Question

我有这个Perl代码：

my $C;

if($someCondition) {
    my $A = My::Package::A::V1->new();
    my $B = My::Package::B::V1->new();
    $C = My::Package::C::V1->new();
} else {
    my $A = My::Package::A::V2->new();
    my $B = My::Package::B::V2->new();
    $C = My::Package::C::V2->new();
}
# Have fun with $C

考虑到唯一的区别是基于V1的{{1}}或V2，我发现这很多余。有什么办法可以做类似$someCondition的事情吗？

如果我也有呢？

my $A = My::Package::A::V{$someCondition ? 1 : 2}->new();

为此，我尝试过：

my $C;
if($someCondition) {
    my $C = Some::Package::Z->new(
      myPackageA => &My::Package::A::V1
    );
} else {
    my $C = Some::Package::Z->new(
      myPackageA => &My::Package::A::V2
    );
}

但是我得到了错误：my $api_version; if($someCondition) { $api_version = 'V1'; } else { $api_version = 'V2'; } my $C = Some::Package::Z->new( myPackageA => "&My::Package::A::$api_version" );

Answer 1

您可以将类存储在字符串中，并使用它来引用包：

my $api_version;
if($someCondition) {
    $api_version = 'V1';
} else {
    $api_version = 'V2';
}
die "No API version found" unless $api_version;

my $A = "My::Package::A::$api_version"->new();
my $B = "My::Package::B::$api_version"->new();
my $C = "My::Package::C::$api_version"->new();

# Have fun with $C

如果您还想在名称空间中调用函数（就像您对扩展问题所做的那样），那么简单的方法是关闭strict并仅使用它们的名称来调用函数：

my $make_A;
{
    no strict 'refs';
    $make_A = \&{ "My::Package::A::$api_version" };
}
my $C = Some::Package::Z->new(
  myPackageA => $make_A->(),
);

根据条件不同

1 个答案: