Shell script print out file names when given a folder [duplicate]

Question

I am writing a shell script.. given a a folder as a command line arg it will print out the names of the files/folders in it

#!/bin/sh
folder="$1"
for f in "$folder"
do
  echo "$f"
done

This only prints out some the first file/folder of the given folder.

Answer 1

You need shell globbing operator * after $folder separated by the directory separator /:

for f in "$folder"/*
do
  echo "$f"
done

* would expand to all files inside $folder.

As an aside, i see you have used sh (not bash) as the script interpreter; sh is not bash in all systems (e.g. in Ubuntu), so if you were to use bash (does not make any difference in this case though), explicitly use bash in the Shebang (e.g. #!/usr/bin/env bash).

Answer 2

You can also use a find command:

folder="$1"
find "$folder"

As @Anubis pointed out in comments, the find command is recursive by default: it will also search for files/folders into subdirectories starting from target folder. You can restrict results to the current directory with the -mindepth and -mindepth options:

find "$folder" -mindepth 1 -maxdepth 1

Note that the target directory will be listed too if you omit -mindepth 1