I have an array with strings of dates. The format is "2/27/16 3:47" so "m-d-y H:M". However, DateTime parses this as 0016-27-02T03:47:00. I would like to have the output be: 2016-27-2T03:47:00.
My code is:
map(date-> DateTime(date, "mm/dd/yy HH:MM"), datsub[:date])
Side question: The output type becomes Any. Is this the correct type or should it be DateTime or something similar?
As #akrun mentioned, you should add the year yourself:
Dates.Year(2000) + DateTime(date, "m/d/y H:M")
This is more explicit about exactly what is happening. Otherwise Dates would have to guess what exactly something like 97 means: 1997 or 2097, or actually year [00]97?
It's possible that you might want to come up with a reasonable cutoff for what year to add. You can try the following:
expandyear(date::DateTime) = date + (Dates.year(date) < 25 ? Dates.Year(2000) : Dates.Year(1900))
with whatever cutoff you think makes sense.
The issue with return types with map is a known problem that has been fixed in the latest v0.5 nightlies. Julia v0.5 is likely to be released in the near future, perhaps within several months.
Related
I am using Tableau Server version 10.4.3
I have a dimension rTime which has string value. the entries in rTime is of like this
May 10, 2019 8:59:56.303 PM UTC
I want to check whether the rTime is today or not. I cannot use DateParse since my server doesn't have this functionality.
I created a calculated field CheckrTime with below content :
STR(LEFT(SPLIT([rTime],':',1),LEN(SPLIT([rTime],':',1))-2))
When I am dragging CheckrTime into workspace area, the output is coming in below format which is what I wanted :
May 10, 2019
When I am checking ISDATE("May 10, 2019") (a normal string), it is outputting TRUE as expected but when I am checking ISDATE(CheckrTime) it is outputting FALSE . Why?
The reason I am checking above thing is I am looking to use DATEDIFF function of tableau in this way:
DATEDIFF('day', DATE(CheckrTime), TODAY()) == 0
NOTE
If someone is wondering ,I have taken care of the level of granularity.
If you have a better solution then the one I am following, please do tell me.
This is working for me. I would expect May 10 <> May 16 (today) and therefore return false. However, when I change your example to today's date it does in fact come back as true.
You could also try this formula for the date LEFT([rTime],FINDNTH([rTime],' ',3)). It is slightly less complicated but will give you the same answer.
Calculated field (date type) depends the locale used which defines date format. Are you able to use Date function?
In Tableau website there is a example using english locale
For string 06May2017
DATE (LEFT([Original Date], 2) + "/" + MID([Original Date],3,3) + "/" + RIGHT([Original Date],4))
Above mentioned highligts / character between digits which is depending on locale
I'm currently working with embarcadero c++, this is the first time I'm working with it so it's completely new to me.
What I'm trying to achieve is to get the current date, make sure the date has the "dd/MM/yyyy" format. When I'm sure this is the case I want to add a month to the current date.
So let's say the current date is 08/18/2016 this has to be changed to 18/08/2016 and then the end result should be 18/09/2016.
I've found that there is a method for this in embarcardero however I'm not sure how to use this.
currently I've only been able to get the current date like this.
TDateTime currentDate = Date();
I hope someone will be able to help me out here.
I figured it out.
After I've searched some more I found the way to use the IncMonth method on this page.
The example given my problem is as follows:
void __fastcall TForm1::edtMonthsExit(TObject *Sender)
{
TDateTime StartDate = edtStartDate->Text;
int Months = edtMonths->Text.ToInt();
TDateTime NextPeriod = IncMonth(StartDate, Months);
edtNextPeriod->Text = NextPeriod;
}
After looking at I changed my code accordingly to this
TDateTime CurrentDate = Date();
TDateTime EndDate = IncMonth(CurrentDate, 1);
A date object doesn't have a format like "dd/MM/yyyy". A date object is internally simply represented as a number (or possibly some other form of representation that really isn't your problem or responsibility).
So you don't have to check if it's in this format because no date objects will ever be in this format, they simply don't have a format.
You will have to do additions/subtractions on the Date object that the language or library gives you, THEN (optionally) you can format it to a human-readable string so it looks like 18/08/2016 or 18th of August 2016 or whatever other readable format that you choose.
It might be that the TRANSFER of a date between 2 systems is in a similar format, but then formatting the date like that is entirely up to you.
As for how to do that, the link you posted seems like a possible way (or alternatively http://docwiki.embarcadero.com/Libraries/Berlin/en/System.SysUtils.IncMonth), I'm afraid I can't give you an example as I'm not familiar with the tool/language involved, I'm just speaking generically about Date manipulations and they should ALWAYS be on the raw object.
