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

Question

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 ] 

Answer 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.

Answer 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