Bash find line number

Question

I have this script:

find . -name "$2" -print | xargs grep --colour=auto "$1"

It searches for a $1 word in $2 matching files. How can I make it print the line number on which the word has been found.

Thanks

Answer 1

You don't need to use find and xargs here. You can use recursive grep like this:

grep -RHn --colour=auto "$1" --include='$2' .

Options:

-n  # for printing line numbers
-R  # for recursive grep
-H  # for printing file names

Answer 2

From man grep :

-n, --line-number
     Prefix each line of output with the 1-based line number within its input file.  (-n is specified by POSIX.)

Answer 3

To get only the line numbers, you could use

grep -n 'regex' | sed 's/^\([0-9]\+\):.*$/\1/'

Or you could simply use sed:

sed -n '/regex/=' file