Hi all,
I am trying to zgrep / grep list of files so that it displays only the matching filename:line number and does not display the whole line, like:

(echo "1.txt";echo "2.txt") | xargs zgrep -no STRING

If I use -o option, it displays the matching STRING and if not used, displays the whole matching line. I don't want to display the line content. Just the filename and linenumber needs to be displayed.

I am also trying to play around with awk.

Any response will be highly appreciated.

Thanks in advance,
vvaidyan

I would try something like this:

#!/usr/bin/ksh
for i in `ls *.txt`
do
grep "<STRING>" $i 1>/dev/null 2>&1
if [[ $? = 0 ]]
then
echo -n "${i}: "
grep -n "<STRING>" $i | awk -F: '{printf "%s",$1" "}'
fi
done
echo

Thanks Keelba, from your program, I got it into one liner:

(echo "1.txt";echo "2.txt") | xargs zgrep -n STRING | awk -F: '{printf "%s %s", $1" -", $2 "\n"}'

Later:

xargs zgrep -n STRING | awk -F: '{printf "%s %s", $1" -", $2 "\n"}'

I placed the above line in a file: vv-grep


Then, I can run this script neatly as:

(echo "1.txt";echo "2.txt") | ./vv-grep


The final thing which I want to enhance here is only for the input string to be searched for, so that the program can be executed like:

(echo "1.txt";echo "2.txt") | ./vv-grep STRING


I am trying to find out how I can get input in command line parameter to the script.


Thanks for the great help. I was able to understand your suggestion and could mould it in the way I wanted.

It will be great if you could tell me if you know as how to get the input from comand line parameter to the script.

vvaidyan

Got it...

accessed through $0 / $1 / $2, depending upon the position of the argument.

Thanks,
vvaidyan