awk - How to print the number of characters for the first n lines in a file?

Question

I have a command:

$ awk '{ print length($0); }' /etc/passwd

It prints number of characters of every line in a passwd file:

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

Thor · Accepted Answer · 2017-03-28T19:00:42.343

17

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.

edited Mar 28 '17 at 19:00

answered Mar 28 '17 at 08:57

Thor

score 14 · Answer 2 · answered Mar 28 '17 at 10:30

14

A direct Awk version (not so efficient as @Thor's), but slightly more clear:

awk 'NR <= 3 {print length}' /etc/passwd

answered Mar 28 '17 at 10:30

score 6 · Answer 3 · edited Mar 29 '17 at 06:33

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

3 Answers3