Question

use strict;
use warnings;

my $newPasswd = 'abc123';
my @lines = ( "pwd = abc", "pwd=abc", "password=def", "name= Mike" );

my %passwordMap = (
    'pwd(\\s*)=.*'      => 'pwd\\1= $newPasswd',
    'password(\\s*)=.*' => 'password\\1= $newPasswd',
);

print "@lines\n";

foreach my $line (@lines) {
    while ( my ( $key, $value ) = each(%passwordMap) ) {
        if ( $line =~ /$key/ ) {
            my $cmdStr = "\$line =~ s/$key/$value/";
            print "$cmdStr\n";
            eval($cmdStr);
            last;
        }
    }
}

print "@lines";

运行它会给我正确的结果：

pwd = abc pwd=abc password=def name= Mike
$line =~ s/pwd(\s*)=.*/pwd\1= $newPasswd/
\1 better written as $1 at (eval 2) line 1 (#1)
$line =~ s/password(\s*)=.*/password\1= $newPasswd/
\1 better written as $1 at (eval 3) line 1 (#1)
pwd = abc123 pwd=abc password= abc123 name= Mike

我不想看到警告，试图使用$ 1而不是\ 1，但它不起作用。我该怎么办？非常感谢。

Answer 1

\1是一个正则表达式，意思是“匹配第一组捕获的parens捕获的内容。”在替换表达式中使用它是完全没有意义的。要获取第一组捕获parens捕获的字符串，请使用$1。

$line =~ s/pwd(\s*)=.*/pwd\1= $newPasswd/

应该是

$line =~ s/pwd(\s*)=.*/pwd$1= $newPasswd/

所以

'pwd(\\s*)=.*'      => 'pwd\\1= $newPasswd',
'password(\\s*)=.*' => 'password\\1= $newPasswd',

应该是

'pwd(\\s*)=.*'      => 'pwd$1= $newPasswd',
'password(\\s*)=.*' => 'password$1= $newPasswd',

或更好

qr/((?:pwd|password)\s*=).*/ => '$1= $newPasswd',

Answer 2

我看到你的代码中有很多重复。

假设您使用的是Perl 5.10或更高版本，这就是我编写代码的方式。

use strict;
use warnings;
use 5.010;

my $new_pass = 'abc123';
my @lines = ( "pwd = abc", "pwd=abc", "password=def", "name= Mike" );

my @match = qw'pwd password';
my $match = '(?:'.join( '|', @match ).')';

say for @lines;
say '';

s/$match \s* = \K .* /$new_pass/x for @lines;
# which is essentially the same as:
# s/($match \s* =) .* /$1$new_pass/x for @lines;

say for @lines;

Answer 3

假设您的模式匹配地图的模式保持不变，为什么不摆脱它并简单地说：

$line =~ s/\s*=.*/=$newPassword/

perl正则表达式警告：\ 1最好在（eval 1）第1行写为$ 1

3 个答案: