Moose如何仅在$ undef时更改属性值?

时间:2014-03-17 21:59:26

标签: perl moose

现在有:

has 'id' => (
    is => 'rw',
    isa => 'Str',
    default => sub { "id" . int(rand(1000))+1 }
);

工作正常,:

PKG->new(id => 'some'); #the id is "some"
PKG->new()              #the id is #id<random_number>

在下一个场景中:

my $value = undef;
PKG->new(id => $value);

(当然)有一个错误:

Attribute (id) does not pass the type constraint because: Validation failed for 'Str' with value undef at /Users/me/perl5/perlbrew/perls/perl-5.16.3/lib/site_perl/5.16.3/darwin-thread-multi-2level/Moose/Exception.pm line 37

问题是:

如何在设置为undef之后更改值(仅当它是$ undef时)?所以,

has 'id' => (
    is => 'rw',
    isa => 'Str|Undef',  #added undef to acceptable Type
    default => sub { "id" . int(rand(1000))+1 }
);

现在,它接受$undef,但我不想要$undef,但想要"id" . int(rand(1000))+1。如何在设置之后更改属性值

仅为不是构造函数的访问器调用after。从coercionUndef可能有些奇怪的Str - 但这个属性只有 吗?

Ps:使用PKG->new( id => $value // int(rand(10000)) )是不可接受的解决方案。模块应该接受$undef并且应该静默地将其更改为随机数。

3 个答案:

答案 0 :(得分:6)

Type::Tiny的目标之一是轻松将强制添加到个别属性中。这是一个例子:

use strict;
use warnings;

{
    package Local::Test;
    use Moose;
    use Types::Standard qw( Str Undef );

    my $_id_default = sub { "id" . int(rand(1000)+1) };

    has id => (
        is      => 'rw',
        isa     => Str->plus_coercions(Undef, $_id_default),
        default => $_id_default,
        coerce  => 1,
    );

    __PACKAGE__->meta->make_immutable;
}

print Local::Test->new(id => 'xyz123')->dump;
print Local::Test->new(id => undef)->dump;
print Local::Test->new->dump;

您还可以查看MooseX::UndefTolerant,它会将undef值传递给构造函数,就好像它们被完全省略一样。这不会涵盖将undef传递给访问者;只是构造者。

答案 1 :(得分:2)

这是另一种选择,使用Moose'BRO构造方法,在创建对象后调用它。

#!/usr/bin/perl

package Test;
use Moose;

has 'id' => (
    is => 'rw',
    isa => 'Str|Undef',
);

sub BUILD {
    my $self = shift;
    unless($self->id){
        $self->id("id" . (int(rand(1000))+1));
    }
}
1;

package Main;


my $test = Test->new(id => undef);
print $test->id; ###Prints random number if id=> undef

有关BUILD的更多信息: http://metacpan.org/pod/Moose::Manual::Construction#BUILD

答案 2 :(得分:0)

@choroba在评论中提及triggers。基于此,找到了下一个解决方案。在id=>undef的情况下,触发器被调用两次,否则它会起作用。

use Modern::Perl;

package My;
use namespace::sweep;
use Moose;

my $_id_default = sub { "id" . int(rand(100_000_000_000)+1) };
my $_check_id = sub { $_[0]->id(&$_id_default) unless $_[1] };

has id => (
    is      => 'rw',
    isa => 'Str|Undef',
    default => $_id_default,
    trigger => $_check_id,
);

__PACKAGE__->meta->make_immutable;

package main;

say My->new->id;
say My->new(id=>'aaa')->id;
say My->new(id=>undef)->id;