DateTime::Format::CLDR 解析长捷克日期的问题

DateTime::Format::CLDR issues parsing long Czech dates

我正在尝试使用 DateTime::Locale::cs_CZ 指定的 CLDR 长格式解析捷克语日期。我看不出它应该失败的明显原因。

my $locale2 = DateTime::Locale->load('cs_CZ');
my $cldr5 = DateTime::Format::CLDR->new(
    pattern     => $locale2->date_format_long,
    locale      => $locale2,
    );

print $cldr5->pattern, "\n";
print $cldr5->parse_datetime('17. listopad 1989');

输出为

d. MMMM y

Use of uninitialized value in print at ./test-cldr.pl line 54.

我做错了什么?为什么它不解析日期?

更新

这是我实际用来在命令行上测试的内容:

perl -MData::Dumper -MDateTime -MDateTime::Format::CLDR -wle'
    local $SIG{__DIE__} = sub { print( Carp::longmess (shift) ); };
    $x="19. září 1979";
    my $loc = DateTime::Locale->load("cs_CZ");
    print Dumper($loc), "\n";
    my $cldr = DateTime::Format::CLDR->new(
       pattern=>$loc->date_format_long, locale=>$loc, on_error => "croak");
    print Dumper($cldr->parse_datetime($x));
'

这是我得到的输出:

$VAR1 = bless( {
                 'en_language' => 'Czech',
                 'native_territory' => "\x{10c}esk\x{e1} republika",
                 'en_territory' => 'Czech Republic',
                 'native_language' => "\x{10d}e\x{161}tina",
                 'default_time_format_length' => 'medium',
                 'id' => 'cs_CZ',
                 'native_complete_name' => "\x{10d}e\x{161}tina \x{10c}esk\x{e1} republika",
                 'default_date_format_length' => 'medium',
                 'en_complete_name' => 'Czech Czech Republic'
               }, 'DateTime::Locale::cs_CZ' );


Could not get datetime for 19. září 1979 (Error marked by 'HERE-->'): '19.  HERE-->září 1979' at -e line 7.
 at /usr/share/perl5/DateTime/Format/CLDR.pm line 965.
    DateTime::Format::CLDR::_local_croak(DateTime::Format::CLDR=HASH(0x166bcd8), "Could not get datetime for 19. z\x{c3}\x{a1}\x{c5}\x{99}\x{c3}\x{ad} 1979 (Error marked by "...) called at /usr/share/perl5/DateTime/Format/CLDR.pm line 558
    DateTime::Format::CLDR::__ANON__("z\x{c3}\x{a1}\x{c5}\x{99}\x{c3}\x{ad} 1979") called at /usr/share/perl5/DateTime/Format/CLDR.pm line 582
    DateTime::Format::CLDR::parse_datetime(DateTime::Format::CLDR=HASH(0x166bcd8), "19. z\x{c3}\x{a1}\x{c5}\x{99}\x{c3}\x{ad} 1979") called at -e line 7

Could not get datetime for 19. září 1979 (Error marked by 'HERE-->'): '19.  HERE-->září 1979' at -e line 7.

17。列表 1989

print(Dumper($self->_build_pattern())); 可以一窥模块的期望。在这种情况下它输出以下内容。

$VAR1 = [
          [
            qr/(3[01]|[12]\d|0?[1-9])/,
            'd',
            1
          ],
          '\.',
          '\s+',
          [
            "(listopadu|\x{10d}ervence|prosince|b\x{159}ezna|kv\x{11b}tna|\x{10d}ervna|ledna|\x{fa}nora|dubna|srpna|\x{159}\x{ed}jna|z\x{e1}\x{159}\x{ed})",
            'M',
            4
          ],
          '\s+',
          [
            qr/(-?\d{1,4})/,
            'y',
            1
          ]
        ];

它期望 listopadu,而不是 listopad。我对捷克语一无所知,但模块的行为似乎是 correct。您可以 "fix up" 在解析它们之前的日期。

$date_string =~ s/\blistopad\b\K/u/;

19。扎日 1979

你实际上并没有超过 září ("z\x{e1}\x{159}\x{ed}")。您正在传递其 UTF-8 编码 ("z\x{c3}\x{a1}\x{c5}\x{99}\x{c3}\x{ad}")。添加 use utf8; 以告诉 Perl 源代码使用 UTF-8 编码。