Print selection of line based on line number

Shell Programming and Scripting
1

Hi Unix gurus

Basically i am searching for the pattern and getting the line numbers of the grepped pattern. I am trying to print the series of lines from 7 lines before the grepped line number to the grepped line number.

I am trying to use the following code. but it is not working.

cat filename | grep -n pattern | sed 's/:/ /g' | awk 'NR==$1-7,NR==$1'

Please help

2

Use ggrep for this, ggrep can be found under samba:

/usr/sfw/bin/ggrep "PATTERN" -A 0 -B 7 yourfile

Besides, you can print line number of 4th occurance of a pattern "mohan":

nawk '$0~/mohan/{i++} i==4{print NR;exit}' yourfile

Using these knowledge you can try to solve your issue,

Regards

3

Hi thanks for your reply

Both ggrep and nwak is not available in the system.

Please any other way using awk.

4

Try

$ grep -n pattern file | grep -B 6 .