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I have this one liner:
perl -pe 's|.*?((\d{1,3}\.){3})xxx.*|\1|'
I feed this command with some input, like 192.168.1.xxx, and it works. Now, I want to append a 0 to the output sequence, but of course if I just append the 0 right after the \1 it will be parsed as the tenth capturing group. How can I concatenate then it to the \1 directive?
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Substitution Operator or 's' operator in Perl is used to substitute a text of the string with some pattern specified by the user.
=~ is the Perl binding operator. It's generally used to apply a regular expression to a string; for instance, to test if a string matches a pattern: if ($string =~ m/pattern/) {
Regular Expression (Regex or Regexp or RE) in Perl is a special text string for describing a search pattern within a given text. Regex in Perl is linked to the host language and is not the same as in PHP, Python, etc. Sometimes it is termed as “Perl 5 Compatible Regular Expressions“.
m operator in Perl is used to match a pattern within the given text. The string passed to m operator can be enclosed within any character which will be used as a delimiter to regular expressions.
You should use $1 instead of \1 in substitutions. Then you can use braces to write it unambiguously like this:
perl -pe 's|.*?((\d{1,3}\.){3})xxx.*|${1}0|'
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