How do I reformat the output of a BASH script?-CodePudding

I am trying to find specific files in a directory that contain a string.

Code I've written so far:

for x in $(find "$1" -type f -name "*."$2"") ;
do  
     grep -Hrnw $x -e "$3"
done

The output I get:

./crop.py:2:import torch
./crop.py:3:import torch.backends.cudnn as cudnn

I am trying to get spaces on both sides of the colon like this:

./crop.py : 2 : import torch
./crop.py : 3 : import torch.backends.cudnn as cudnn

I am fairly new to programing in BASH. I've tried using sed command but had not luck with it.

CodePudding user response：

I am trying to get spaces on both sides of the colon

I've tried using sed command but had not luck with it.

sed 's/:/ : /g'

CodePudding user response：

Why are you using a for-loop for browsing through the find results, while you can use a find ... -exec?

Like this:

find "$1" -type f -name "*."$2"" -exec grep -Hrnw {} -e "$3" \;

(I didn't test this, it might contain some bugs)

CodePudding user response：

Suggesting to try one liner command:

 grep -Hrnw "$3" $(find "$1" -type f -name "*.$2") | sed 's/:/ : /g'