Re: [Vala] Date format

[Reid Thompson <Reid Thompson ateb com>]

From the bottom of this page
http://www.gnu.org/s/hello/manual/libc/The-Elegant-and-Fast-Way.html

you would just need to query for the default format using
 nl_langinfo (D_T_FMT)
check it, and then configure your output to match except with YYYY in
place of YY.

An example of nl_langinfo usage is a function which has to print a given date and time in a locale-specific 
way. At first one might think that, since strftime internally uses the locale information, writing something 
like the following is enough:

size_t
     i18n_time_n_data (char *s, size_t len, const struct tm *tp)
     {
       return strftime (s, len, "%X %D", tp);
     }

The format contains no weekday or month names and therefore is
internationally usable. Wrong! The output produced is something like
"hh:mm:ss MM/DD/YY". This format is only recognizable in the USA. Other
countries use different formats. Therefore the function should be
rewritten like this:

size_t
     i18n_time_n_data (char *s, size_t len, const struct tm *tp)
     {
       return strftime (s, len, nl_langinfo (D_T_FMT), tp);
     }

Now it uses the date and time format of the locale selected when the
program runs. If the user selects the locale correctly there should
never be a misunderstanding over the time and date format. 


reid
On Tue, 2011-10-25 at 22:06 +0200, rastersoft wrote:
> Thanks, but that doesn't solve my problem. I already got access to that, 
> and tested format("%x"); unfortunately, it puts the year as a two-digit 
> number.
>
> I want the same than format("%x") but with four-digit years; the problem 
> is how to know if the current locale mandates to write month/day/year or 
> day/month/year.
>
> Currently I do a check at program startup, doing format("%x") for date 
> day 1, month 3, year 2005, and check the output: if there's a 3 as the 
> first number, then the order is month/day/year; if not, it's 
> day/month/year. Unfortunately, that's an extremely ugly hack. So I want 
> to know if is there a straightforward way of doing this.
>
> Thanks.
>
> El 25/10/11 15:35, Reid Thompson escribió:
> > On Mon, 2011-10-24 at 22:33 +0200, rastersoft wrote:
> > > I want to print a date in the format dd/mm/yyyy or mm/dd/yyyy,
> > > acording
> > slightly modified from http://live.gnome.org/Vala/TimeSample
> >
> > void main () {
> >
> >      // A DateTime from a Unix timestamp
> >      int64 timestamp = 1234151912;
> >      var time = new DateTime.from_unix_utc (timestamp);
> >      // convert back to Unix timestamp
> >      assert (time.to_unix () == timestamp);
> >
> >      // A DateTime from year, month, day, hour, minute, second
> >      time = new DateTime.utc (2010, 10, 22, 9, 22, 0);
> >
> >      // The current time in local timezone
> >      var now = new DateTime.now_local ();
> >      print ("Is daylight savings time: %s\n", now.is_daylight_savings () ? "yes" : "no");
> >      print ("The timezone abbreviation is: %s\n", now.get_timezone_abbreviation ());
> >
> >      // returns time in RFC 3339 format: 2010-10-21T23:48:03+0200
> >      string date_string = now.to_string ();
> >      print ("Current time in RFC 3339 format: %s\n", date_string);
> >
> >      // for example, according to the current locale
> >      print ("According to the current locale: %s\n", now.format ("%x %X"));
> >
> >      print ("Day of month: %d\n", now.get_day_of_month ());
> >      print ("Week of year: %d\n", now.get_week_of_year ());
> >
> >      // Add one day, three hours and five minutes to a DateTime:
> >      var future = now.add_days (1).add_hours (3).add_minutes (5);
> >      print (@"Plus one day, three hours and five minutes: $future\n");
> >
> >
> >      print ("According to the current locale: %s\n", now.format ("%d/%m/%Y"));
> >      print ("According to the current locale: %s\n", now.format ("%m/%d/%Y"));
> >
> > }