Question

是否可以使用条件项创建数组？

my @a = (1, ($condition) ? 2 : "no-op", 3);

这样"no-op"是有效的功能，如果$condition为假，那么我得到列表(1, 3)，但如果$condition为真，我得到{{1} }

背景

(1, 2, 3)

在某些情况下，我想加入

行

use strict;
use warnings;
use File::Find::Rule;

my $rule = File::Find::Rule->new();
$rule->or(
    $rule->new->name('*.cfg')->prune->discard,
    $rule->directory->name("_private.d")->prune->discard,
    $rule->new->name('*.t')->prune->discard,
    $rule->new->name('*.bak')->prune->discard,
    $rule->new->name('.*.bak')->prune->discard,
    $rule->new->name('.#*')->prune->discard,
);

my @files = $rule->in(".");

在其他情况下，我不想排除目录$rule->directory->name("_private.d")->prune->discard ..

Answer 1

您可以使用空列表()跳过第二个元素

my @a = (1, ($condition ? 2 : ()), 3);

Answer 2

一般来说，您可以使用

获得一些可读性增益

my @a;
push @a, 1;
push @a, 2 if $condition;
push @a, 3;

在上下文中，那将是

my @rules;
push @rules, $rule->new->name('*.cfg')->prune->discard;
push @rules, $rule->directory->name("_private.d")->prune->discard;
push @rules, $rule->new->name('*.t')->prune->discard;
push @rules, $rule->new->name('*.bak')->prune->discard;
push @rules, $rule->new->name('.*.bak')->prune->discard;
push @rules, $rule->new->name('.#*')->prune->discard;

my $rule = File::Find::Rule->new()->or(@rules);
my @files = $rule->in(".");

使用条件项生成列表

2 个答案: