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

Question

I would like to search for a pattern in a file and prints 5 lines after finding that pattern.

I need to use awk in order to do this.

Example:

File Contents:

. . . . ####PATTERN####### #Line1 #Line2 #Line3 #Line4 #Line5 . . . 

How do I parse through a file and print only the above mentioned lines? Do I use the NR of the line which contains "PATTERN" and keep incrementing upto 5 and print each line in the process. Kindly do let me know if there is any other efficient wat to do it in Awk.

Answer 1

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 2

awk ' {      if (lines > 0) {         print;         --lines;     } }  /PATTERN/ {     lines = 5 }  ' < input 

This yields:

#Line1 #Line2 #Line3 #Line4 #Line5