perl - Moose 从单个参数构造对象

Question

我已经涉足 Moose 大约七个月，而 Perl 只是稍微长一点，但无法弄清楚如何通过为每个属性提供一个参数而不是它们的整个 hashref 来构造一个类中的多个属性。我已经广泛搜索了文档和网络，但我要么在寻找错误的词，要么遗漏了一些东西。

我已经调整了设计，使其更通用。使用以下基本设置:

package First;
use Moose;
use Second::Type1;
use Second::Type2;
has 'type1' => (
  is => 'rw',
  isa => 'Second::Type1',
  default => sub {Second::Type1->new(name => 'random')}
);

has 'type2' => (
  is => 'rw',
  isa => 'Second::Type2',
  default => sub {Second::Type2->new(name => 'random')}
);

package Second::Type1;
use Moose;
use This;
has 'name' => (
  is => 'rw',
  isa => 'Str',
  required => 1,
);
has 'this' => (
  is => 'rw',
  isa => 'This',
  default => sub {This->new()}
);
# package has more attributes, but you get the idea
__PACKAGE__->meta->make_immutable();
no Moose;
1;

package Second::Type2;
use Moose;
use That;
has 'name' => (
  is => 'rw',
  isa => 'Str',
  required => 1,
);
has 'that' => (
  is => 'rw',
  isa => 'That',
  default => sub {That->new()}
);
# package has more attributes, but you get the idea
__PACKAGE__->meta->make_immutable();
no Moose;
1;

我希望能够通过说:

use First;
my $first = First->new(type1 => 'foo', type2 => 'bar');

其中“foo”等于 Second::Type1 的“name”属性的值，“bar”等于 Second::Type2 的“name”属性的值。

现在至于我自己的解决方案，我已经(成功地)制作了一个 Moose::Role，它只包含一个“围绕 BUILDARGS”的子，然后使用一个 Factory 类(其内容在 IMO 中不相关):

package Role::SingleBuildargs;

use Moose::Role;
use Factory::Second;
requires 'get_supported_args';

around BUILDARGS => sub {
my ($class, $self, %args) = @_;
my @supported_args = $self->get_supported_args;
my $factory = Factory::Second->new();
    my @errors = ();
    foreach my $arg (sort {$a cmp $b} keys %args) {
        if (grep {$_ eq $arg} @supported_args) {
            my $val = $args{$arg};
            if (!ref $val) {    # passed scalar init_arg
                print "$self (BUILDARGS): passed scalar\n";
                print "Building a Second with type '$arg' and name '$val'\n";
                $args{$arg} = $factory->create(type => $arg, name => $val)
            } elsif (ref $val eq 'HASH') {  # passed hashref init_arg
                print "$self (BUILDARGS): passed hashref:\n";
                my %init_args = %$val;
                delete $init_args{name} unless $init_args{name};
                $init_args{type} = $arg;
                $args{$arg} = $factory->create(%init_args);
            } else {    # passed another ref entirely
                print "$self (BUILDARGS): cannot handle reference of type: ", ref $val, "\n";
                die;
            }
        } else {
            push @errors, "$self - Unsupported attribute: '$arg'";
        }
    }
    if (@errors) {
        print join("\n", @errors), "\n";
        die;
    }
    return $self->$class(%args);
    };

no Moose;
1;

然后我在 First 类和其他类(如 First)中使用该角色。

我也尝试通过以下方式进行强制:

package Role::Second::TypeConstraints;
use Moose::Util::TypeConstraints

subtype 'SecondType1', as 'Second::Type1';
subtype 'SecondType2', as 'Second::Type2';
coerce 'SecondType1', from 'Str', via {Second::Type1->new(name => $_};
coerce 'SecondType2', from 'Str', via {Second::Type2->new(name => $_};

no Moose::Util::TypeConstraints;
1;

并修改了第一个包(仅列出更改):

use Role::Second::TypeConstraints;
has 'type1' => (
   isa => 'SecondType1',
   coerce => 1,
);
has 'type2' => (
   isa => 'SecondType2',
   coerce => 1,
);    

然而这并没有奏效。如果有人能解释原因，那就太好了。

至于实际问题:在您的类(class)中获得这种行为的最佳方法是什么？真的没有比修改 BUILDARGS 更好的方法，还是我错过了什么(关于 Moose::Util::TypeConstraints，也许)？ TMTOWTDI 和所有，但我的似乎根本没有效率。

编辑:为一致性进行编辑(混合通用类名)

Answer 1

您可以完全按照您描述的方式使用强制转换

添加

use Moose::Util::TypeConstraints;

至 First , Second::Type1和 Second::Type2

向 Second::Type1 添加强制操作

coerce 'Second::Type1'
    => from 'Str'
        => via { Second::Type1->new( name => $_ ) };

并到 Second::Type2

coerce 'Second::Type2'
    => from 'Str'
        => via { Second::Type2->new( name => $_ ) };

为 type1 启用强制转换和 type2 First 的属性

has 'type1' => (
  is      => 'rw',
  isa     => 'Second::Type1',
  default => sub { Second::Type1->new },
  coerce  => 1,
);

has 'type2' => (
  is      => 'rw',
  isa     => 'Second::Type2',
  default => sub { Second::Type2->new },
  coerce  => 1,
);

然后你可以创建一个 First完全按照你说的反对

my $first = First->new(type1 => 'foo', type2 => 'bar');