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Accepted
Mario Palumbo
Asked: 2020-12-21 02:24:37 +0800 CST

bash perl \z doesn't work

  • 772

I have this text file or sdout:

text1
text2
text1
text2
text1
text2

I have this code:

perl -pe "s/text2\n\z/text3/s" text.txt #Note the modifier "/s"

I with \z expect him to understand that this is the last line before the eof and consequently I expect:

text1
text2
text1
text2
text1
text3

But instead it returns:

text1
text3text1
text3text1
text3 #without final newline

What am I doing wrong?

bash regex perl

3 Answers

  1. The command line option -p means that the file is processed line by line, i.e. what you do is basically:

    while (<>) {
     s/text2\n\z/text3/s;
     print $_
}

    Therefore the \z does not match the end of all input data but the end of each line.

    What you likely need is not to use -p but instead read the full file and replace only the end. Note that the regex can be simplified by just using $ and /s and \z are not needed:

    perl -e 'local $/; (my $x = <>) =~s{text2$}{text3}; print $x' text.txt
    • 6

  2. There is no need to slurp in the entire file. Use this Perl one-liner:

    perl -lpe 's{text2}{text3} if eof ' in_file

    Prints:

    text1
text2
text1
text2
text1
text3
    • 4
  3. Best Answer

    The solution that I have adopted is:

    perl -0pe 's/text2\s*$/text3\n/' text.txt
    • 0