Linux – Quick unix command to display specific lines in the middle of a file

bashlinuxtext;unix

Trying to debug an issue with a server and my only log file is a 20GB log file (with no timestamps even! Why do people use System.out.println() as logging? In production?!)

Using grep, I've found an area of the file that I'd like to take a look at, line 347340107.

Other than doing something like

head -<$LINENUM + 10> filename | tail -20

… which would require head to read through the first 347 million lines of the log file, is there a quick and easy command that would dump lines 347340100 – 347340200 (for example) to the console?

update I totally forgot that grep can print the context around a match … this works well. Thanks!

Best Answer

I found two other solutions if you know the line number but nothing else (no grep possible):

Assuming you need lines 20 to 40,

sed -n '20,40p;41q' file_name

or

awk 'FNR>=20 && FNR<=40' file_name

When using sed it is more efficient to quit processing after having printed the last line than continue processing until the end of the file. This is especially important in the case of large files and printing lines at the beginning. In order to do so, the sed command above introduces the instruction 41q in order to stop processing after line 41 because in the example we are interested in lines 20-40 only. You will need to change the 41 to whatever the last line you are interested in is, plus one.

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