Bash script with named parameter containing equal sign - bash

Question

I am trying to pass some values to my bash script using named parameters similar to the following:
./script.sh --username='myusername' --password='superS3cret!' --domainou="OU=Groups with Space,OU=subou,DC=mydomain,DC=local"
I have the following code:
#!/bin/bash
while [ "$1" != "" ]; do
PARAM=`echo $1 | awk -vFPAT='([^=]*)|("[^"]+")' -vOFS="=" '{print $1}'`
VALUE=`echo $1 | awk -vFPAT='([^=]*)|("[^"]+")' -vOFS="=" '{print $2}'`
case $PARAM in
-u | --username)
username=$VALUE
;;
-p | --password)
password=$VALUE
;;
-ou | --domainou)
domainou=$VALUE
;;
*)
echo "ERROR: unknown parameter \"$PARAM\""
exit 1
;;
esac
shift
done
echo $username
echo "$password"
echo "$domainou"
What I get when I run my script is:
myusername
superS3cret!
OU
Now the first two lines are correct but obviously I don't want OU...
I want:
OU=Groups with Space,OU=subou,DC=mydomain,DC=local
Awk seems to be matching the = inside the quote. As best as I can tell the way to solve that is using
-vFPAT='([^=]*)|("[^"]+")' -vOFS="="
But clearly that's not working so I am just wondering if any awk gurus can help me understand what's wrong with my awk statement.
Thanks
Brad

Answer 1

You can do it like this:
#!/bin/bash
while [ $# -gt 0 ]; do
case "$1" in
-u=* | --username=*)
username="${1#*=}"
;;
-p=* | --password=*)
password="${1#*=}"
;;
-ou=* | --domainou=*)
domainou="${1#*=}"
;;
*)
printf "Error: unknown option: $1\n"
exit 1
esac
shift
done
printf "username: $username\n"
printf "password: $password\n"
printf "domainou: $domainou\n"

Answer 2

For parsing command line options that include both long and short optoins, consider using GNU getopt, which has support for long options. While it is possible to build-your-own parser replacement, using the getopt provides for more robust parsing:
Abbreviation of options (e.g., accepting --user for --username).
Checking for required/optional values
Error handling
See also: Using getopts to process long and short command line options
set $(getopt --long 'username:,password:,ou:,domain:' -o 'u:p:' -- "$0" "$#")
while [ "$#" -gt 0 ] ; do
OP=$1
shift
case "$OP" in
--) PROG=$1 ; shift ; break ;;
-u | --username) username=$1 ; shift ;;
-p | --password) password=$1 ; shift ;;
--ou | --domain) domainou=$1 ; shift ;;
esac
done
# Positional arguments are set ...

Answer 3

Below is what ultimately worked best for me.
#dash-o definitely got me pointed in the right direction but the script you provided was printing out extraneous info:
set: usage: set [-abefhkmnptuvxBCHP] [-o option-name] [--] [arg ...]
I believe the offending line was this:
set --long 'username:,password:,ou:,domain:' -o 'u:p:' -- "$0" "$#"
Here's the code that accomplished what I needed. I can't take credit for this. I stole it from here Using getopts to process long and short command line options but I never would have found that if not for dash-o so a big thank you!
#!/bin/bash
die() { echo "$*" >&2; exit 2; } # complain to STDERR and exit with error
needs_arg() { if [ -z "$OPTARG" ]; then die "No arg for --$OPT option"; fi; }
while getopts ab:c:-: OPT; do
# support long options: https://stackoverflow.com/a/28466267/519360
if [ "$OPT" = "-" ]; then # long option: reformulate OPT and OPTARG
OPT="${OPTARG%%=*}" # extract long option name
OPTARG="${OPTARG#$OPT}" # extract long option argument (may be empty)
OPTARG="${OPTARG#=}" # if long option argument, remove assigning `=`
fi
case "$OPT" in
u | username ) needs_arg; username="$OPTARG" ;;
p | password ) needs_arg; password="$OPTARG" ;;
o | domainou ) needs_arg; domainou="$OPTARG" ;;
??* ) die "Illegal option --$OPT" ;; # bad long option
\? ) exit 2 ;; # bad short option (error reported via getopts)
esac
done
shift $((OPTIND-1)) # remove parsed options and args from $# list
echo "$username"
echo "$password"
echo "$domainou"