Perl 使用变量搜索和替换，字符串包含点

Question

I'm using a variable to search and replace a string using Perl. I want to replace the string 23.0 with 23.0.1 , so I tried this:

  my $old="23.0";
  my $new="23.0.1";
  $_ =~ s/$old/$new/g;         

The problem is that it also replaced the string 2310 , so I tried:

  my $old="23\.0"

and also /ee . But can't get the correct syntax for it to work. Can someone show me the correct syntax?

Answer 1

There are two things that will help you here:

The quotemeta function - that will escape meta characters. And also the \\Q and \\E regex flags, that stop regex interpolation.

print quotemeta "21.0";

Or:

my $old="23.0";
my $new="23.0.1";
my $str = "2310";
$str =~ s/\Q$old\E/$new/g;
print $str;

Answer 2

只需使用单引号并转义点即可。

my $old='23\.0';

Answer 3

To complement Sobrique's excellent answer, let me note that the reason your attempt with "23\\.0" didn't work is that "23\\.0" and "23.0" evaluate to the same string: in a double-quoted string literal, the backslash escape sequence \\. simply evaluates to . .

There are several things you could do to avoid this:

• If you indeed want to match a fixed string, and don't need or want to include any special regexp metacharacters in it, you can do as Sobrique suggest and use quotemeta or \\Q to escape them.

  In particular, this is almost always the correct solution if the string to be matched comes from user input. If you do want to allow some limited set of non-literal metacharacters, you can unescape those after running the pattern through quotemeta . For a simple example, here's a quick-and-dirty way to turn a basic glob -like pattern (using the metacharacters ? and * for "any character" and "any string of characters" repectively) into an equivalent regexp:

   my $regexp = "^\\Q$glob\\E\\$"; # quote and anchor the pattern $regexp =~ s/\\\\\\?/./g; # replace "?" (escaped to "\\?" by \\Q) with "." $regexp =~ s/\\\\\\*/.*/g; # replace "*" (escaped to "\\*" by \\Q) with ".*"

• Conversely, if you want to have a literal regexp pattern in your code, without immediately matching it against something, you can use the qr// regexp-like quote operator, like this:

   my $old = qr/\\b23\\.0(\\.0)?\\b/; # match 23.0 or 23.0.0 (but not 123.012!) my $new = "23.0.1"; # just a literal string s/$old/$new/g; # replace any string matching $old in $_ with $new

  Note that qr// has other effects beyond just allowing you to use regexp syntax in a string literal: it actually pre-compiles the pattern into a special Regexp object, so that it doesn't need to be recompiled every time it's used later. In particular, as a side effect, the string representation of a qr// regexp literal will usually not exactly match the original content, although it will be equivalent as a regexp. For example, say qr/\\b23\\.0(\\.0)?\\b/ will, on my Perl version, output (?^u:\\b23\\.0(\\.0)?\\b) .

• You could also just use a normal double-quoted string literal, and double any backslashes in it, but that's (usually) less efficient than using qr// , and also less readable due to leaning toothpick syndrome .

  Using a single-quoted string literal would be slightly better, since backslashes in a single-quoted string are only special when followed by another backslash or a single quote. Even so, readability can still suffer if you happen to need to match any literal backslashes in your regexp, not to mention that it's easy to create subtle bugs if you forget to double a backslash in those rare places where it's still needed.