Shell script with parameters? - bash

I have a shell script that takes parameters, below is the code..
Right now it will only accept parameters if passed if called like this: script.sh --mode=load (or -m=load)
Is there a way to modify this so that it can be called with or without the "=" sign, so that I can call: script.sh --mode load (or -m load)
Ideally needs to work in pure bash as I don't have access to install additional tools, etc.
for i in "$#"
do
case $i in
-m=*|--mode=*)
MODE="${i#*=}"
if [[ $MODE =~ ^(dump|load)$ ]]; then
echo "" > /dev/null
else
bark "Invalid --mode set, set this to dump or load.";
exit 1
fi
;;
-p=*|--db-path=*)
DBPATH="${i#*=}"
;;
-d=*|--dump-dir=*)
DUMPDIR="${i#*=}"
;;
-l=*|--list-file=*)
TABLES="${i#*=}"
# check if file exists on disk
if [ -e $TABLES ]
then
echo "" >> /dev/null
else
bark "Table file not found!";
exit 1
fi
;;
-t=*|--tenant-name=*)
TENANT="${i#*=}"
# check if tenant is correct
if [[ $TENANT =~ ^($TENANT_LIST)$ ]]; then
echo "" >> /dev/null
else
bark "Tenant name does not match, aborting.";
exit 1
fi
;;
-s|--shared)
SHARED=YES
;;
*) usage # unknown option
;;
esac
done
My bash version:
bash --version
GNU bash, version 4.3.22(1)-release (powerpc-ibm-aix5.1.0.0)

Loop on $#. When $1 is "-m", do a shift. So in the next loop $1 will now be the argument to the -m option.
script.sh --mode load
# FIRST LOOP
$# is "--mode load"
$1 is "--mode"
shift
# SECOND LOOP
$# is "load"
$1 is "load"
This is also useful if you can specify many arguments instead of just one like you have right now. Error checking should be done to validate your argument values, and if a user did script.sh --mode with no other argument.

Don't reinvent the wheel.
If you're OK with just 1 character options, use the bash builtin getopts
#!/bin/bash
while getopts :m:p:d:l:t:s opt; do
case $opt in
m) mode=$OPTARG ;;
p) dbpath=$OPTARG ;;
d) dumpdir=$OPTARG ;;
l) tables=$OPTARG
# test file existence
;;
t) tenant=$OPTARG
# test tenant
;;
s) shared=YES ;;
:) echo "Missing argument for option -$OPTARG" >&2
exit 2
;;
*) echo "Invalid option -$OPTARG" >&2
exit 2
;;
esac
done
shift $((OPTIND - 1))
cat << SHOW_VARS
I have:
mode=$mode
dbpath=$dbpath
dumpdir=$dumpdir
tables=$tables
tenant=$tenant
shared=$shared
rest of args=$*
SHOW_VARS
Otherwise, you may be able to use the external getopt program to help parse your args. I don't have an AIX box to test on, so YMMV
tempargs=$(
getopt \
-o m:d:l:t:s \
--long mode:,db-path:,dump-dir:,list-file:,tenant-name:,shared \
-- "$#"
)
if [[ $? -ne 0 ]]; then echo "Error..." >&2; exit 2; fi
eval set -- "$tempargs"
while true; do
case $1 in
-m|--mode) mode=$2; shift 2;;
-p|--db-path) dbpath=$2; shift 2;;
-d|--dump-dir) dumpdir=$2; shift 2;;
-l|--list-file) tables=$2
# test file existence
shift 2
;;
-t|--tenant-name) tenant=$2
# test tenant
shift 2
;;
-s|--shared) shared=YES; shift;;
--) shift; break ;;
*) echo "Error..." >&2; exit 2 ;;
esac
done

