Bash Shell Script - Check for a flag and grab its value

LinuxBashShellDebian

Linux Problem Overview


I am trying to make a shell script which is designed to be run like this:

script.sh -t application

Firstly, in my script I want to check to see if the script has been run with the -t flag. For example if it has been run without the flag like this I want it to error:

script.sh

Secondly, assuming there is a -t flag, I want to grab the value and store it in a variable that I can use in my script for example like this:

FLAG="application"

So far the only progress I've been able to make on any of this is that $@ grabs all the command line arguments but I don't know how this relates to flags, or if this is even possible.

Linux Solutions


Solution 1 - Linux

You should read this getopts tutorial.

Example with -a switch that requires an argument :

#!/bin/bash
 
while getopts ":a:" opt; do
  case $opt in
    a)
      echo "-a was triggered, Parameter: $OPTARG" >&2
      ;;
    \?)
      echo "Invalid option: -$OPTARG" >&2
      exit 1
      ;;
    :)
      echo "Option -$OPTARG requires an argument." >&2
      exit 1
      ;;
  esac
done

Like greybot said(getopt != getopts) :

> The external command getopt(1) is never safe to use, unless you know > it is GNU getopt, you call it in a GNU-specific way, and you ensure > that GETOPT_COMPATIBLE is not in the environment. Use getopts (shell > builtin) instead, or simply loop over the positional parameters.

Solution 2 - Linux

Use $# to grab the number of arguments, if it is unequal to 2 there are not enough arguments provided:

if [ $# -ne 2 ]; then
   usage;
fi

Next, check if $1 equals -t, otherwise an unknown flag was used:

if [ "$1" != "-t" ]; then
  usage;
fi

Finally store $2 in FLAG:

FLAG=$2

Note: usage() is some function showing the syntax. For example:

function usage {
   cat << EOF
Usage: script.sh -t <application>

Performs some activity
EOF
   exit 1
}

Solution 3 - Linux

Here is a generalized simple command argument interface you can paste to the top of all your scripts.

#!/bin/bash

declare -A flags
declare -A booleans
args=()

while [ "$1" ];
do
    arg=$1
    if [ "${1:0:1}" == "-" ]
    then
      shift
      rev=$(echo "$arg" | rev)
      if [ -z "$1" ] || [ "${1:0:1}" == "-" ] || [ "${rev:0:1}" == ":" ]
      then
        bool=$(echo ${arg:1} | sed s/://g)
        booleans[$bool]=true
        echo \"$bool\" is boolean
      else
        value=$1
        flags[${arg:1}]=$value
        shift
        echo \"$arg\" is flag with value \"$value\"
      fi
    else
      args+=("$arg")
      shift
      echo \"$arg\" is an arg
    fi
done


echo -e "\n"
echo booleans: ${booleans[@]}
echo flags: ${flags[@]}
echo args: ${args[@]}

echo -e "\nBoolean types:\n\tPrecedes Flag(pf): ${booleans[pf]}\n\tFinal Arg(f): ${booleans[f]}\n\tColon Terminated(Ct): ${booleans[Ct]}\n\tNot Mentioned(nm): ${boolean[nm]}"
echo -e "\nFlag: myFlag => ${flags["myFlag"]}"
echo -e "\nArgs: one: ${args[0]}, two: ${args[1]}, three: ${args[2]}"

By running the command:

bashScript.sh firstArg -pf -myFlag "my flag value" secondArg -Ct: thirdArg -f

The output will be this:

"firstArg" is an arg
"pf" is boolean
"-myFlag" is flag with value "my flag value"
"secondArg" is an arg
"Ct" is boolean
"thirdArg" is an arg
"f" is boolean


booleans: true true true
flags: my flag value
args: firstArg secondArg thirdArg

Boolean types:
	Precedes Flag(pf): true
	Final Arg(f): true
	Colon Terminated(Ct): true
	Not Mentioned(nm): 

Flag: myFlag => my flag value

Args: one => firstArg, two => secondArg, three => thirdArg

Basically, the arguments are divided up into flags booleans and generic arguments. By doing it this way a user can put the flags and booleans anywhere as long as he/she keeps the generic arguments (if there are any) in the specified order.

Allowing me and now you to never deal with bash argument parsing again!

You can view an updated script here

This has been enormously useful over the last year. It can now simulate scope by prefixing the variables with a scope parameter.

Just call the script like

replace() (
  source $FUTIL_REL_DIR/commandParser.sh -scope ${FUNCNAME[0]} "$@"
  echo ${replaceFlags[f]}
  echo ${replaceBooleans[b]}
)

Doesn't look like I implemented argument scope, not sure why I guess I haven't needed it yet.

Solution 4 - Linux

Try shFlags -- Advanced command-line flag library for Unix shell scripts.

http://code.google.com/p/shflags/

It is very good and very flexible.

FLAG TYPES: This is a list of the DEFINE_*'s that you can do. All flags take a name, default value, help-string, and optional 'short' name (one-letter name). Some flags have other arguments, which are described with the flag.

DEFINE_string: takes any input, and intreprets it as a string.

DEFINE_boolean: typically does not take any argument: say --myflag to set FLAGS_myflag to true, or --nomyflag to set FLAGS_myflag to false. Alternately, you can say --myflag=true or --myflag=t or --myflag=0 or --myflag=false or --myflag=f or --myflag=1 Passing an option has the same affect as passing the option once.

DEFINE_float: takes an input and intreprets it as a floating point number. As shell does not support floats per-se, the input is merely validated as being a valid floating point value.

DEFINE_integer: takes an input and intreprets it as an integer.

SPECIAL FLAGS: There are a few flags that have special meaning: --help (or -?) prints a list of all the flags in a human-readable fashion --flagfile=foo read flags from foo. (not implemented yet) -- as in getopt(), terminates flag-processing

EXAMPLE USAGE:

-- begin hello.sh --
 ! /bin/sh
. ./shflags
DEFINE_string name 'world' "somebody's name" n
FLAGS "$@" || exit $?
eval set -- "${FLAGS_ARGV}"
echo "Hello, ${FLAGS_name}."
-- end hello.sh --

$ ./hello.sh -n Kate
Hello, Kate.

Note: I took this text from shflags documentation

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJimmyView Question on Stackoverflow
Solution 1 - LinuxGilles QuenotView Answer on Stackoverflow
Solution 2 - LinuxVegerView Answer on Stackoverflow
Solution 3 - Linuxjozsef morrisseyView Answer on Stackoverflow
Solution 4 - LinuxAnandkumarView Answer on Stackoverflow