I would like to convert Julian Date value to normal date. Can anyone please help me?
I have tried JulianDateToDateTime(somedouble value) but it raised exception.
Then I tried ModifiedJulianDateToDateTime(some double value) it posted a date but its totaly wrong.
For example my Julian Value is 226. It says that it must show 14 th August. But I couldnt convert it usig delphi.
Thanks
System.DateUtils.JulianDateToDateTime converts a Julian Date to a TDateTime value.
From documentation:
The Julian date is the number of days, including fractional days, since 4713 BC January 1, Greenwich noon.
And for System.DateUtils.ModifiedJulianDateToDateTime:
The modified Julian date is the number of days, including fractional days, since Greenwich midnight on November 17, 1858. Modified Julian dates are based on Julian dates, but adjusted to use midnight rather than noon as a starting point and they use a more recent date as a starting point.
So you have to enter a Julian Date based on either of those starting points.
You can use the reverse functions to get a correct Julian Date: DateTimeToJulianDate/DateTimeToModifiedJulianDate.
In comments it seems as the date you have is the day number of a certain year (sometimes referred to as the Julian Day in meteorological data logging systems).
Then use this:
myDT := EncodeDate(2014,1,1) + yourJulianDay - 1; // If the year is 2014
If your "Julian Date" really is "Julian Day" (ie. the number of days into a given year, a.k.a. Ordinal Date), then you can use the following function to convert it to a TDate (you specify the year the Julian Day should be considered to be in):
uses DateUtils;
FUNCTION JulianDay(Year,Day : Cardinal) : TDate;
BEGIN
Result:=IncDay(EncodeDate(Year,1,1),PRED(Day))
END;
This will return August 14th for JulianDay(2014,226)
Related
Informix is not my normal environment and the way it handles datetime values is throwing me for a loop. I can't imagine this is difficult, but for the life of me I'm not yet able to figure it out.
This is the SQL:
SELECT agentid,
extension As Ext,
resourcefirstname As FirstNm,
resourcelastname As LastNm,
Min(eventdatetime) As FirstIn
FROM agentstatedetail AS asdr Join
resource As r On asdr.agentid = r.resourceid
WHERE asdr.eventdatetime BETWEEN '2016-10-20 04:00:00' AND '2016-10-21 03:59:59'
AND eventtype = 3
AND assignedteamid = 14
Group By agentid, extension, resourcefirstname, resourcelastname
Order By Min(eventdatetime)
Everything works as is, but the dates in the Between clause are currently entered manually- not optimal. I just need some way to describe "yesterday at 4:00 AM" and "Today at 4:00 AM" Will somebody please clue me in?
Using Informix version 12.10.FC6DE, I can do this:
SELECT
TODAY::DATETIME YEAR TO SECOND AS today_zerohour
, TODAY::DATETIME YEAR TO SECOND - '20:00:00'::INTERVAL HOUR TO SECOND AS yesterday_dawn
, TODAY::DATETIME YEAR TO SECOND + '04:00:00'::INTERVAL HOUR TO SECOND AS today_dawn
FROM
systables
WHERE
tabid = 1;
And it returns:
today_zerohour yesterday_dawn today_dawn
2016-10-21 00:00:00 2016-10-20 04:00:00 2016-10-21 04:00:00
So, what is happening here:
The operator TODAY returns the system date as a DATE type. The DATE type does not have the precision I want (it only has year, month and day), so I cast the value (cast operator is ::) to a DATETIME with precision from year to second (the hour, minutes and seconds are set to zero):
TODAY::DATETIME YEAR TO SECOND
In Informix, for an addition or subtraction with a DATETIME value to return another DATETIME value, I need to add or subtract an INTERVAL value. So I created 2 INTERVAL values.
One INTERVAL of 20 hours to subtract from the today value (again the cast operator :: is used, this time to cast from a string to an INTERVAL):
'20:00:00'::INTERVAL HOUR TO SECOND
One INTERVAL of 4 hours to add to the today value:
'04:00:00'::INTERVAL HOUR TO SECOND
If I have week 7 in 2017 what week date is the Monday in that week in Google Sheets?
=DATE(B9,1,1)-WEEKDAY(DATE(B9,1,1),3)+7*(WEEKDAY(DATE(B9,1,1),3)>3)+7*(A9-1)
is the least complicated formula I know which works for week numbers in Sweden (i.e. Monday first day of week, ISO rules for what is week 1).
Short answer (A1==Week, B1==Year):
=DATE(B1;1;1)+((A1-1)*7)-WEEKDAY(DATE(B1;1;1);3)
Long answer:
DATE(<year>;1;1) // days since 1970 until the frist day of the year
plus
((<week number>-1)*7) // how many days into the year is this week
minus
WEEKDAY(DATE(<year>;1;1);3) // how many extra days from previous year in first week
PS:
This assumes monday as the first day of week you have to change the arguments for WEEKDAY to change it to sunday
Because of this definition (https://en.wikipedia.org/wiki/Week) the 4th of January must be used instead the 1st. The 4th of January is the first day which is always in the week 1.
