I'm trying to manipulate a date value to go back in time exactly 1 ISO-8601 year.
The following does not work, but best describes what I want to accomplish:
date_add(date '2018-01-03', interval -1 isoyear)
I tried string conversion as an intermediate step, but that doesn't work either:
select parse_date('%G%V%u',safe_cast(safe_cast(format_date('%G%V%u',date '2018-01-03') as int64)-1000 as string))
The error provided for the last one is "Failed to parse input string "2017013"". I don't understand why, this should always resolve to a unique date value.
Is there another way in which I can subtract an ISO year from a date?
This gives the corresponding day of the previous ISO year by subtracting the appropriate number of weeks from the date. I based the calculation on the description of weeks per year from the Wikipedia page:
CREATE TEMP FUNCTION IsLongYear(d DATE) AS (
-- Year starting on Thursday
EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 5 OR
-- Leap year starting on Wednesday
(EXTRACT(DAY FROM DATE_ADD(DATE(EXTRACT(YEAR FROM d), 2, 28), INTERVAL 1 DAY)) = 29
AND EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 4)
);
CREATE TEMP FUNCTION PreviousIsoYear(d DATE) AS (
DATE_SUB(d, INTERVAL IF(IsLongYear(d), 53, 52) WEEK)
);
SELECT PreviousIsoYear('2018-01-03');
This returns 2017-01-04, which is the third day of the 2017 ISO year. 2018-01-03 is the third day of the 2018 ISO year.
Related
I have a variable in a SAS dataset that has a number of dates (e.g. 01APR21). What I'm looking to do is create a new variable that shows the date of the first Monday of that week. So using the above example of 01APR21, the output would be 29/03/2021 as that what was when the Monday in that week was. I'm assuming it's using intnx, but I can't get my head around it.
data test;
format date date8.;
format first_day date10.;
date = '01APR21'd;
first_day = ?;
run;
INTNX Parameters:
Interval : WEEK
Increment: 0 (same week)
Alignment: Beginning
(Sunday)
Then add 1 to get to Monday instead of Sunday. You could probably play with the SHIFT INDEX parameter as well.
Monday = intnx('week', dateVariable, 0, 'B') + 1
How can I get the number of days passed in the current quarter?
For example, if today is 1/2/2021, it will return 2.
If today is 2/5, it will return 36.
If today is 4/2, it will return 2.
Use date_trunc() to get the start of the quarter and subtract dates:
WITH cte(day) AS (
VALUES
(date '2021-01-02')
, (date '2021-02-05')
, (date '2021-04-02')
)
SELECT day
, day - date_trunc('quarter', day)::date + 1 AS days_passed_in_quarter
FROM cte;
day | days_passed_in_quarter
------------+------------------------
2021-01-02 | 2
2021-02-05 | 36
2021-04-02 | 2
+ 1 to fix off-by-one error as you clearly want to include the current day as "passed".
Always use unambiguous ISO 8601 date format (YYYY-MM-DD - 2021-02-05), which is the default in Postgres and always unambiguous, or you depend on the current datestyle setting (and may be in for surprises). Also avoids misunderstandings general communication.
Related:
PostgreSQL: between with datetime
I'm trying to find out the weekday i.e Mon, Tue, Wed etc. from a date-range formatted as yyyy mm dd
I tried to use the formula format(day(Date Table),"ddd"), but the weekday is wrong. In my example, the output of 2020.01.01 gives Sunday, but it should be Wednesday.
I think your formula is wrong:
Instead of
format(day(Date Table),"ddd")
Use
format(<Target Table>[<date column>],"ddd")
I.e. Omit the DAX DAY call. This is resulting in the day of the month (1..31) being passed to the format function.
When you use the DAY function in DAX, it returns the day of the month (1 through 31).
Thus DAY ( DATE ( 2020, 1, 1) ) = 1 which means you're trying to format the number 1 as a date. Integers are interpreted as days since 1899/12/30 when treated as a date, so 1 corresponds to 1899/12/31, which happened to be a Sunday. Thus FORMAT(1, "ddd") = "Sun".
There's no reason to get DAY involved here. You can simply write
Day = FORMAT ( 'Calendar'[Date], "ddd" )
LibreOffice Calc spreadsheet offers a function ISOWEEKNUM to return the standard ISO 8601 week number of the specified date.
I want the opposite.
➠ Given a standard week number, give me the date of the first day of that week (the Monday date).
Passing integers is acceptable. Also nice if able to pass a string in standard format.
Like this:
DATE_OF_ISOWEEKNUM( 2017 , 42 ) ➝ date of Monday of week 42 in week-based year 2017
DATE_OF_ISOWEEKNUM( "2017-W42" ) ➝ date of Monday of week 42 in week-based year 2017
Ideally, I would be able to pass a number 1-7 for Monday-Sunday to specify the day-of-week for which I want a date. Something like this:
DATE_OF_ISOWEEKNUM( 2017 , 42 , 1 ) ➝ date of Monday of week 42 in week-based year 2017
DATE_OF_ISOWEEKNUM( "2017-W42-1" ) ➝ date of Monday of week 42 in week-based year 2017
DATE_OF_ISOWEEKNUM( 2017 , 42 , 7 ) ➝ as above, but Sunday
DATE_OF_ISOWEEKNUM( "2017-W42-7" ) ➝ as above, but Sunday
Example:
Formula:
=DATE(B$1,1,$A4*7)+(2-WEEKDAY(DATE(B$1,1,$A4*7)))-7*(ISOWEEKNUM(DATE(B$1,1,1))=1)
Calculate the date of day (weeknumber * 7) in the year.
Correct the day to be weekday Monday.
Correct to 7 days before, if the first day of the year is in the
first ISO weeknumber.
In my quarterly report Im trying to validate the two parameters StartDate and EndDate.
I first check if the difference between the dates is 2 months:
Switch(DateDiff(
DateInterval.Month, Parameters!StartDate.Value, Parameters!EndDate.Value) <> 2,
"Error message")
Then I try to add whether the StartDate is the first day of month AND EndDate is last day of month:
And (Day(Parameters!StartDate.Value) <> 1
And Day(DATEADD(DateInterval.Day,1,Parameters!EndDate.Value)))
So the whole expression looks like this:
Switch(DateDiff(DateInterval.Month, Parameters!StartDate.Value, Parameters!EndDate.Value) <> 2
And
Parameters!IsQuarterly.Value = true
And
Day(Parameters!StartDate.Value) <> 1
And
Day(DATEADD(DateInterval.Day,1,Parameters!EndDate.Value))<>1),
"Error: Quarterly report must include 3 months")
But It works wrong when the difference between dates is still 2 months, but StartDate and EndDate are not first and last day of the whole period.
I'd appreciate any help :)
I would say just change the implementation Add another two Parameter With Quarter and Year
Quarter like Q1,Q2,Q3 & Q4 with Value 1,2,3 & 4 respectively and year 2012,2013,2014 & so on
Now based on the parameter selected Qtr & Year set Default value of start & End Date
=DateSerial(Parameters!Year.Value), (3*Parameters!Qtr.Value)-2, 1) --First day of Quarter
=DateAdd("d",-1,DateAdd("q",1,Parameters!Year.Value, (3*Parameters!Qtr.Value)-2, 1))) --Last day of quarter
Doing this no need to do any validation bcz its always get the correct Date Difference.
Other Reference
First day of current quarter
=DateSerial(Year(Now()), (3*DatePart("q",Now()))-2, 1)
Last day of current quarter
=DateAdd("d",-1,DateAdd("q",1,DateSerial(Year(Now()), (3*DatePart("q",Now()))-2, 1)))