How to write a bash script that takes optional input arguments?
I want my script to be able to take an optional input,
eg currently my script is
#!/bin/bash
somecommand foo
but I would like it to say:
#!/bin/bash
somecommand [ if $1 exists, $1, else, foo ]
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, eg not if it's an empty string), use this syntax instead:
somecommand ${1-foo}
Again from Bash Reference Manual - 3.5.3 Shell Parameter Expansion:
Omitting the colon results in a test only for a parameter that is unset. Put another way, if the colon is included, the operator tests for both parameter's existence and that its value is not null; if the colon is omitted, the operator tests only for existence.
You can set a default value for a variable like so:
somecommand.sh
#!/usr/bin/env bash
ARG1=${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
if [ ! -z $1 ]
then
: # $1 was given
else
: # $1 was not given
fi