I would like to show 6 months worth of data. For example from the 1st of September to 1st March inclusive.
The code below starts from the 1st of September but finishes on today
date >= date_trunc('month' , now() - ('6 months' :: interval))
I am working on Postgres.
Related
If I run this query on my database
SELECT EXTRACT('epoch' FROM age('2021-01-01'::date, '2019-12-01'::date))
The result is 34149600.
But then if I try with 2019-12-02 (one day more)
SELECT EXTRACT('epoch' FROM age('2021-01-01'::date, '2019-12-02'::date))
The result is exactly the same 34149600!
As if the seconds passed from 02 december 2019 to 01 january 2021 are the same as the seconds passed from 01 december 2019 to 01 january 2021.
Why is this? I've already tried the above code with timezones on 00:00:00+00 timezone for all dates (for 1st january 2021, 1st december 2019 and 2nd december 2021), and it gives the same result
Obviously, I would have expected the epoch to be different, around 3600*24 of difference (seconds in a day).
The similar results come from the age() function which returns an interval with years, months, days. In such an interval, 1 month = 30 days, so their conversions in seconds are similar.
You will get the expected result with
SELECT extract(epoch from ('2021-01-01'::timestamp - '2019-12-01'::timestamp)) => 34300800.000000
SELECT extract(epoch from ('2021-01-01'::timestamp - '2019-12-02'::timestamp)) => 34214400.000000
I am trying to get week numbers ( resetting at 1 for each month) as per ISO format for each month in 2019.For example I am interested in getting
All dates in July 2019: week 1 to 4,
All dates in Aug 2019 : week 1 to 4 and so on.
I first created the calculated field (Week_Number_ISO) to get the overall week number in year 2019.I used the following formula;
DATEPART('iso-week',[ Date]) which works as intended.
To get the monthly week number I used the following formula
INT((DATEPART('day',[Created Date])-DATEPART('iso-weekday',[Created Date])+7)/7)+1.
(Idea was to calculate the date of the first day of each week & then divide by 7 and take the integer part)
As per the ISO format, shouldn't July 29 to 31st be a part of week 4 for July?But the formula is showing it as week 5 for July 2019.I feel I am missing something in the formula or am missing something about ISO week number resetting at 1 for each month.
Can someone help me?
Here is an example of the dates in July 2019 and the associated week numbers.
Why would July 28th-July 31st 2019 be considered week 4?
Would it be possible to get same day of week last year using Excel? please below example:
Input: Monday 9 Nov 2015 | Output: 10 Nov 2014
Thanks
Simply subtract 52 full weeks with 7 days = 364 days. So if the date is in A1, the formula =A1-364 will get the date exactly 52 weeks before, which is the same day of week in the year before.
To show that it works even for leap years, try the following:
You see the formula date - 364 (=A2-364, =A3-364, ...) always gets the same day of week a year before. That is because it gets the day minus 52 full weeks (52 * 7 days) before. In leap years it gets a different day but the same day of week.
Try this:
=DATE(YEAR(A1)-1,MONTH(A1),DAY(A1))+WEEKDAY(A1)-WEEKDAY(DATE(YEAR(A1)-1,MONTH(A1),DAY(A1)))
It returns the closest date within a week. A1 is the cell with this year's date.
In postgresql extract(week from '2014-12-30'::timestamp) gives week number 1 of 2015. How do I extract the associated year that corresponds to the week number? Using extract(year ... gives 2014
Since week is the ISO-defined week, you want the isoyear:
#= select extract(isoyear from '2014-12-30'::timestamp);
date_part
-----------
2015
(1 row)
From the docs:
By definition, ISO weeks start on Mondays and the first week of a year
contains January 4 of that year. In other words, the first Thursday of
a year is in week 1 of that year.
In the ISO week-numbering system, it is possible for early-January
dates to be part of the 52nd or 53rd week of the previous year, and
for late-December dates to be part of the first week of the next year.
For example, 2005-01-01 is part of the 53rd week of year 2004, and
2006-01-01 is part of the 52nd week of year 2005, while 2012-12-31 is
part of the first week of 2013. It's recommended to use the isoyear
field together with week to get consistent results.
According to the postgresql documentation for extract week:
The number of the week of the year that the day is in. By definition
(ISO 8601), the first week of a year contains January 4 of that year.
(The ISO-8601 week starts on Monday.) In other words, the first
Thursday of a year is in week 1 of that year. (for timestamp values
only)
Based on this, you might have to do a little logic:
case when extract(week from '2014-12-30'::timestamp) = 1
and extract(month from '2014-12-30'::timestamp) = 12
then extract(year from '2014-12-30'::timestamp)+1
else extract(year from '2014-12-30'::timestamp)
end
The MS SQL DateDiff function counts the number of boundaries crossed when calculating the difference between two dates.
Unfortunately for me, that's not what I'm after. For instance, 1 June 2012 -> 30 June 2012 crosses 4 boundaries, but covers 5 weeks.
Is there an alternative query that I can run which will give me the number of weeks that a month intersects?
UPDATE
To try and clarify exactly what I'm after:
For any given month I need the number of weeks that intersect with that month.
Also, for the suggestion of just taking the datediff and adding one, that won't work. For instance February 2010 only intersects with 4 weeks. And the DateDiff calls returns 4, meaning that simply adding 1 would leave me the wrong number of weeks.
Beware: Proper Week calculation is generally trickier than you think!
If you use Datepart(week, aDate) you make a lot of assumptions about the concept 'week'.
Does the week start on Sunday or Monday? How do you deal with the transition between week 1 and week 5x. The actual number of weeks in a year is different depending on which week calculation rule you use (first4dayweek, weekOfJan1 etc.)
if you simply want to deal with differences you could use
DATEDIFF('s', firstDateTime, secondDateTime) > (7 * 86400 * numberOfWeeks)
if the first dateTime is at 2011-01-01 15:43:22 then the difference is 5 weeks after 2011-02-05 15:43:22
EDIT: Actually, according to this post: Wrong week number using DATEPART in SQL Server
You can now use Datepart(isoww, aDate) to get ISO 8601 week number. I knew that week was broken but not that there was now a fix. Cool!
THIS WORKS if you are using monday as the first day of the week
set language = british
select datepart(ww, #endofMonthDate) -
datepart(ww, #startofMonthDate) + 1
Datepart is language sensistive. By setting language to british you make monday the first day of the week.
This returns the correct values for feburary 2010 and june 2012! (because of monday as opposed to sunday is the first day of the week).
It also seems to return correct number of weeks for january and december (regardless of year). The isoww parameter uses monday as the first day of the week, but it causes january to sometimes start in week 52/53 and december to sometimes end in week 1 (which would make your select statement more complex)
SET DATEFIRST is important when counting weeks. To check what you have you can use select ##datefirst. ##datefirst=7 means that first day of week is sunday.
set datefirst 7
declare #FromDate datetime = '20100201'
declare #ToDate datetime = '20100228'
select datepart(week, #ToDate) - datepart(week, #FromDate) + 1
Result is 5 because Sunday 28/2 - 2010 is the first day of the fifth week.
If you want to base your week calculations on first day of week is Monday you need to do this instead.
set datefirst 1
declare #FromDate datetime = '20100201'
declare #ToDate datetime = '20100228'
select datepart(week, #ToDate) - datepart(week, #FromDate) + 1
Result is 4.