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Shell Programming and Scripting

Extract 10th line in the file

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# 8  
Old 05-25-2010
Quote:
Originally Posted by Sekar1
Thanks for your reply,

But i received below error message

Code:
grep -A10 -B10 "A00001" source1.txt |  sed '2,20d'
grep: illegal option -- A
grep: illegal option -- 1
grep: illegal option -- 0
grep: illegal option -- B
grep: illegal option -- 1
grep: illegal option -- 0
Usage: grep -hblcnsviw pattern file . . .

Please correct me if i am wrong in giving the command
nothing wrong in the way given, but only gnu grep supports these options.
# 9  
Old 05-25-2010
Oh.... Thanks for your response. With your help, i framed the below code. There are two files. "file" contains the list of words to be searched in "file1". from file1, i need to extract 2nd line above the matched string and 10th line below the matched string. The below code is not working. Please correct my mistake and update me.
Thanks in advance

Code:
targetfile="/apps/file.txt"
targetfile1="/apps/file1.txt"
targetfile3="/apps/finallist.txt"
totalrecs=`sed -n $= $targetfile`
tot=1
cat $targetfile1 | while read c_line
do
ARRAY14[$tot]=`echo "$c_line"`
tot=$i+1
done
fet=1
while [ $fet -le $tot ]
do
l=`echo $(grep -n $ARRAY14[$fet] "$targetfile1" | cut -d":" -f1)`
i-=2
j+=10
sed -n "$j"p filename >> targetfile3
sed -n "$i"p filename >> targetfile3
fet=$fet+1
done
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