Related

Parsing optional and not optional arguments

I am new with bash and after reading and trying a lot about how to parse arguments I cannot what I really want to do I want to parse optional and not optional arguments. More specifically I want to parse 3 arguments, first (a fastaq file) second (a second optional fastaq file) a third argument that will be a directory.
my_script.sh -f1 file1.fasta --f2 file2.fasta -d/home/folder1/folder2
or
my_script.sh -f1 file1.fasta -d /home/folder1/folder2
I have tried to do this in many ways but I dont know how to let the program identifies when there are two fasta files and a directory and, when there is only one fasta file and a directory.
With this arguments I want to save them in variables because they will be used later by third parties.
I have tried this:
for i in "$#"; do
case $i in
-f1=|-fasta1=)
FASTA1="${i#=}"
shift # past argument=value
;;
-d) DIRECTORY=$2
shift 2
;;
-d=|-directory=) DIRECTORY="${i#=}"
shift # past argument=value
;;
--f2=|-fasta2=) FASTA2="${i#*=}"
shift # past argument=value
;;
*)
;;
esac
done
But I just got this
scripts_my_first_NGS]$ ./run.sh -f1 fasta.fasta -d /home/folder1
FASTA1 =
DIRECTORY =
FASTA2 =
Never parse command line options on your own!
Instead either use the Bash function getopts, if you do not need GNU style long options or use use the GNU program getopt otherwise.
The following examples uses an array for FASTA. FASTA1 is ${FASTA[0]} and FASTA2 is ${FASTA[1]}. In case of getopts this makes it possible to use just one option character (-f) multiple times.
Using getopts with only one-character options:
#! /bin/bash
FASTA=()
DIRECTORY=
while getopts 'f:d:' option; do
case "$option" in
f)
FASTA+=("$OPTARG")
;;
d)
DIRECTORY="$OPTARG"
;;
*)
printf 'ERROR: Invalid argument\n' >&2
exit 1
;;
esac
done
shift $((OPTIND-1))
if [[ -z ${FASTA[0]} ]]; then
printf 'ERROR: FASTA1 missing\n' >&2
exit 1
fi
if [[ -z $DIRECTORY ]]; then
printf 'ERROR: DIRECTORY missing\n' >&2
exit 1
fi
printf 'FASTA1 = %s\n' "${FASTA[0]}"
printf 'FASTA2 = %s\n' "${FASTA[1]}"
printf 'DIRECTORY = %s\n' "$DIRECTORY"
Usage:
run -f file1.fasta -f file2.fasta -d /home/folder1/folder2
Using getopt with one-character and GNU style long options mixed:
#! /bin/bash
FASTA=()
DIRECTORY=
options=$(getopt -o d: -l f1: -l f2: -- "$#") || {
printf 'ERROR: Invalid argument\n' >&2
exit 1
}
eval set -- "$options"
while true; do
case "$1" in
--f1)
FASTA[0]="$2"
shift 2;;
--f2)
FASTA[1]="$2"
shift 2;;
-d)
DIRECTORY="$2"
shift 2;;
--)
shift
break;;
*)
break;;
esac
done
if [[ -z ${FASTA[0]} ]]; then
printf 'ERROR: FASTA1 missing\n' >&2
exit 1
fi
if [[ -z $DIRECTORY ]]; then
printf 'ERROR: DIRECTORY missing\n' >&2
exit 1
fi
printf 'FASTA1 = %s\n' "${FASTA[0]}"
printf 'FASTA2 = %s\n' "${FASTA[1]}"
printf 'DIRECTORY = %s\n' "$DIRECTORY"
Usage:
run --f1 file1.fasta --f2 file2.fasta -d /home/folder1/folder2
Basically you need to add a separate parser for versions of the options where they aren't used with the equal sign.
Also your shift commands are useless since you're processing a for loop. So convert it to to a while [[ $# -gt 0 ]]; do loop instead.
I also added a few modifications which I suggest be added.
while [[ $# -gt 0 ]]; do
case $1 in
-f1|-fasta1)
FASTA1=$2
shift
;;
-f1=*|-fasta1=*)
FASTA1=${1#*=}
;;
-d|-directory)
DIRECTORY=$2
shift
;;
-d=*|-directory=*)
DIRECTORY=${1#*=}
;;
-f2|fasta2)
FASTA2=$2
shift
;;
-f2=*|-fasta2=*)
FASTA2=${1#*=}
;;
-*)
echo "Invalid option: $1" >&2
exit 1
;;
--)
# Do FILES+=("${#:2}") maybe
break
;;
*)
# TODO
# Do FILES+=("$1") maybe
;;
esac
shift
done
The "parser" for the with-equal and non-with-equal versions of the options can also be unified by
using a helper function:
function get_opt_arg {
if [[ $1 == *=* ]]; then
__=${1#*=}
return 1
elif [[ ${2+.} ]]; then
__=$2
return 0 # Tells that shift is needed
else
echo "No argument provided to option '$1'." >&2
exit 1
fi
}
while [[ $# -gt 0 ]]; do
case $1 in
-d|-directory|-d=*|-directory=*)
get_opt_arg "$#" && shift
DIRECTORY=$__
;;
-f1|-fasta1|-f1=*|-fasta1=*)
get_opt_arg "$#" && shift
FASTA1=$__
;;
-f2|fasta2|-f2=*|-fasta2=*)
get_opt_arg "$#" && shift
FASTA2=$__
;;
-*)
echo "Invalid option: $1" >&2
exit 1
;;
--)
# Do FILES+=("${#:2}") maybe
break
;;
*)
# TODO
# Do FILES+=("$1") maybe
;;
esac
shift
done
Update
I found a complete solution to command-line parsing without relying on getopt[s] and it does it even more consistentlty: https://konsolebox.io/blog/2022/05/14/general-command-line-parsing-solution-without-using-getopt-s.html

