How to print 5 consecutive lines after a pattern in file using awk [duplicate]

Answer 1

grep "PATTERN" search-file -A 5 will do the job for you if decide to give grep a chance.

Edit: You can call grep using system() function from inside your awk script as well.

Answer 2

Another way to do it in AWK:

awk '/PATTERN/ {for(i=1; i<=5; i++) {getline; print}}' inputfile

in sed:

sed -n '/PATTERN/{n;p;n;p;n;p;n;p;n;p}' inputfile

in GNU sed:

sed -n '/PATTERN/,+7p' inputfile

or

sed -n '1{x;s/.*/####/;x};/PATTERN/{:a;n;p;x;s/.//;ta;q}' inputfile

The # characters represent a counter. Use one fewer than the number of lines you want to output.

Answer 3

Quick&dirty solution: awk '/PATTERN/ {i=0} { if (i<=5) {print $0}; i+=1} BEGIN {i=6}'

Answer 4

As limited as Johnsyweb's solution but using a different approach and GNU awk's advanced feature allowing full RegEx record separators resulting in a piece of code that, IMHO, is very awk-ish:

awk -F\\\n '{NF=5}NR-1' RS="PATTERN[^\n]*." OFS=\\\n

So if you want a stable solution that is also readable, go along with Dennis Williamson's answer (+1 for that one) but maybe you too simply enjoy the beauty of awk when looking at lines like this.

Answer 5

awk '
{ 
    if (lines > 0) {
        print;
        --lines;
    }
}

/PATTERN/ {
    lines = 5
}

' < input

This yields:

#Line1
#Line2
#Line3
#Line4
#Line5

Answer 6

Edit: didn't notice PATTERN shouldn't be part of the output.

cat /etc/passwd | awk '{if(a-->0){print;next}} /qmaild/{a=5}'

or

cat /etc/passwd | awk ' found && NR-6 < a{print} /qmaild/{a=NR;found=1}'

The shortest I can come up with is:

cat /etc/passwd | awk 'a-->0;/qmaild/{a=5}'

Read as a tends to 0. /qmaild/ sets a to 5 :-)