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grep for a particular pattern and remove few lines above top and bottom of the patter

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# 15  
Old 03-03-2008
grep+sed
Hi,

Try this one, first use grep to find the line number of 'pattern', then delete as expected.

Code:
line=`grep -n pattern filename | cut -d":" -f1`
a1=`expr $line - 3`
a2=`expr $line - 1`
b1=`expr $line + 1`
b2=`expr $line + 5`
sed "${a1},${a2}d;${b1},${b2}d" filename

# 16  
Old 07-23-2008
ghostdog74,

Thanks a lot, I am sorry to update it now. Yours one worked perfect, although I dont say other suggestions did not work.
# 17  
Old 07-25-2008
line=`cat -n a | grep pat | awk '{print $1}'`
nawk -v t="$line" '{
if(NR<t-5 || NR>t+6)
print
}' file
# 18  
Old 07-25-2008
If what you really actually want is for anything between two occurrences of ?> matching your pattern to be removed, rather than static numbers of lines of text, perhaps something like this would work.

Code:
perl -0777 -pe 's%\?>.*Host: googlesindication\.cn.*.*\?>%?>%gs' 123.txt

Extending it to also cover anything between beginning of file and the first occurrence of ?> would not be very hard.
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