regex - 如何重置 $1 perl 变量？

Question

我尝试进行正则表达式匹配:

$address =~ s/$re_region//;
$new_address_h->{ region     } =  $1  //'';

# $1 =  undef; # ERROR: Modification of a read-only value attempted at ...
$address =~ s/$re_city//;
$new_address_h->{ city       } =  $1;

当城市不匹配时，我希望它用undef填充。但是当 city 不匹配时，$1 具有之前匹配的值。为什么？我希望当匹配失败时，它的输出没有值，甚至没有特殊变量。像 eval 刷新 $@

没有额外的 if 是否可行？

Answer 1

记录在案的行为 ( perlvar ):

Perl sets these variables when it has a successful match, so you should check the match result before using them.

您可以使用 do使用三元运算符填充哈希:

$new_address_h->{region} = do { $address =~ s/$re_region// ? $1 : undef };
$new_address_h->{city}   = do { $address =~ s/$re_city//   ? $1 : undef };

或者，用 undefs 填充散列，然后在匹配时用值替换它们:

my $new_address_h = { region => undef, city => undef };
$address =~ s/$re_region// and $new_address_h->{region} = $1;
$address =~ s/$re_city//   and $new_address_h->{city}   = $1;

常见的方式是在列表赋值中只匹配，而不是替换和填充:

($new_address_h->{region}) = $address =~ /$re_region/;
($new_address_h->{city})   = $address =~ /$re_city/;