Best way to parse command line args in Bash?

Question

After several days of research, I still can\'t figure out the best method for parsing cmdline args in a .sh script. According to my references the getopts cmd is the way to

Answer 1

If you want to avoid using getopt you can use this nice quick approach:

• Defining help with all options as ## comments (customise as you wish).
• Define for each option a function with same name.
• Copy the last five lines of this script to your script (the magic).

Example script: log.sh

#!/bin/sh
## $PROG 1.0 - Print logs [2017-10-01]
## Compatible with bash and dash/POSIX
## 
## Usage: $PROG [OPTION...] [COMMAND]...
## Options:
##   -i, --log-info         Set log level to info (default)
##   -q, --log-quiet        Set log level to quiet
##   -l, --log MESSAGE      Log a message
## Commands:
##   -h, --help             Displays this help and exists
##   -v, --version          Displays output version and exists
## Examples:
##   $PROG -i myscrip-simple.sh > myscript-full.sh
##   $PROG -r myscrip-full.sh   > myscript-simple.sh
PROG=${0##*/}
LOG=info
die() { echo $@ >&2; exit 2; }

log_info() {
  LOG=info
}
log_quiet() {
  LOG=quiet
}
log() {
  [ $LOG = info ] && echo "$1"; return 1 ## number of args used
}
help() {
  grep "^##" "$0" | sed -e "s/^...//" -e "s/\$PROG/$PROG/g"; exit 0
}
version() {
  help | head -1
}

[ $# = 0 ] && help
while [ $# -gt 0 ]; do
  CMD=$(grep -m 1 -Po "^## *$1, --\K[^= ]*|^##.* --\K${1#--}(?:[= ])" log.sh | sed -e "s/-/_/g")
  if [ -z "$CMD" ]; then echo "ERROR: Command '$1' not supported"; exit 1; fi
  shift; eval "$CMD" $@ || shift $? 2> /dev/null
done

Testing

Running this command:

./log.sh --log yep --log-quiet -l nop -i -l yes

Produces this output:

yep
yes

By the way: It's compatible with posix!