Issue 79 - cnetjavascript - DatePicker bugs with %d/%m/%y format (Year 09)

Issue 79:	DatePicker bugs with %d/%m/%y format (Year 09)
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Status:	WontFix
Owner:	anutron
Type-Defect
Priority-Medium

Reported by cristiandavilla, Feb 25, 2009

This is the case 
<form>
<table>
<tr>
<td>Date input 2 (dd/mm/yy): </td>
<td>
<input type=\"text\" name=\"date\" value=\"\" id=\"my_date\"/>

<script type=\"text/javascript\">

   new DatePicker('my_date', {
        format: '%d/%m/%y'
   });

</script>

</td>
</tr>
</table>
</form>

The first click is ok, you see on the top of The Datepicker "February
2009".
Datepicker return "19/02/09"  (Great).

But...
But if you click again you see the Year "February 9" on the top and
Datepicker return me "19/02/" (No Great). 


How to solve:
You can see the problem (format year wrong with %d/%m/%y) when
Datepicker works with $parsePatterns (we use Date.$culture="GB").

In this case if you use
   format: '%a, %m %d %y' // Mon, Jan 01 07  (This is ok,      not use
$parsePatterns)

In this case if you use
   format: '%d/%m/%y'     // 24/02/09        (This is not ok, this use
$parsePatterns)

So i solve this problem with a little code
in the $parsePatterns.

This is the patch code:
   if(d.get('year') < 100 ){
      d.increment('year', 2000);
   }

Here we go... with the code:
$parsePatterns: [
   {
   //"12.31.08", "12-31-08", "12/31/08", "12.31.2008", "12-31-2008",
"12/31/2008"
   re: /^(\d{1,2})[\.\-\/](\d{1,2})[\.\-\/](\d{2,4})$/,
   handler: function(bits){
   var d = new Date();
   var culture = Date.$cultures[Date.$culture];
   d.set('year' , bits[Date.$cIndex('year')]);

   // Patch Begin
   if(d.get('year') < 100 ){
      d.increment('year', 2000);
   }
   // Patch End

   d.set('date' , bits[Date.$cIndex('date')]);
   d.set('month', bits[Date.$cIndex('month')] - 1);
   return Date.fixY2K(d);
   }

},

Comment 1 by cristiandavilla, Mar 02, 2009

This not work on month != February 2009.
This seem a bugs of d.increment.

I don't Know why but if your system's date is 
 on feb 2009 d.increment('year', 2000); 
    -->d.increment return 09 + 2000 = 2009 [ok this work well]
But
 on mar 2009 d.increment('year', 2000); 
    -->d.increment return 09 + 2000 = 2007 [ko no good]


So try to solve whit this code and it's work:
{
   //"12.31.08", "12-31-08", "12/31/08", "12.31.2008", "12-31-2008", "12/31/2008"
   re: /^(\d{1,2})[\.\-\/](\d{1,2})[\.\-\/](\d{2,4})$/,
   handler: function(bits){
	var d = new Date();
	var culture = Date.$cultures[Date.$culture];

        // criscd begin
        if( bits[Date.$cIndex('year')] < 100){
           d.set('year' , '20' + bits[Date.$cIndex('year')]);
        } else {
           d.set('year' , bits[Date.$cIndex('year')] );
        }
        // criscd end

	d.set('date' , bits[Date.$cIndex('date')]);
        d.set('month', bits[Date.$cIndex('month')] - 1);
        return Date.fixY2K(d);
	}
},

Comment 2 by ssgoodman, Mar 18, 2009

There are actually a few bugs involved with this ticket.

#1:
The class does not adjust for the culture correctly.
DatePicker.clickCalendar function, first line:
     if(this.options.format == "%d/%m/%Y") Date.$culture = "GB";
Should be:
     if(this.options.format.contains('%d/%m')) Date.$culture = "GB";

As long as the day comes before the month, the culture should be European, even the
format chosen is %X%d%m.  [In this ticket, the person was using: "%d/%m/%y", which
did not match. ]

Comment 3 by ssgoodman, Mar 18, 2009

#2:
The Date.fixY2K() function does not handle two digit dates at all.

Where it says: 
if (newDate.get('year') < 2000 && d.toString().indexOf(newDate.get('year')) < 0) {
	newDate.increment('year', 100);
}

It should really be something like:
     var yr = newDate.get('year') < 2000;
     if(yr < 2000) new Date.set('year', (yr < 45 ? 2000 : 1900)+yr);    
a) I did not understand the second condition you had, and it did not affect my tests.
 Add it back.
b) I switched to "set" instead of "increment" as increment is a relatively expensive
function [there is no need to test for a leap year 2000 times], and is not needed
here.  Increment is where a person is adding a year.  Here, we don't need to add 2000
years to the year 0014 - we need to start with the year 2014! 
c) I assume that 45 is a good turnover date for a two digit year to be past 2000

#3:
Christian refers to a bug with increment in which it drops a day for every leap year,
and falls a month when using increment over 1000 years.  
a)I did not use increment [so it does not affect this issue]  
b) do not understand why when you increment the year you adjust for leap years 

Once the above two changes are made, this issue is solved.

Comment 4 by anutron, May 07, 2009

All tickets are now on github:

http://github.com/anutron/clientcide/issues/

Status: WontFix