For my project I need to have an absolute numerical correspondence between days of the week and 1...7 values.
As you probably know the association between days and numbers can vary according to the locale, for example in Germany Monday is 1 and Sunday is 7, while in US Monday is 2 while Sunday is 1.
So, searching for a solution, I found the following code which seems working regardless of the locale, assigning Monday=1...Sunday=7:
1 + TRUNC (date) - TRUNC (date, 'IW')
Can someone explain me how does it work? In particular I just can't understand what this instruction:
TRUNC (date, 'IW')
exactly does.
TRUNC(DATE,'IW') returns the first day of the week. For me TRUNC(SYSDATE,'IW) returns Monday. Today is Tuesday Feb 21. Subtract from that TRUNC(SYSDATE,'IW') which would be Monday the 20th, and you'll get 1 (because 21-20=1). Add 1 onto that as you do in the beginning of your equation and you get 2, which we associate with Tuesday.
The very basic concept of ISO week is to make it NLS territory independent.
From documentation,
Week of year (1-52 or 1-53) based on the ISO standard.
A week starts on a Monday and ends on a Sunday.
Related
I have a weird situation where I need to make a DueDate to be the following Friday.
So if the date is Monday, March 11th 2019 then the DueDate is Friday March 22nd.
I am able to do this easily with:
TRUNC(Next_Day(FilingPeriod, 'FRI')) + 7 as DueDate
My issue comes when the day is Friday March 15th 2019 ad it should also have a DueDate of Friday March 22nd, instead it has a DueDate of Friday March 29th.
And I get why. I have been looking for a way to simplify this or throw in a check to determine if the current date is a Friday and go from there.
I have looked for a bit and read similar questions but I still can not seem to find it.
Thanks for any help
Just subtract 1 from the date before using NEXT_DAY():
TRUNC(Next_Day(FilingPeriod - 1, 'FRI')) + 7 as DueDate
Get back to Monday, the first day of the ISO week. Then go plus 4, then plus 7 days:
TRUNC(FilingPeriod, 'iw') + 11
On a sidenote: I'd avoid NEXT_DAY whenever possible. The day name depends on the current language setting and NEXT_DAY has no parameter to override this. This renders the function rather dangerous, because it makes a query fail-prone.
I have spent days trying to figure this out to no avail, so hopefully someone can help me. I have a queried date set which contains several fields including a column of dates. What I want to do is create a new field in my query that tells what the Monday and Friday is for the week of that row's particular date.
So for example; if the date in one of my rows is "1/16/18",
the new field should indicate "1/15/18 - 1/19/18".
So basically I need to be able to extract the Monday date (1/15/18) and the Friday date (1/19/18) of the week of 1/16/18 and then concatenate the two with a dash ( - ) in between. I need to do this for every row.
How on earth do I do this? I've been struggling just to figure out how to find the Monday or Friday of the given date...
Assuming that your column is of type date, you can use trunc to get the first day of the week (monday) and then add 4 days to get the friday.
For example:
with yourTable(d) as (select sysdate from dual)
select trunc(d, 'iw'), trunc(d, 'iw') + 4
from yourTable
To format the date as a string in the needed format, you can use to_char; for example:
with yourTable(d) as (select sysdate from dual)
select to_char(trunc(d, 'iw'), 'dd/mm/yy') ||'-'|| to_char(trunc(d, 'iw') + 4, 'dd/mm/yy')
from yourTable
gives
15/01/2018-19/01/18
There may be a simpler, canonical Oracle method to this but you can still reduce it to a simple calculation on your own either way. I'm going to assume you're dealing with only dates falling Monday through Friday. If you do need to deal with weekend dates then you might have to be more explicit about which logical week they should be attached to.
<date> - (to_char(<date>, 'D') - 2) -- Monday
<date> + (6 - to_char(<date>, 'D')) -- Friday
In principle all you need to do is add/subtract the appropriate number of days based on the current day of week (from 1 - 7). There are some implicit casts going on in there and it would probably be wise to handle those better. You might also want to check into NLS settings to make sure you can rely on to_char() using Sunday as the first day of week.
https://docs.oracle.com/cd/B19306_01/server.102/b14200/sql_elements004.htm
You can also use the NEXT_DAY function, as in:
SELECT TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '7' DAY AS PREV_MONDAY,
TRUNC(NEXT_DAY(SYSDATE, 'FRI')) AS NEXT_FRIDAY
FROM DUAL;
Note that using the above, on weekends the Monday will be the Monday preceding the current date, and the Friday will be the Friday following the current date, i.e. there will be 11 days between the two days.
You can also use
SELECT TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '7' DAY AS PREV_MONDAY,
TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '3' DAY AS NEXT_FRIDAY
FROM DUAL;
in which case the Monday and Friday will always be from the same week, but if SYSDATE is on a weekend the Monday and Friday returned will be from the PREVIOUS week.
I have 70.000 rows of data, including a date time column (YYYY-MM-DD HH24-MM-SS.).
