如何在不设置变量的情况下执行 perl 内联正则表达式? [英] How to do perl inline regex without setting to a variable?

问题描述

通常，如果您想使用正则表达式更改变量，请执行以下操作:

Normally if you wish to change a variable with regex you do this:

$string =~ s/matchCase/changeCase/; 

但是有没有一种方法可以简单地进行内联替换而不将其设置回变量?

But is there a way to simply do the replace inline without setting it back to the variable?

我希望在这样的地方使用它:

I wish to use it in something like this:

my $name="jason";
print "Your name without spaces is: " $name => (/\s+/''/g);

类似的东西，有点像 PHP 中的 preg_replace 函数.

Something like that, kind of like the preg_replace function in PHP.

推荐答案

针对 Perl 5.14 进行了修订.

Revised for Perl 5.14.

自 5.14 起，使用 /r 标志来返回替换，您可以这样做:

Since 5.14, with the /r flag to return the substitution, you can do this:

print "Your name without spaces is: [", do { $name =~ s/\s+//gr; }
    , "]\n";

<小时>

您可以使用 map 和一个词法变量.

my $name=" jason ";

print "Your name without spaces is: ["
    , ( map { my $a = $_; $a =~ s/\s+//g; $a } ( $name ))
    , "]\n";

现在，您必须使用词法，因为 $_ 会别名，从而修改您的变量.

Now, you have to use a lexical because $_ will alias and thus modify your variable.

输出是

Your name without spaces is: [jason]
# but: $name still ' jason '

诚然do 也能正常工作(也许更好)

Admittedly do will work just as well (and perhaps better)

print "Your name without spaces is: ["
    , do { my ( $a = $name ) =~ s/\s+//g; $a }
    , "]\n";

但是词法复制仍然存在.my 中的赋值是一些人(不是我)喜欢的缩写.

But the lexical copying is still there. The assignment within in the my is an abbreviation that some people prefer (not me).

对于这个习语，我开发了一个我称之为 filter 的操作符:

For this idiom, I have developed an operator I call filter:

sub filter (&@) { 
    my $block = shift;
    if ( wantarray ) { 
        return map { &$block; $_ } @_ ? @_ : $_;
    }
    else { 
       local $_ = shift || $_;
       $block->( $_ );
       return $_;
    }
}

你这样称呼它:

print "Your name without spaces is: [", ( filter { s/\s+//g } $name )
    , "]\n";

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