I've been trying to do a timeline of dates with vis.js-timeline that are BC or before 0.
Tried start: '-0050-01-1' and start: '-50-01-1' and they don't show.
Couldn't find documentation on this.
So reading the object of Expanded Years from ECMAScript 2022 Language Specification found that:
In the simplified ECMAScript format, such an expanded year
representation shall have 6 digits and is always prefixed with a + or - sign.
Tried '-000460-04-20' and it shows as -460 (BC).
Related
This question already has answers here:
Get dates from AUT?
(2 answers)
Closed 5 years ago.
I have a VBScript that runs on the developer machine, in which the following line of code
CDate("01/09/2017")
returns the date as 1 September 2017.
But when deployed on certain clients the same line of code returns 9 January 2017 as the date.
How can I control this?
This has been answered before in detail;
Use SetLocale() to choose how you want VBScript to interpret the value.
SetLocale(1106) 'Set to United Kingdom
WScript.Echo CDate("01/09/2017")
For valid Locale ID values see Microsoft Locale ID Values (you also appear to able to use valid IETF language tag codes as well like en-us etc).
Most likely the date string is parsed according to the regional settings of the respective system. For stable results across systems with different regional settings you probably need to parse the date yourself, e.g. like this:
s = "01/09/2017"
a = Split(s, "/")
d = DateSerial(a(2), a(1), a(0))
So I would like to know what now-1y is producing as a result. So is it:
now - 365days
now - 365.25days
now with the year part subtracted, so 2014-12-23T08:46:00
I've read this documentation but can't find any more information.
After some testing it appears that the third proposition is correct (ie "now with the year part subtracted, so 2014-12-23T08:46:00"
If 2 was correct the following example should have return 2 results.
example1
As 2012 is a leap year. The following example should return 0 result if 1 was correct.
example2
Note: example only worked as expected when I wrote this answer, in order to make it work for you, you have to update the date field of the indexed document.
I am trying to get the difference between two dates in months for an xPath.
I have no problems getting it in days (1127)
days-from-duration(xs:date('2012-06-17')-xs:date('2009-05-17'))
When I try doing it in months I get 0
months-from-duration(xs:date('2012-06-17')-xs:date('2009-05-17'))
I did notice that this comes back as days only ("P1126D") so that maybe the problem just not sure how to fix it.
xs:date('2012-06-17')-xs:date('2009-05-17')
Thanks for any help!
So the best thing I can seem to do is manually calculate it.
(year-from-date(xs:date('2012-06-17')) - year-from-date(xs:date('2012-05-18')))*12 + month-from-date(xs:date('2012-06-17')) -month-from-date(xs:date('2012-05-18')) + (if (day-from-date(xs:date('2012-06-17')) < day-from-date(xs:date('2012-05-18')) ) then -1 else 0)
This function was not included in the spec because we couldn't agree semantics for it. We heard arguments that the difference between 31 March 2015 and 30 Sept 2015 was six months, and arguments that it was five months. (Advice: negative differences seem to be even more problematic than positive differences.) You can easily find that a date plus 6 months minus 6 months is not the date where you started. You'll have to define what you think the answer should be, and implement it using lower-level facilities such as month-from-date and year-from-date.
The problem is that days-from-duration returns the days field -- not the number of days the duration includes. Likewise for months and years.
Now, you can divide by days, but not by months (since months have uneven boundaries):
(: arg1 here is a dayTimeDuration, which can't be easily converted to a yearMonthDuration :)
let $arg1 := xs:date('2012-06-17')-xs:date('2009-05-17')
return xs:dayTimeDuration($arg1) div xs:dayTimeDuration("P30D")
I am trying to strip some repeated text out of my Kindle clippings that look like this:
The starting point,obviously,is a thorough analysis ofthe intellectual property portfolio,the contents ofwhich can be broadly divided into two categories:property that is in use and property that is not in use
==========
Essentials of Licensing Intellectual Property (Alexander I. Poltorak, Paul J. Lerner)
- Highlight on Page 25 | Added on Friday, 25 November 11 10:53:36 Greenwich Mean Time
commentators (a euphemism for prolific writers with little experience
==========
Essentials of Licensing Intellectual Property (Alexander I. Poltorak, Paul J. Lerner)
- Highlight on Page 26 | Added on Friday, 25 November 11 10:54:29 Greenwich Mean Time
I am trying to strip out everthing between "Essentials" and "Time". The regexp I am playing with right now looks like this:
Essentials([^,]+)Time
But obviously it is not working:
http://rubular.com/r/gwSJFgOQai
Any help for this nuby would be massively appreciated!
You need the /m modifier which makes . match a newline:
/Essentials(.*?)Time/m
See it working here:
http://rubular.com/r/qgmkWnLzW6
Why don't you use this:
/Essentials(.*?)Time/m
Updated. Forgot the m for multiline.
Regex are powerful, but you'll find they also often add needless complexity to a problem.
This is how I'd go about the problem:
text = <<EOT
The starting point,obviously,is a thorough analysis ofthe intellectual property portfolio,the contents ofwhich can be broadly divided into two categories:property that is in use and property that is not in use
==========
Essentials of Licensing Intellectual Property (Alexander I. Poltorak, Paul J. Lerner)
- Highlight on Page 25 | Added on Friday, 25 November 11 10:53:36 Greenwich Mean Time
commentators (a euphemism for prolific writers with little experience
==========
Essentials of Licensing Intellectual Property (Alexander I. Poltorak, Paul J. Lerner)
- Highlight on Page 26 | Added on Friday, 25 November 11 10:54:29 Greenwich Mean Time
EOT
text.each_line do |l|
l.chomp!
next if ((l =~ /\AEssentials/) .. (l =~ /Time\z/))
puts l
end
Which outputs:
The starting point,obviously,is a thorough analysis ofthe intellectual property portfolio,the contents ofwhich can be broadly divided into two categories:property that is in use and property that is not in use
==========
commentators (a euphemism for prolific writers with little experience
==========
This works because the .., AKA range operator, gains new capability when used with an if, and turns into what we call the flip-flop operator. In operation what happens is ((l =~ /\AEssentials/) .. (l =~ /Time\z/)) returns false, until (l =~ /\AEssentials/) matches. From then until (l =~ /Time\z/) matches it returns true. Once the final regex matches it returns to returning false.
This behavior works really well for extracting sections from text.
If you are aggregating text, for subsequent output, replace the puts l with something to append l to a buffer, then output that buffer at the end of your run.
Is there a way to get a 2 character day-name of the week such as MO/TU/WE/TH/FR/SA/SU?
Currently I only know of using FormatDateTime():
"ddd" returns "Fri"
"dddd" returns "Friday"
The main reason is that I want to obtain localized version of the 1 or 2 character day names:
Say FRIDAY in "ddd" would return:
French Windows = "Vendredi", the 2 char would be "VE", note it's the 1st and 2nd char.
Chinese Windows = "星期五", the char would be "五", note it's the 3rd char.
Japanese Windows = "金曜日", the char would be "金", note it's the 1st char.
Edit1:
Currently using Delphi, but i think applies to other languages too.
Edit2:
Simply put, I'm looking to obtain the shorter version of "ShortDayName" through the use of some functions or constants, so that I don't have to build a table of constants containing the 7 day "Shorter" day names for every possible windows language.
I wonder if such functions really exist.
Maybe the calendar 1 or 2 char day names in Outlook are hard-coded themselves, right?
You can get the local names for the days of the week with ShortDayNames and LongDayNames, and you can use DayOfWeek to get the numeric value for the day.
ShortDayNames[Index]; //Returns Fri
or
LongDayNames[Index]; //Returns Friday
The only way I know to shorten them to two chars would be to trim the resulting string
LeftStr(LongDayNames[Index],2);//Returns Fr
So today's Day would be
LeftStr(LongDayNames[DayOfWeek(date)],2); //Returns Fr
Click Here
Depicts the standards in custom date formatting.
You may also use the 'ddd' standard and trim it.
Delphi's routines does nothing special - they just ask OS.
Here is how to to it: Retrieving Time and Date Information. I looked through MSDNs docs and found this.
Note, that there is no really such thing as "2 character day-name" or "3 character day-name" here. There are: native ("long" in Delphi), abbreviated ("short" in Delphi) or short (Vista and above, not present in Delphi) formats.
For example, abbreviated name of the day of the week for Monday: Mon (3 chars, en-US), Пн (2 chars, ru-RU).
So, you probably look for LOCALE_SSHORTESTDAYNAMEX format (which is called "short" by MSDN and doesn't appear in Delphi), but it is availavle only on Vista and above.
For example, the following code:
const
LOCALE_SSHORTESTDAYNAME1 = $60;
procedure TForm1.Button1Click(Sender: TObject);
begin
SetThreadLocale($409);
ShowMessage(
GetLocaleStr(GetThreadLocale, LOCALE_SSHORTESTDAYNAME1, '') + #13#10 +
GetLocaleStr(GetThreadLocale, LOCALE_SABBREVDAYNAME1, '')
);
end;
will show you:
Mo
Mon
But doing this for Russian will output:
Пн
Пн
Hope my edits make answer more clear ;)