Shell script in /usr/local/bin not working unless explicitly sourced in terminal before use

Question

I have a shell script in /usr/local/bin

#!/bin/sh

extract() {
 if [ -z "$1" ]; then
    # display usage if no parameters given
    echo "Usage: extract <path/file_name>.<zip|rar|bz2|gz|tar|tbz2|tgz|Z|7z|xz|ex|tar.bz2|tar.gz|tar.xz>"
 elif [ -f "$1" ]; then
        case $1 in
          *.tar.bz2)   tar xvjf "$1"    ;;
          *.tar.gz)    tar xvzf "$1"    ;;
          *.tar.xz)    tar xvJf "$1"    ;;
          *.lzma)      unlzma "$1"      ;;
          *.bz2)       bunzip2 "$1"     ;;
          *.rar)       unrar x "$1"     ;;
          *.gz)        gunzip "$1"      ;;
          *.tar)       tar xvf "$1"     ;;
          *.tbz2)      tar xvjf "$1"    ;;
          *.tgz)       tar xvzf "$1"    ;;
          *.zip)       unzip "$1"       ;;
          *.Z)         uncompress "$1"  ;;
          *.7z)        7z x "$1"        ;;
          *.xz)        unxz "$1"        ;;
          *.exe)       cabextract "$1"  ;;
          *)           echo "extract: '$1' - unknown archive method" ;;
        esac
        echo "extraction successful: $1"
    else
        echo "$1 - file does not exist"
fi
}

However, it doesn't run properly unless I do source /usr/local/bin/extract in the current terminal. I know /usr/local/bin is in my path.

I have tried to change owner of the script to root:root to see if that makes a difference. File permissions are 755. I have tried to source /usr/local/bin again in my .zshenv but it was already in $PATH anyway. The only way I can make it work is if I use source /usr/local/bin/extract directly. I have tried rebooting after various changes like this. I checked the file for bashisms in shellcheck. I tried with different shebangs, #!/usr/bin/env bash and #!/bin/sh. my /bin/sh is linked to dash.

I expected any script or binary in /usr/local/bin to work without sourcing it for each terminal sessions.

Not sure why the question was downvoted. I provided a legitimate question after searching for answers. I also provided the solution to the problem after experimenting for a while. Furthermore, I provided the problem, what I tried, and what I expected. It was a simple mistake, but I feel like it could happen to anyone who hasn't already experienced such an issue or wasn't previously educated on how to handle this properly. — Brandon Johnson, Jul 29 '23 at 07:06
If this is really is the **whole** script, it does not do anything, and I wonder why it ended up in `/usr/local/bin`. It definitely shouldn't be there. I suggest that you search the culprit who placed it in this directory and kindly ask him to leave the company. Basically, the script just defines a function, but that's all. So when you call it as a child process, there is no observable behaviour (except eating some CPU cyles). If you **source** the script, it still doesn't do anything, but at least it places the function `extract` into your shell process, so you can call `extract` afterwards. — user1934428, Jul 31 '23 at 07:48
_Not sure why the question was downvoted._ I don't know for sure, but you tagged the question as _bash_ and _zsh_, but don't explain how it is related to those two languages. More importantly, you just said "it does not run properly", but you don't explain what you would expect from a "proper" execution. — user1934428, Jul 31 '23 at 07:50

Brandon Johnson · Answer 1 · 2023-07-29T06:29:56.747

I realized the problem shortly after posting this. All the code was inside of a function and it was never called. It worked when I sourced the file because the function came into the local scope of my current shell. I deleted the function wrapper and put it as raw code in the shell script e.g.

if [ -z "$1" ]; then
    # display usage if no parameters given
    echo "Usage: extract <path/file_name>.<zip|rar|bz2|gz|tar|tbz2|tgz|Z|7z|xz|ex|tar.bz2|tar.gz|tar.xz>"
 elif [ -f "$1" ]; then
        case $1 in
          *.tar.bz2)   tar xvjf "$1"    ;;
          *.tar.gz)    tar xvzf "$1"    ;;
          *.tar.xz)    tar xvJf "$1"    ;;
          *.lzma)      unlzma "$1"      ;;
          *.bz2)       bunzip2 "$1"     ;;
          *.rar)       unrar x "$1"     ;;
          *.gz)        gunzip "$1"      ;;
          *.tar)       tar xvf "$1"     ;;
          *.tbz2)      tar xvjf "$1"    ;;
          *.tgz)       tar xvzf "$1"    ;;
          *.zip)       unzip "$1"       ;;
          *.Z)         uncompress "$1"  ;;
          *.7z)        7z x "$1"        ;;
          *.xz)        unxz "$1"        ;;
          *.exe)       cabextract "$1"  ;;
          *)           echo "extract: '$1' - unknown archive method" ;;
        esac
        echo "extraction successful: $1"
    else
        echo "$1 - file does not exist"
fi

and now it works fine. I would like to know if there is a way to keep this as a function? I think I'll test to see if I can just call extract() at the end of the script.

EDIT:

Thanks to Gordon's comment, I figured out how to keep it a function and call it properly. I should not have named the function the same name as the script, and I learned how to call the function properly and pass the arguments from the script call to the function.

#!/bin/sh
extractfunc() {
if [ -z "$1" ]; then
   # display usage if no parameters given
   echo "Usage: extract <path/file_name>.<zip|rar|bz2|gz|tar|tbz2|tgz|Z|7z|xz|ex|tar.bz2|tar.gz|tar.xz>"
elif [ -f "$1" ]; then
       case $1 in
         *.tar.bz2)   tar xvjf "$1"    ;;
         *.tar.gz)    tar xvzf "$1"    ;;
         *.tar.xz)    tar xvJf "$1"    ;;
         *.lzma)      unlzma "$1"      ;;
         *.bz2)       bunzip2 "$1"     ;;
         *.rar)       unrar x "$1"     ;;
         *.gz)        gunzip "$1"      ;;
         *.tar)       tar xvf "$1"     ;;
         *.tbz2)      tar xvjf "$1"    ;;
         *.tgz)       tar xvzf "$1"    ;;
         *.zip)       unzip "$1"       ;;
         *.Z)         uncompress "$1"  ;;
         *.7z)        7z x "$1"        ;;
         *.xz)        unxz "$1"        ;;
         *.exe)       cabextract "$1"  ;;
         *)           echo "extract: '$1' - unknown archive method" ;;
       esac
       echo "extraction successful: $1"
   else
       echo "$1 - file does not exist"
fi
}
extractfunc "$@"

The differences are that I renamed the function to extractfunc() and called it with extractfunc "$@" so all arguments when calling the script will be passed to the function call.

Don't give the function the same name as the script itself, that's just asking for more confusion. If you do want to function-ize it within the script, you'd put something like `extractfunc "$@"` to pass on the script's arguments (see [this](https://stackoverflow.com/questions/4824590/propagate-all-arguments-in-a-bash-shell-script) or [this](https://stackoverflow.com/questions/3811345/how-to-pass-all-arguments-passed-to-my-bash-script-to-a-function-of-mine)). — Gordon Davisson, Jul 29 '23 at 05:23
@GordonDavisson okay thank you, I will read those! I was playing with this for a while and couldn't figure it out, and couldn't find anything about it. — Brandon Johnson, Jul 29 '23 at 05:25
It worked! I renamed the function to extractfunc() and added `extractfunc "$@"` to the end. Thank you so much! — Brandon Johnson, Jul 29 '23 at 05:34
In fact, don't go sticking random scripts into `/usr/local/bin`. You want this to be a function, which is in your `~/.bashrc` or whatever profile file. — Kaz, Jul 29 '23 at 05:52
@Kaz I put it in /usr/local/bin so it would be available to anyone on the PC, not just myself — Brandon Johnson, Jul 29 '23 at 05:54

Shell script in /usr/local/bin not working unless explicitly sourced in terminal before use

1 Answers1