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I want to see line number n and some lines before ...

I have a log file with command invocations and output. The interesting result is on line number $lno. I want to see this line and the preceding $lcount lines. How can I do this with head and tail?



head -$lno $logfile | tail -$lcount

This extracts the first $lno lines from $logfile and show the last $lcount lines of this.

Simple - isn't it?


It could still be simpler!

To get the line number, you need something like grep -n .... In addition, the line number must be extracted with cut or some shell mechanics:

lno=$( set -- $(grep $pattern $logfile | head -1); echo $1)

(This runs in ksh or bash only; with the Bourne shell, you need plain ol' backticks and backslashes to protect the inner pair.)

That's a lot of work, only to get the number, so that the head / tail sequence works.

To simplify the whole procedure, forget the line number, and parse your file with sed:

sed <$logfile "/$pattern/q" | tail -$lcount

That's it!

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