Is there a way in bash script to have an option to give an argument but it shouldn't a must?

I have a scenario where i would like to assign an option a default value but a user can decide to give it another argument:
Here is an example
check_param() {
for arg in "$#"; do
shift
case "$arg" in
"--force") set -- "$#" "-f" ;;
"--type") set -- "$#" "-t" ;;
"--help") set -- "$#" "-h" ;;
"--"*) echo "Unknown parameter: " $arg; show_help; exit 1 ;;
*) set -- "$#" "$arg"
esac
done
# Standard Variables
force=0
type="daily"
OPTIND=1
while getopts "hft:v" opt
do
case "$opt" in
"f") force=1 ;;
"t") type=${OPTARG} ;;
"h") show_help; exit 0 ;;
"?") show_help; exit 1 ;;
esac
done
shift $(expr $OPTIND - 1) # remove options from positional parameters
From the above example, i would like when the user gives the parameter -t without any argument to apply the default value which is daily , and the user can also use parameter -t with any other argument and that will be checked later in code.
The problem is now the parameter -t must be given an argument due to the colon, but i kinda need for it to do both, with or without argument.
Thanks in advance for any explanations or links to any article that can help.
So according to a suggestion i got Here is the test result
check_param() {
## Standard Variablen der Parameter
force=0
type="daily.0"
## Break down the options in command lines for easy parsing
## -l is to accept the long options too
args=$(getopt -o hft::v -l force,type::,help -- "$#")
eval set -- "$args"
## Debugging mechanism
echo ${args}
echo "Number of parameters $#"
echo "first parameter $1"
echo "Second parameter $2"
echo "third parameter $3"
while (($#)); do
case "$1" in
-f|--force) force=1; ;;
-t|--type) type="${2:-${type}}"; shift; ;;
-h|--help) show_help; exit 0; ;;
--) shift; break; ;;
*) echo "Unbekannter Parameter"; exit 1; ;;
esac
shift
done
echo ${type}
}
check_param $#
echo ${type}
The output:
sh scriptsh -t patch.0
-t '' -- 'patch.0'
Number of parameters 4
first parameter -t
Second parameter
third parameter --
daily.0
daily.0
It still didn't assign the value patch to the variable type
Is there a way in bash script to have an option to give an argument but it shouldn't a must?
Yes, there is a way.
getopts does not supports optional arguments. So... you can:
roll your own bash library for parsing arguments or
use another tool that has support for optional arguments.
A common tool is getopt that should be available on any linux.
args=$(getopt -o hft::v -l force,type::,help -- "$#")
eval set -- "$args"
while (($#)); do
case "$1" in
-f|--force) force=1; ;;
-t|--type) type="${2:-default_value}"; shift; ;;
-h|--help) echo "THis is help"; exit; ;;
--) shift; break; ;;
*) echo "Error parsgin arguments"; exit 1; ;;
esac
shift
done
getopt handles long arguments and reorders arguments, so you can ./prog file1 -t opt and ./prog -t opt file1 with same result.

