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brianegge
Asked: 2010-02-11 15:20:14 +0800 CST

How do I extract the matching part of a regex on Solaris

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GNU's grep has the option --only-matching, which prints just the matching region of a regular expression. I'm on a Solaris 5.10 box without any GNU tools installed, and I'm trying to achieve the same thing. Example:

grep -Eo "[0-9]+ ms" *.log

Is there a sed or awk expression that can do the same thing?

solaris regex grep

3 Answers

  1. Best Answer
    sed -n 's/.*[^0-9]\([0-9]\+ ms\).*/\1/p' *.log

    A possible refinement for versions of sed which support this type of alternation:

    sed -n 's/\(^\|.*[^0-9]\)\([0-9]\+ ms\).*/\2/p' *.log

    In OS X, I had to use extended regular expressions because I couldn't get basic enhanced regular expressions to work (this would work in GNU sed, too, if you change -E to -r, but the basic enhanced version works there, too):

    sed -En 's/(^|.*[^0-9])([0-9]+ ms).*/\2/p' *.log

    These two latter examples work even if the sequence you're searching for appears at the beginning of the line with no leading characters.

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  2. Try using /usr/sfw/bin/ggrep

    /usr/sfw/bin/ggrep -V grep (GNU grep) 2.5

    Copyright 1988, 1992-1999, 2000, 2001 Free Software Foundation, Inc. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

    >

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  3. Solaris 5.10 includes bash 3.00, which has a modern regex engine. The follow command works, though it's quite a bit more verbose than a grep or sed solution. 

    cat *.log | while read line; 
do 
  if [[ $line =~ "([0-9]+) ms" ]]; then 
    echo "${BASH_REMATCH[1]}"; 
  fi;
done

    (Formatted on multiple lines for readability)

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