我想导出一个函数,该函数取决于导出到的类的名称。我认为这应该很容易,Sub::Exporter
但不幸的是,into
密钥没有传递给生成器。我最终得到了那些丑陋的示例代码:
use strict;
use warnings;
package MyLog;
use Log::Log4perl qw(:easy get_logger);
use Sub::Exporter -setup => {
exports => [
log => \&gen_log,
audit_log => \&gen_log,
],
groups => [ default => [qw(log audit_log)] ],
collectors => ['category'],
installer => \&installer, # tunnel `into` value into generators
};
if ( not Log::Log4perl->initialized() ) {
#easy init if not initialised
Log::Log4perl->easy_init($ERROR);
}
sub gen_log {
my ( $class, $name, $arg, $global ) = @_;
my $category = $arg->{category};
$category = $global->{category}{$name} unless defined $category;
return sub { # return generator
my $into = shift; # class name passed by `installer`
$category = $name eq 'audit_log' ? "audit_log.$into" : $into
if !defined $category; # set default category
# lazy logger
my $logger;
return sub {
$logger or $logger = get_logger($category);
};
};
}
sub installer {
my ( $args, $todo ) = @_;
# each even value is still generator thus generate final function
my $i;
1 & $i++ and $_ = $_->( $args->{into} ) for @$todo;
Sub::Exporter::default_installer(@_);
}
1;
有没有更好的方法可以在不牺牲所有这些丰富Sub::Exporter
能力的情况下做到这一点?
例如,我想使用其中之一:
use MyLog category => { log => 'foo', audit_log => 'bar' };
use MyLog -default => { -prefix => 'my_' };
use MyLog
audit_log => { -as => 'audit' },
log => { -as => 'my_log', category => 'my.log' };
编辑:增加Sub::Exporter
了对问题的能力要求。
Edit2:添加了使用示例。