Grep a directory and return list with line numbers

awkfindgrepxargs

I'm currently trying to learn more about bash scripting and all of that fun stuff, and I pieced together this little command:

find $path | xargs grep -n $pattern | awk '{print $1}'

While this DOES work, I was wondering if I was reinventing the wheel. Is there a better way to search through a directory, grep the files for a pattern, and return a list with line numbers?

Best Answer

Many grep variants implement a recursive option. E.g., GNU grep

-R, -r, --recursive
          Read all files under each directory, recursively; this is equivalent to the -d recurse option.

You can then remove find:

grep -n -r $pattern $path | awk '{ print $1 }'

but this keeps more than the line number. awk is printing the first column. This example

src/main/package/A.java:3:import java.util.Map;
src/main/package/A.java:5:import javax.security.auth.Subject;
src/main/package/A.java:6:import javax.security.auth.callback.CallbackHandler;

will be printed as

src/main/package/A.java:3:import
src/main/package/A.java:5:import
src/main/package/A.java:6:import

notice the :import in each line. You might want to use sed to filter the output.

Since a : could be present in the file name you can use the -Z option of grep to output a nul character (\0) after the file name.

grep -rZn $pattern $path | sed -e "s/[[:cntrl:]]\([0-9][0-9]*\).*/:\1/"

with the same example as before will produce

src/main/package/A.java:3
src/main/package/A.java:5
src/main/package/A.java:6
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