AWK to print every nth line from a file

问题

i want to print every Nth line of a file using AWK. i tried modifying the general format :-

awk '0 == NR % 4' results.txt

to:-

awk '0 == NR % $ct' results.txt

where 'ct' is the number of lines that should be skipped. its not working . can anyone please help me out? Thanks in advance.

回答1:

You may want to use:

awk -v patt="$ct" 'NR % patt' results.txt

Explanation

Given a file like the following:

$ cat -n a
     1  hello1
     2  hello2
     3  hello3
     4  hello4
     5  hello5
     ...
    37  hello37
    38  hello38
    39  hello39
    40  hello40

These are equivalent:

$ awk 'NR % 7 == 0' a
hello7
hello14
hello21
hello28
hello35
$ ct=7
$ awk -v patt="$ct" 'NR % patt == 0' a
hello7
hello14
hello21
hello28
hello35

Or even

$ awk -v patt="$ct" '!(NR % patt)' a

Note that the syntax NR % n == 0 means: number of line is multiple to n. If we say !(NR % patt), then this is true whenever NR % patt is false, ie, NR is multiple of patt.

Update

As you comment you are using Solaris, instead of default awk use the following:

/usr/xpg4/bin/awk

回答2:

How about this:

awk -v n="$ct" '0 == NR % n' results.txt

or a bit shorter

awk -v n="$ct" '!(NR % n)' results.txt

回答3:

This do also work, but its not a good practice.

awk '!(NR%'$ct')' results.txt

This is ok to use:

awk '!(NR%n)' n="$ct" results.txt

来源：https://stackoverflow.com/questions/18352181/awk-to-print-every-nth-line-from-a-file