'How to write a bash script that takes optional input arguments?

I want my script to be able to take an optional input,

e.g. currently my script is 

#!/bin/bash
somecommand foo

but I would like it to say:

#!/bin/bash
somecommand  [ if $1 exists, $1, else, foo ]
bashargumentsparameter-passing


Solution 1:[1]

You can set a default value for a variable like so:

somecommand.sh

#!/usr/bin/env bash

ARG1=${1:-foo}
ARG2=${2:-'bar is'}
ARG3=${3:-1}
ARG4=${4:-$(date)}

echo "$ARG1"
echo "$ARG2"
echo "$ARG3"
echo "$ARG4"

Here are some examples of how this works:

$ ./somecommand.sh
foo
bar is
1
Thu 19 May 2022 06:58:52 ADT

$ ./somecommand.sh ez
ez
bar is
1
Thu 19 May 2022 06:58:52 ADT

$ ./somecommand.sh able was i
able
was
i
Thu 19 May 2022 06:58:52 ADT

$ ./somecommand.sh "able was i"
able was i
bar is
1
Thu 19 May 2022 06:58:52 ADT

$ ./somecommand.sh "able was i" super
able was i
super
1
Thu 19 May 2022 06:58:52 ADT

$ ./somecommand.sh "" "super duper"
foo
super duper
1
Thu 19 May 2022 06:58:52 ADT

$ ./somecommand.sh "" "super duper" hi you
foo
super duper
hi
you

Solution 2:[2]

if [ ! -z $1 ] 
then 
    : # $1 was given
else
    : # $1 was not given
fi

Solution 3:[3]

You can check the number of arguments with $#

#!/bin/bash
if [ $# -ge 1 ]
then
    $1
else
    foo
fi

Solution 4:[4]

please don't forget, if its variable $1 .. $n you need write to a regular variable to use the substitution

#!/bin/bash
NOW=$1
echo  ${NOW:-$(date +"%Y-%m-%d")}

Solution 5:[5]

This allows default value for optional 1st arg, and preserves multiple args.

 > cat mosh.sh
   set -- ${1:-xyz} ${@:2:$#} ; echo $*    
 > mosh.sh
   xyz
 > mosh.sh  1 2 3
   1 2 3

Solution 6:[6]

For optional multiple arguments, by analogy with the ls command which can take one or more files or by default lists everything in the current directory:

if [ $# -ge 1 ]
then
    files="$@"
else
    files=*
fi
for f in $files
do
    echo "found $f"
done

Does not work correctly for files with spaces in the path, alas. Have not figured out how to make that work yet.

Solution 7:[7]

It's possible to use variable substitution to substitute a fixed value or a command (like date) for an argument. The answers so far have focused on fixed values, but this is what I used to make date an optional argument:

~$ sh co.sh
2017-01-05

~$ sh co.sh 2017-01-04
2017-01-04

~$ cat co.sh

DAY=${1:-$(date +%F -d "yesterday")}
echo $DAY

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 sorpigal
Solution 3
Solution 4 Vadzim
Solution 5 mosh
Solution 6 Jesse Glick
Solution 7 Garren S

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