Super User is a question and answer site for computer enthusiasts and power users. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a number of log files that look like this:

arbitrary number of lines
blah blah
blah blah

Using simple Bash commands (preferably sed, not awk), how would I retrieve only the header lines (ideally including the comment markers)?

I've RTFM and tried googling, also found some hints, but not enough to get me started.


share|improve this question
Another related link: That does the opposite of what I need though. – AnC Sep 6 '09 at 15:32
up vote 2 down vote accepted

If you confirm the following things, the script in this answer will work for you.

  1. Files start with the "/*" characters
  2. There may be multi-line C-Syntex comment block that is not nested
  3. The comment block ends with no additional C-Statements on the same line
sed -n '/^\/\*/,/\*\// p' file.c
#        - - -    - - 

This will match all lines from the start of the file to end of the comment block.
The second line (with a "#" in the start) highlights the match being searched for.
The "-n" at the start and the "p" at the end tell sed to print only the matching part.

share|improve this answer
Thanks, much appreciate the detailed response! – AnC Sep 6 '09 at 16:49
@AnC, I think there is a problem in my solution too. It does not stop like Richard's command after the first match! So, you'll get all the comments starting at new lines (requirement 1 is not limited to the start of the file) in a file. – nik Sep 6 '09 at 16:54
Just realized that - so I'm now combining both of your solutions: cat $LOGFILE | sed -e '/*\//q' | sed -n '/^\/*/,/*\// p' Sadly, I can't properly credit you both. – AnC Sep 6 '09 at 17:09
@AnC, you could do: sed -n '/^\/*/,/*\// p' $LOGFILE | sed -e '/*\//q' and skip the cat. But, I am not yet happy about it. – nik Sep 6 '09 at 17:32

This look at every file with an extension .log and if the first line is "/*" only, print everything until a line with "*/" only.

for file in *.log; do
      head -n1 $file | grep  -q '^/*' &&
      sed '/^\*\/$/q' $file; 
share|improve this answer
That does the job - thanks! One caveat: It won't work if there's anything above the header, or if there a multiple comment blocks. However, I shouldn't worry about that until I actually need it. – AnC Sep 6 '09 at 16:21
And if there is no header, it will print the whole file! There are a lot of failure modes. Perhaps you can ask the question "How can I print comments in a file?" (And please let us use awk.) – Richard Hoskins Sep 6 '09 at 16:28
You're right - nik's solution seems a little safer in that regard, so I'll use that for now. Thanks again! – AnC Sep 6 '09 at 16:48
I have updated my answer to get rid of most of its worst problems. – Richard Hoskins Sep 6 '09 at 19:32
Interesting approach. Now I'm torn between this and the combination of both of your solutions (see comments above)... – AnC Sep 7 '09 at 12:07
cat *.log | sed -e 's/*\/\*//g' -e 's/\*\///g' >> smt.log

That might be it

share|improve this answer
I'm afraid that only removes the "*/" line. I've experimented with similar expressions, but didn't get anywhere - I think it's because sed works on a line-by-line basis. – AnC Sep 6 '09 at 16:06

In case, if like me you stumbled across this question while trying to retrieve Doxygen style comments, the sed command is:

sed -n '/^\/\*/p; /^ \*/p' < file

For a more detailed explanation, I blogged about it.

share|improve this answer

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .