samizdat-shell-help.bash

#!/bin/bash
###
### my-script — does one thing well
###
### Usage:
###   my-script <input> <output>
###
### Options:
###   <input>   Input file to read.
###   <output>  Output file to write. Use '-' for stdout.
###   -h        Show this message.

help() {
    sed -rn 's/^### ?//;T;p' "$0"
}

if [[ $# == 0 ]] || [[ "$1" == "-h" ]]; then
    help
    exit 1
fi

echo Hello World

That's a good idea. Awesome!

@brucewoodward 🙏

Glad someone proved this is a good strategy to write helper message, I also did it in my commonly-used script:
https://gist.github.com/typebrook/b0d2e7e67aa50298fdf8111ae7466b56#file-gist

what I did is simply print the comment at the top(except the first line to # --):
sed -Ene "/^#/ !q; 1,/^# --/ d; s/^# //p

@kovetskiy
@benhutchins

This doesn't actually work cross-platform, the -r option does not exist on macOS. You can do the same with awk though, just use:

-E works on BSD sed, so you can make it portable with sed -Ene 's/^### ?//;T;p' "$0" both on mac(BSD sed) or Linux(GNU sed).

666

Why even use the extended version?

sed -n 's/^###//p' achieves the same. Just keep the leading ### format to the help only.

sed -n '/^###/ s///p' this one's faster but a tiny bit more verbose.

and sed '/^###/!d' if you don't mind hashes. Or grep "^###"

My version, that doesn't require multiple # (but allows more than one if you want it for stylistic reasons):

#!/bin/bash
# Usage:
#   test_usage.sh ...

print_usage_and_exit() {
  # The sed script below:
  # - Deletes the first line (shebang).
  # - Quits on the first line that doesn't start with #.
  # - Removes the '#+ ?' prefix from the string.
  sed -En '1d; /^[^#]/ q; s/^#+ ?//p;' < "$0" >&2
  exit 1
}