Grep with continuation lines

awkgrepsedtext processing

How can I grep/awk/sed a file looking for some pattern, and print the entire line (including continuation lines if the matched line ends with \?

File foo.txt contains:

something
whatever
thisXXX line \
    has a continuation line
blahblah
a \
multipleXXX \
continuation \
line

What should I execute to get (not necessarily in one line, not necessarily removing multiple spaces):

thisXXX line has a continuation line
a multipleXXX continuation line

BTW I'm using bash and fedora21, so it does not need to be POSIX-compliant (but I'll appreciate a solution if it is POSIX)

Best Answer

Another approach using perl to remove newlines that are preceded by \ and whitespace:

$ perl -pe 's/\\\n/ /' file | grep XXX
thisXXX line      has a continuation line
a  multipleXXX  continuation  line

To remove extra spaces, pass it through sed:

$ perl -pe 's/\\\n/ /' file | grep XXX | sed 's/  */ /g'
thisXXX line has a continuation line
a multipleXXX continuation line

Related Solutions

How to strip multiple spaces to one using sed

The use of grep is redundant, sed can do the same. The problem is in the use of * that match also 0 spaces, you have to use \+ instead:

iostat | sed -n '/hdisk1/s/ \+/ /gp'

If your sed do not supports \+ metachar, then do

iostat | sed -n '/hdisk1/s/  */ /gp'

awk – Multiline Regexp with Grep, Sed, Awk, and Perl

You can do this with Awk by setting the "Record Separator" variable to be a regex matching at least two consecutive newline characters:

awk -v RS='\n\n+' '/1.*2.*3/' file.txt

You can also set the "Field Separator" to be a single newline character:

awk -v RS='\n\n+' -F '\n' '$1 == "LINE OF TEXT 1" && $2 == "LINE OF TEXT 2" && $3 == "LINE OF TEXT 3"' file.txt

Broken up for readability:

awk -v RS='\n\n+' -F '\n' '
  $1 == "LINE OF TEXT 1" &&
  $2 == "LINE OF TEXT 2" &&
  $3 == "LINE OF TEXT 3"
' file.txt

With your requirement of only printing the filename if a match is found, you can do this like so:

awk -v RS='\n\n+' -F '\n' '
  $1 == "LINE OF TEXT 1" &&
  $2 == "LINE OF TEXT 2" &&
  $3 == "LINE OF TEXT 3" {
    match++
  }
  END {
    if (match) {
      print FILENAME
    }
' file.txt

But considering you are talking about using find in combination with awk, I'd recommend just using Awk for the exit status and using find for the printing:

find . -type f -exec awk -v RS='\n\n+' -F '\n' '
  $1 ~ /LINE OF TEXT 1/ &&
  $2 ~ /LINE OF TEXT 2/ &&
  $3 ~ /LINE OF TEXT 3/ {
    exit 0
  }
  END { exit 1 }
' {} \; -print

That way, if you want to do something else before printing (some other find primary), you're already set up to do so.

Best Answer

Related Solutions

How to strip multiple spaces to one using sed

awk – Multiline Regexp with Grep, Sed, Awk, and Perl

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