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 ]

Solution 1:

You could use the default-value syntax:

somecommand ${1:-foo}

The above will, as described in Bash Reference Manual - 3.5.3 Shell Parameter Expansion [emphasis mine]:

If parameter is unset or null, the expansion of word is substituted. Otherwise, the value of parameter is substituted.

If you only want to substitute a default value if the parameter is unset (but not if it's null, e.g. not if it's an empty string), use this syntax instead:

somecommand ${1-foo}

Again from Bash Reference Manual - 3.5.3 Shell Parameter Expansion:

Omitting the colon results in a test only for a parameter that is unset. Put another way, if the colon is included, the operator tests for both parameter’s existence and that its value is not null; if the colon is omitted, the operator tests only for existence.

Solution 2:

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

somecommand.sh

#!/usr/bin/env bash

ARG1=${1:-foo}
ARG2=${2:-bar}
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
1
Thu Mar 29 10:03:20 ADT 2018

$ ./somecommand.sh ez
ez
bar
1
Thu Mar 29 10:03:40 ADT 2018

$ ./somecommand.sh able was i
able
was
i
Thu Mar 29 10:03:54 ADT 2018

$ ./somecommand.sh "able was i"
able was i
bar
1
Thu Mar 29 10:04:01 ADT 2018

$ ./somecommand.sh "able was i" super
able was i
super
1
Thu Mar 29 10:04:10 ADT 2018

$ ./somecommand.sh "" "super duper"
foo
super duper
1
Thu Mar 29 10:05:04 ADT 2018

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

Solution 3:

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