Need one liner to search pattern and print everything expect 6 lines from where pattern match made

Shell Programming and Scripting
1

i need to search for a pattern from a big file and print everything expect the next 6 lines from where the pattern match was made.

2

Hi,

Try this one,

awk -v pat="YourPatternStr" -v cnt=6 '{a[NR]=$0;}END{for(i=1;i<=NR;i++){print a;if(a ~ pat ){i=i+cnt;}}}' file

Cheers,
Ranga:)

3

Hi ,

It works but i am trying something like this.. and it removes the original file. why so ?

ssh $i "/usr/xpg4/bin/awk -v pat="CMDS=" -v cnt=6 '{a[NR]=$0;}END{for(i=1;i<=NR;i++){print a;if(a ~ pat ){i=i+cnt;}}}'  sudofile > sudofilecopy"

when i check it empties sudofile content

4 
$ cat -n a.txt
     1	one
     2	two
     3	three
     4	four
     5	five
     6	six
     7	seven
     8	eight
     9	nine
    10	ten
    11	eleven
$ awk '/two/{for(i=0;i<=5;i++){getline}}1' a.txt
one
eight
nine
ten
eleven
1 Like
5

i would like to to include the matched pattern line as well in the output . how to achieve that

6 
$ awk '/two/{print;for(i=0;i<=5;i++){getline}}1' a.txt
one
two
eight
nine
ten
eleven
7

can you please provide some explanation of your oneliner.. i am trying to learn awk.

---------- Post updated at 10:19 AM ---------- Previous update was at 10:17 AM ----------

rangarasan .. sorry that was problem with my script that nullified the file.

---------- Post updated at 10:19 AM ---------- Previous update was at 10:19 AM ----------

rangarasan .. sorry that was problem with my script that nullified the file.

8 
$ awk '
# Run this code block whenever we see a line containing the string 'two'
/two/{
        # Print the entire unmodified line
        print;
        # Read 6 more lines.
        for(i=0;i<=5;i++){getline}
        # Print the very last of the 6, but not the 5 before.
}1' a.txt
9 
awk 'c-->0{next} $0~pat{c=6}1' pat=two infile

The awks with getline : watch what happens when you look for "nine" for example...