Is there a (POSIX command line) way to print all of a file EXCEPT the last n lines? Use case being, I will have multiple files of unknown size, all of which contain a boilerplate footer of a known size, which I want to remove. I was wondering if there is already a utility that does this before writing it myself.
9 Answers
cat <filename> | head -n -10 # Everything except last 10 lines of a file
cat <filename> | tail -n +10 # Everything except 1st 10 lines of a file
tac file.txt | tail +[n+1] | tac
This answer is similar to user9645's, but it avoids the tail -r command, which is also not a valid option many systems. See, e.g., https://ubuntuforums.org/showthread.php?t=1346596&s=4246c451162feff4e519ef2f5cb1a45f&p=8444785#post8444785 for an example.
Note that the +1 (in the brackets) was needed on the system I tried it on to test, but it may not be required on your system. So, to remove the last line, I had to put 2 in the brackets. This is probably related to the fact that you need to have the last line ending with regular line feed character(s). This, arguably, makes the last line a blank line. If you don't do that, then the tac command will combine the last two lines, so removing the "last" line (or the first to the tail command) will actually remove the last two lines.
My answer should also be the fastest solution of those listed to date for systems lacking the improved version of head. So, I think it is both the most robust and the fastest of all the answers listed.
head -n $((`(wc -l < Windows_Terminal.json)`)) Windows_Terminal.json
This will work on Linux and on MacOs, keep in mind Mac does not support a negative value. so This is quite handy.
n.b : replace Windows_Terminal.json with your file name
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