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Accepted
Mario Palumbo
Asked: 2021-01-12 03:35:00 +0800 CST

replace text with part of text using regex with bash perl

  • 772

For example I have this output:

string1 anynameveryveryverylong string2
string1 othernameveryveryverylong   string2

I want truncate the name to the first ten characters:

string1 anynamever  string2
string1 othernamev  string2

a pseudo regex can be:

perl -pe "s/([^\t]+\t)([^\t]+)\t/\1\2{10}\t/g"

How do i get this?

bash

2 Answers

  1. Best Answer
    perl -pe 's/^(\S+\s+)(\S{10})\S*/$1$2/'
    • ^ matches at the start of the string
    • \S means non-whitespace
    • + means repeated at least once
    • \s means whitespace
    • {10} means repeated 10 times

    I.e. Keep the first word and the first 10 characters of the following word while forgetting the remaining characters of the second word.

    Your pseudoregex has one substantial problem: the {10} is placed in the replacement part, but the replacement is just a string. The regex happens in the pattern part only.

    • 4

  2. Some more choices:

    1. Perl with autosplitting on tabs:

      $ perl -F"\t" -lae '$F[1]=substr($F[1],0,10); print join "\t",@F' file 
string1  anynamever  string2
string1  othernamev  string2

    2. awk

      $ awk -F"\t" -vOFS="\t" '{$2=substr($2,1,10)}1' file 
string1  anynamever  string2
string1  othernamev  string2

    3. sed

      $ sed -E 's/(\S+\t\S{10})[^\t]+/\1/' file 
string1  anynamever  string2
string1  othernamev  string2

    4. One more Perl

      $ perl -pe 's/(\S+\t\S{10})[^\t]+/\1/' file 
string1  anynamever  string2
string1  othernamev  string2
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