Bash script cannot list file in a directory. What I am doing wrong?

Question

Here is my script, as simple as possible:

#!/bin/bash

for file in /home/pi/watch-dir/*.torrent; do
   echo "$file"
   sleep 2
done

Base on this answer How to iterate over files in a directory with Bash? I believe the for-in should see /home/pi/watch-dir/*.torrent as a list.

Inside the watch-dir directory:

pi@raspberry:~/watch_dir $ ls -al
total 68
drwxrwx---  2 pi pi  4096 Sep 15 18:59 .
drwxr-xr-x 12 pi pi  4096 Sep 15 19:03 ..
-rwxrwx---  1 pi pi 19065 Sep 15 18:59 asdf.torrent
-rwxrwx---  1 pi pi 19065 Sep 15 18:59 qwer.torrent
-rwxrwx---  1 pi pi 19065 Sep 15 18:59 zxcv.torrent

And this is the only-one-line-output after run the scipt, I really don't know what I am doing wrong...

pi@raspberry:~ $ ./test.sh
/home/pi/watch-dir/*.torrent

Answer 1

You've got watch-dir in your loop and watch_dir is the folder!

- != _

Answer 2

This is obviously nothing related to Raspberry Pi but to Linux and Bash scripting.

The for-in operator operates on a space separated (line separated? You can change it with IFS=<your separator>) lists.

IFS=" "
for x in alice bob cindy d e f; do
  echo $x
done

will output

alice
bob
cindy
...

There is no globbing or thatsoever. So instead of your glob you will have to get a listing of the actual files. Something like $(find . -type f -name "*.torrent")

So you should be able to achieve your intent with

#!/bin/bash

for file in $(find /home/pi/watch-dir -type f -name "*.torrent"); do
   echo "$file"
   sleep 2
done

$(ls *.torrent) should also work if you run the script from within the same directory or change into the dirctory within your script.