How can I accept long arguments using getopts in Bash?

I'm trying to have my getops function run with multiple flags and arguments but instead of short (-f style) flag, I want to accept a long one (--flag style). For example:
if [ $# -lt 1 ]; then
usage >&2
exit 1
else
while $1 "hf:" opt; do
case $opt in
h)
echo "Here is the help menu:"
usage
;;
f)
ls -l $OPTARG >&2
;;
\?)
echo "Invalid option: -$OPTARG" >&2
;;
:)
echo "Option -$OPTARG requires an argument" >&2
exit 1
;;
esac
done
fi
I would like the -h and -f to be --help and --file respectively.
How do I do that?
getopt will do this for you. It handles short options, long options, options with and without arguments, -- to end option parsing, and more.
Boilerplate usage looks like:
options=$(getopt -o hf: -l help,file: -n "$0" -- "$#") || exit
eval set -- "$options"
while [[ $1 != -- ]]; do
case $1 in
-h|--help) echo "help!"; shift 1;;
-f|--file) echo "file! $2"; shift 2;;
*) echo "bad option: $1" >&2; exit 1;;
esac
done
shift
# Process non-option arguments.
for arg; do
echo "arg! $arg"
done

Why do I have error with multi long options using getopt in bash?

I just wrote a script in bash, which work expect for multi long option:
#!/bin/bash
OPTS=`getopt -q -o fdhl: -l free,df,help,log: -- "$*"`
#Check if error with getopt
if [ $? != 0 ]
then
echo -e "error: parameter could not be found\n\nUsage:\n supervision [options]\n\n Try 'supervision --help'\n or 'supervision -h'\n for additional help text." ;
exit 1
fi
eval set -- "$OPTS"
while true ; do
case "$1" in
-f|--free)
free -h ;
shift;;
-d|--df)
df -h ; #Run df system command
shift;;
-l|--log)
case "$2" in
"") echo "miss file" ;
shift 2;; #No file passed as parameter
*)
df -h >> "$2" ;
shift 2;;
esac ;;
-h|--help) #Display help
shift;;
--) #End of parsed parameters list
shift ; break ;;
*)
break ;;
esac
done
I don't get why i'm supposed to, when I use more than 1 long option, for example:
sh myscript --free --df
And when I use --log:
sh myscript --log logfile
Both case exit on the if [ $? != 0 ], seems like the element which follow the 1st long option doesn't get parsed.
Ok, I figured out and it's all due to the using of "$*" instead of "$#" in the getopt call. I don't exactly why, i guessed both do the same thing, but it turns out to be the one which causes the problem.

How to read inline parameters using command line args in bash script?

I have a bash script which takes few command line args and a filename as inline parameter. I am not able to read the inline parameter.
sh test.sh -a a -b b -c c < pwd.txt
test.sh has
if [ $# = 0 ]
then
echo $USAGE >&2
exit $STATUS_ERROR_FAIL
fi
# Parse command line options.
while getopts a:b:c: OPT;
do
case "$OPT" in
a)
a="$OPTARG"
;;
b)
b="$OPTARG"
;;
c)
c="$OPTARG"
;;
\?)
# getopts issues an error message
echo $USAGE
exit $STATUS_ERROR_FAIL
;;
esac
done
shift $((OPTIND-1))
echo "1=$1"
your script is working fine, your error is
echo "1=$1"
if you want to see your parameter you should add an echo/print in your case
#!/bin/bash
if [ $# = 0 ]
then
echo $USAGE >&2
exit $STATUS_ERROR_FAIL
fi
# Parse command line options.
while getopts a:b:c: OPT;
do
case "$OPT" in
a)
a="${OPTARG}"
echo "a[$a]"
;;
b)
b="${OPTARG}"
echo "b[$b]"
;;
c)
c="${OPTARG}"
echo "c[$c]"
;;
\?)
# getopts issues an error message
echo $USAGE
exit $STATUS_ERROR_FAIL
;;
esac
done
shift $((OPTIND-1))
or
you can add the echo/print at the end of the script.. it's depend by your needs
output
[shell] ➤ ./t -a 1 -b 2 -c 3
a[1]
b[2]
c[3]
Regards
Claudio

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