Perl的正则表达式警告：\ 1更好地写成$ 1日（EVAL 1）1号线

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"; 

来看，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 

我不希望看到的警告，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是正则表达式这意味着“匹配捕获到的第一个副本的内容”。在替换表达式中使用它绝对没有意义。要获得由第一组捕获配对捕获的字符串，请使用$1。

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

应该

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

所以

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

应该

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

或更好，但

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

Answer 2

假设你的模式匹配的地图格局保持不变，为什么不能摆脱它，简单地说：

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

Answer 3

我看到你的代码有很多重复的。

假设您使用的是Perl 5.10或更高版本，这就是我将如何编写代码。

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;