Question

I read this post, but I cannot figure out how to make what I need.
I have a variable, which is a path, so I would like to have the name of the file (last column if I separate by /).

So I tried several combinations, such as:

#!/bin/sh

source=$1
target=$2


for i in "$source"/*
do
    $name = awk -F/ -v '{ print $NF }' $i
    echo $name
done

But no success, could anybody help me?? Thanks in advance

Answer 1

you can just use the shell to extract the file names without external tools

for file in $source/*
do
  echo ${file##*/}
done