I have reviewed the post Creating a DateTime object with a specific UTC DateTime in PowerShell, but it does not seem to directly answer the question I am asking:
What is the most direct method in PowerShell (3.0) to return a sortable string representing "now" as UTC?
I expected the correct answer to be:
Get-Date -Format (Get-Culture).DateTimeFormat.UniversalSortableDateTimePattern
OR
get-date -format u
but this is not the case.
Example: At 1300 hrs (1pm) on September 1st, 2016 in the Pacific Time Zone during DST, I get the response:
2016-09-01 13:00:00Z (the local time with a "Z" appended)
when I was expecting:
2016-09-01 20:00:00Z (correct UTC/GMT time)
So basically, my code is just getting a string representing the "local" time and appending a "Z".
Now, I know I can manipulate to get to that point, but I'm looking for the minimal (simplest, cleanest) way to get here.
Bonus Points (as if they existed): How do I get that same, sortable result, but displaying with "UTC" and/or "GMT" as the suffix. Same minimal requirement.
Probably something like this:
[DateTime]::UtcNow.ToString('u')
Which is equivalent to:
[DateTime]::UtcNow.ToString((Get-Culture).DateTimeFormat.UniversalSortableDateTimePattern)
For the bonus, I think the most straightforward way is just to replace Z with UTC:
[DateTime]::UtcNow.ToString('u').Replace('Z','UTC')
I'm assuming you'll always want UTC since that what it seems like from your question. There doesn't appear to be a format string to get just the 3 letter time zone.
I tried this, and it also gives the result I want:
"[DateTime]::UtcNow.ToString('yyyyMMdd_HHmmss_UTC')"
It is showing time in the format 20180108_152407_UTC
so you can play with the date/time formatting as you wish basically
With DateJS, you'd add e.g. six months to the current date like this:
Date.today().addMonths(6);
However, I need to add 24 months not to today's date, but to a date which a user has typed into a date field. So the today() should in principle be replaced by something like this.getField('begin_date').value.
The result shall be written into another data form field.
I tried hard, but couldn't make it. Can anyone help me out?
Providing the input value is a textual representation of a date, you need to convert it into a Date object at the first place. Then you can work with it as you want.
DateJS has a pretty smart parse() function which does exactly that, so you'd achieve it like this:
Date.parse(this.getField('begin_date').value).addMonths(24)
When a specific date format is needed, like DD.MM.YYYY commonly used in Europe, you can use parseExact() and specify the format. Like this:
Date.parseExact(dateToParse, 'dd.MM.yyyy') // leading zeroes required; parses 01.04.2014 but not 1.4.2014
Date.parseExact(dateToParse, 'd.M.yyyy') // leading zeroes not required; parses both 01.04.2014 and 1.4.2014
Here is a solution that I found for my problem, using DateJS as well:
start = this.getField('begin_date').value;
var d1 = util.scand("dd.mm.yyyy", start);
var myDate = new Date(d1);
result = myDate.addMonths(24);
This works pretty fine, also spanning leap years, except for the 28th of February, 2014/2018/2022 ... ; the result then will be the 28th of February, 2016/2020/2024 ... and not the 29th of February 2016/2020/2024... In these cases it's up to the user to accept the 28th or to manually change the date to the 29th.
I would like to calculate precisely the months between two dates to achieve this I do something like :
DateTimeZone ZONE = DateTimeZone.forID("Europe/London");
String DATE_FORMAT = "dd/MM/yyyy";
DateTimeFormatter FORMATTER = DateTimeFormat.forPattern(DATE_FORMAT).withZone(ZONE);
LocalDate dateTime = FORMATTER.parseLocalDate("28/05/2013");
LocalDate dateTime6MonthAfter = FORMATTER.parseLocalDate("28/02/2014");
Period todayUntilEndOfContract = new Period(dateTime,dateTime6MonthAfter);
todayUntilEndOfContract.getMonths() +"M/"+ todayUntilEndOfContract.getWeeks() +"W/"+ todayUntilEndOfContract.getDays() +"D/");
So this give me precisely 9 month between 28/05/2013 and 28/02/2014 BUT!!!
when I calculate the dates (29, 30, 31)/05/2013 with 28/02/2014 it always give me 9 month normally it should say 8M/3W/(6,5,4)D/ why is it always 9M/0W/0D please...?
Thanks a lot
Your issue is that you are expecting something a little different than what is provided. If I ask you the question "what is 30th January plus one month?" then there are a number of different answers which are valid under different assumptions. In Joda's case the answer is "28th February" (or 29th if a leap year).
Although you are asking for month-based information I would suggest that you obtain the number of days instead and use that as a basis, as it is probably closer to what you need:
int days = Days.daysBetween(dateTime, dateTime6MonthAfter).getDays();
You can always use this number to feed back in to your code and obtain different values to fit your requirements.