I want to split this data into 3 separate columns; Hour, day and Week number.
The date time column name is 'REGISTRATIONDATE' from the table 'CONTRACTS'.
This is what I have so far for the day and hour columns:
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour"
FROM CONTRACTS;
I have seen the options to get a week number for specific dates, this assignment concerns 70.000 dates so this is not an option.
You (the OP) still have to explain what week number to assign to the first few days in a year, until the first Monday of the year. Do you assign a week number for the prior calendar year? In a Comment I asked about January 1, 2017, as an example; that was a Sunday. The week from January 2 to January 8 of 2017 is "week 1" according to your definition; what week number do you assign to Sunday, January 1, 2017?
The straightforward calculation below assigns to it week number 0. Other than that, the computation is trivial.
Notes: To find the Monday of the week for any given date dt, we can use trunc(dt, 'iw'). iw stands for ISO Week, standard week which starts on Monday and ends on Sunday.
Then: To find the first Monday of the year, we can start with the date January 7 and ask for the Monday of the week in which January 7 falls. (I won't explain that one - it's easy logic and it has nothing to do with programming.)
To input a fixed date, the best way is with the date literal syntax: date '2017-01-07' for January 7. Please check the Oracle documentation for "date literals" if you are not familiar with it.
So: to find the week number for any date dt, compute
1 + ( trunc(dt, 'iw') - trunc(date '2017-01-07', 'iw') ) / 7
This formula finds the Monday of the ISO Week of dt and subtracts the first Monday of the year - using Oracle date arithmetic, where the difference between two dates is the number of days between them. So to find the number of weeks we divide by 7; and to have the first Monday be assigned the number 1, instead of 0, we need to add 1 to the result of dividing by 7.
The other issue you will have to address is to convert your strings into dates. The best solution would be to fix the data model itself (change the data type of the column so that it is DATE instead of VARCHAR2); then all the bits of data you need could be extracted more easily, you would make sure you don't have dates like '2017-02-29 12:30:00' in your data (currently, if you do, you will have a very hard time making any date calculations work), queries will be a lot faster, etc. Anyway, that's an entirely different issue so I'll leave it out of this discussion.
Assuming your REGISTRATIONDATE if formatted as 'MM/DD/YYYY'
the simples (and the faster ) query is based ond to to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW')
(otherwise convert you column in a proper date and perform the conversio to week number)
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour",
to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW') as "Week"
FROM CONTRACTS;
This is messy, but it looks like it works:
to_char(
to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS') +
to_number(to_char(trunc(to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS'),'YEAR'),'D'))
- 2,
'WW')
On the outside you have the solution previous given by others but using the correct date format. In the middle there is an adjustment of a certain number of days to adjust for where the 1st Jan falls. The trunc part gets the first of Jan from the date, the 'D' gets the weekday of 1st Jan. Since 1 represents Sunday, we have to use -2 to get what we need.
EDIT: I may delete this answer later, but it looks to me that the one from #mathguy is the best. See also the comments on that answer for how to extend to a general solution.
But first you need to:
Decide what to do dates in Jan before the first Monday, and
Resolve the underlying problems in the date which prevent it being converted to dates.
On point 1, if assigning week 0 is not acceptable (you want week 52/53) it gets a bit more complicated, but we'll still be able to help.
As I see it, on point 2, either there is something systematically wrong (perhaps they are timestamps and include fractions of a second) or there are isolated cases of invalid data.
Either the length, or the format, or the specific values don't compute. The error message you got suggests that at least some of the data is "too long", and the code in my comment should help you locate that.
Why when I run
select (EXTRACT(WEEK FROM current_date)::int )
The output is 6 - why?
Today is 2016-02-14 which is the 8th week since the start of this year.
Am I getting this result wrong?
I'm looking for a function which I give it date and it tells me what week of the year this date is.
The documentation is pretty clear on the calculation:
week
The number of the ISO 8601 week-numbering week of the year. 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.
Weeks start on a Monday, so Sunday is the end of a week (and "today" is Sunday where I am and in most of the world at this particular time). Also, the first week depends on the when the year starts.
I'm calculating the week for a specific date in SQL for example
'2016-01-20' (yyyy-mm-dd) but SQL returns week: 4, and that is wrong because this year the first week started on '2016-01-04' the result must be week: 3.
I think the issue is generatad because 2015 was a year with 53 weeks, any solution to that? Thank you and I'm sorry for my bad English
In tSQL the DATEPART() is returning the correct data based on US and Most of Europe as well as UK See here
You can use SET DATEFIRST to adjust the start position however.
The ISO 8601 definition for week 01 is the week with the year's first Thursday in it. I am using Intersystems cache which apparently does not account for that either. So I have used this to address that
CASE WHEN 7-datepart(dw,dateadd(dd,1,dateadd(yy,datediff(yy,0,getdate())-1,0))) < 2
THEN datepart(wk,getdate())-1 ELSE datepart(wk,getdate()) END as WeekNum