=DATE(B1;1;4)+((A1-1)*7)-WEEKDAY(DATE(B1;1;4);3)
If you are using ISO weeks, the accepted answer doesn't account for weeks overlapping on 2 technical years like 2020-w53, which is from 28 Dec 2020 until 3 Jan 2021.
Therefore I'm using this formula instead:
=DATE(K2,1,1)-WEEKDAY(DATE(K2,1,1),2)+7*(WEEKDAY(DATE(K2,1,1),2)>3)+7*(L2-1) +1
Where K is the Year, and L is the Week number (split in 2 columns from yyyy-ww)
to have it in an arrayformula:
=ArrayFormula(if(K2:K="",, DATE(K2:K,1,1)-WEEKDAY(DATE(K2:K,1,1),2)+7*(WEEKDAY(DATE(K2:K,1,1),2)>3)+7*(L2:L-1) +1 ))
You can use =ArrayFormula(if(E2:E="",,split(E2:E,"-"))) to split yyyy-ww in two columns.
NOTE: This formula would return the Monday (Which is the first day of the week in international standard, ISO)
Worked this up for 2023. It will work through end of 2024 too .. that said the AND logic is flawed .. feel free to suggest something to make this better
=IFS(
AND(ISOWEEKNUM(A8)=52,YEAR(A8)<>YEAR(A7)),
DATE(YEAR(A8-1),1,1)-WEEKDAY(DATE(YEAR(A8-1),1,1),3)+7*(WEEKDAY(DATE(YEAR(A8-1),1,1),3)>3)+7*(ISOWEEKNUM(A8)-1),
DATE(YEAR(A8),1,1)-WEEKDAY(DATE(YEAR(A8),1,1),3)+7*(WEEKDAY(DATE(YEAR(A8),1,1),3)>3)+7*(ISOWEEKNUM(A8)-1)
)
I'm using the week_field helper to generate a week picker in a form. When I choose a week and submit, my controller gets the correct serialized week (e.g. '2014-W03') which I can turn into a date object. That all works, but when the date is serialized again it is always decremented by 1 (e.g. it'll be '2014-W02'). I looked at the source code for the week_field helper and it serializes as
def format_date(value)
value.try(:strftime, "%Y-W%W")
end
but this doesn't seem to be the encoding when the date is parsed. Furthermore, parsing and then serializing a date yeilded this wonky result:
irb > Date.parse('2014-W03').strftime('%Y-W%W')
=> "2014-W02"
Any ideas as to what's going on here or how I can do this in a way that makes sense? I'd hate to have an extra +1 on the week number or change the week_field format_date definition if there's a cleaner route.
This is from http://apidock.com/ruby/DateTime/strftime :
%W - Week number of the year. The week starts with Monday. (00..53)
It seems you have the good old - start with 0 or start with 1 - problem. Strftime will start counting weeks with 0.
But maybe %V is the right thing for you:
ISO 8601 week-based year and week number:
The week 1 of YYYY starts with a Monday and includes YYYY-01-04.
The days in the year before the first week are in the last week of
the previous year.
%G - The week-based year
%g - The last 2 digits of the week-based year (00..99)
%V - Week number of the week-based year (01..53)
I want to store date in my Postgres database.
The only problem is that this date can have optional day or even month.
Example:
User provides time period when he was employed - not necessary full date (day + month + year), but only start year and end year.
However there are users, who worked only from may to october in the same year so month have to be provided too.
How to handle this kind of optional date parts?
Use a proper date type anyway. Do not store text or multiple columns. That would be more expensive and less reliable.
Use the function to_date(), which is fit to deal with your requirements out of the box. For instance, if you call it with a pattern 'YYYYMMDD' and the actual string is missing characters for day, or month and day, it defaults to the first month / day of the year / month:
db=# SELECT to_date('2001', 'YYYYMMDD');
to_date
------------
2001-01-01
db=# SELECT to_date('200103', 'YYYYMMDD');
to_date
------------
2001-03-01
You could store a precision flag indicating year / month / day in addition if you need that.
While the accepted answer is a good one, there is another alternative.
ISO 8601
The ISO 8601 standard defines sensible formats for textual representations of various kinds of date-time values.
A year is represented in the obvious manner, a four-digit number: 2014
A year-month is represented with a required hyphen: 2014-01Note that in other ISO 8601 formats, the hyphen is optional. But not for year month, to avoid ambiguity.
A full date is similar: 2014-08-21 or without optional hyphens: 20140821. I recommend keeping the hyphens.
So you could store the values as text. The length of text would tell you whether it is year-only, year-month, or date.
As the question states, what date format is "ds1248083197360" ? Is this a standard date format or a custom one to an application?
ds1248083197360 is 20/07/2009 (or in US format, 07/20/2009)
this might be miliseconds after 1.january year 1970 GMT 00:00. In this case, your example is 20.07.2009 10:46:37,360
It is probably the Unix time format, specified i milliseconds, meaning that it's the number of milliseconds since January 1st 1970. Wikipedia's entry on Unix time.
1248083197.360 is Monday the 20th of July 2009 09:46:37.360 GMT.