.month_of_year;
--
last_week_of_month
52
Here's one example of a calendar table, but it's pretty thin on CHECK constraints.
If you're using SQL Server, you can perform a calculation by using a master table, without creating a calendar table. This fellow gives you a very good explanation, which I recommend that you read. His SQL for calculating the first and last Sundays of each month can be adapted for your use:
declare #year int
set #year =2011
select min(dates) as first_sunday,max(dates) as last_sunday from
(
select dateadd(day,number-1,DATEADD(year,#year-1900,0))
as dates from master..spt_values
where type='p' and number between 1 and
DATEDIFF(day,DATEADD(year,#year-1900,0),DATEADD(year,#year-1900+1,0))
) as t
where DATENAME(weekday,dates)='sunday'
group by DATEADD(month,datediff(month,0,dates),0)
Edit: Once you have the date of the Thursday, you can get the week number from that date like this:
DECLARE #Dt datetime
SELECT #Dt='02-21-2008'
SELECT DATEPART( wk, #Dt)