I have a command:
$ awk '{ print length($0); }' /etc/passwd It prints number of characters of every line in a passwd file:
52 52 61 48 81 58 etc. How can I print the number of characters for only the first n lines?
For example - for the first 3 lines it would give something like:
52 52 61 3 Answers
Tell awk to quit when enough lines have been read:
awk '$0 = length; NR==3 { exit }' /etc/passwd Note that this solution ignores empty lines, although not for the line count.
3A direct Awk version (not so efficient as @Thor's), but slightly more clear:
awk 'NR <= 3 {print length}' /etc/passwd You can execute it with awk only command, as nicely described by @Thor, and @JJoao (+1 from me)
You can combine awk and head with parameter -n follows by the number of lines as described below:
Thanks for @Maerlyn suggestion to execute in this order: head | awk
e.g. You will get the first 3 lines using:
head -n3 /etc/passwd | awk '{ print length($0); }' -n, --lines=[-]K print the first K lines instead of the first 10; with the leading '-', print all but the last K lines of each file 
