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up vote 1 down vote favorite

I have a script that uses diff -c then puts the output on a text file. What I want is to remove the line that does not have the "!" and display the lines with the exclamation mark. Is this possible? Can the cut command do the trick?

I wanted to use diff -c because it separates the files from directory1 to directory2.

example:

*** 1,3 ****
! 3856715355 /home/dir
  4294967277 /home/dir/file1 <---remove this line
! 154272340 /home/dir/file5
--- 1,4 ----
! 1765342654 /home/dir
  4294967277 /home/dir/file1 <--- remove this line
! 803775803 /home/dir/file4
! 2580902204 /home/dir/file99
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awk '/^!/' <filler> –  HalosGhost Jan 12 at 19:01

2 Answers 2

up vote 1 down vote accepted

  • with grep:

    diff -c file1 file2 | grep '^[-!*]'`

  • with sed:

    diff -c file1 file2 | sed '/^[-!*]/!d'
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but then it also removes the separation from dir1 to dir2 –  midnight27 Jan 12 at 18:49
    
@midnight27 ... but then you need to put more input in your question. –  peterph Jan 12 at 18:53
    
@midnight27 see the update –  jimmij Jan 12 at 18:54
up vote 1 down vote

With grep:

diff -c file1 file2 | grep -v '^  '

none of the other lines start with two spaces: not the ones starting with !, and not the line indications.

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