On 10/23/06, Dave Rolsky <[EMAIL PROTECTED]> wrote:
0.52 2006-10-22
- The names_in_country() method no longer sorts the zone
names. Instead, it returns them in order from most- to
least-populated.
In 0.55 it's still sorted by names.
% perl -MDateTime::TimeZone -le 'print $DateTime::TimeZone::VERSION;
print [ DateTime::TimeZone->names_in_country("us") ]->[0]'
0.55
America/Adak
In 0.52 you removed the sort() in the wrong place. Here's a patch and
an unit test against 0.55.
=== t/15catalog.t
==================================================================
--- t/15catalog.t (revision 5779)
+++ t/15catalog.t (local)
@@ -8,7 +8,7 @@
use DateTime::TimeZoneCatalog;
-plan tests => 29;
+plan tests => 30;
{
my @all = DateTime::TimeZone::all_names();
@@ -100,3 +100,10 @@
[ 'America/Santiago', 'Pacific/Easter' ],
'zones for Chile are America/Santiago and Pacific/Easter' );
}
+
+{
+ my @zones = DateTime::TimeZone::names_in_country('us');
+ is( $zones[0], 'America/New_York', 'I <3 New York' );
+}
+
+
=== tools/parse_olson
==================================================================
--- tools/parse_olson (revision 5779)
+++ tools/parse_olson (local)
@@ -431,12 +431,12 @@
$countries{UK} = $countries{GB};
my $countries = '';
- # We explicitly do not sort these because the order in zones.tab
- # is by population.
- for my $c ( keys %countries )
+ for my $c ( sort keys %countries )
{
$countries .= qq| '\L$c' => [ qw(\n|;
- $countries .= join "\n", map { " $_" } sort @{ $countries{$c} };
+ # We explicitly do not sort these because the order in zones.tab
+ # is by population.
+ $countries .= join "\n", map { " $_" } @{ $countries{$c} };
$countries .= "\n) ],\n";
}
--
Tatsuhiko Miyagawa
