768

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 ]
3

10 Answers 10

1094

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.

9
  • 81
    Please note the semantic difference between the above command, "return foo if $1 is unset or an empty string", and ${1-foo}, "return foo if $1 is unset".
    – l0b0
    Commented Feb 21, 2012 at 15:12
  • 17
    Can you explain why this works? Specially, what's the function/purpose of the ':' and '-'?
    – jwien001
    Commented Sep 5, 2014 at 21:11
  • 12
    @jwein001: In the answer submitted above, a substitution operator is used to return a default value if the variable is undefined. Specifically, the logic is "If $1 exists and isn't null, return its value; otherwise, return foo." The colon is optional. If it's omitted, change "exists and isn't null" to only "exists." The minus sign specifies to return foo without setting $1 equal to 'foo'. Substitution operators are a subclass of expansion operators. See section 6.1.2.1 of Robbins and Beebe's Classic Shell Scripting [O'Reilly] (shop.oreilly.com/product/9780596005955.do)
    – Jubbles
    Commented Sep 23, 2014 at 20:36
  • 6
    @Jubbles or if you don't want to buy an entire book for a simple reference... tldp.org/LDP/abs/html/parameter-substitution.html Commented Feb 8, 2017 at 14:39
  • 3
    This answer would be even better if it showed how to make the default be the result of running a command, as @hagen does (though that answer is inelegant).
    – sautedman
    Commented Mar 17, 2017 at 17:47
555

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
6
  • 8
    Ah, ok. The - confused me (is it negated?).
    – khatchad
    Commented Mar 28, 2018 at 20:58
  • 16
    Nope - that's just a weird way bash has of doing the assignment. I'll add some more examples to clarify this a bit... thanks!
    – Brad Parks
    Commented Mar 29, 2018 at 11:09
  • It looks this is the limit of bash argument. Using python terminology, this is not optional, it is positional. It is just a positional argument with default values. Is there no really optional argument in bash? Commented Feb 17, 2022 at 8:20
  • you may be able to use getopts to get what you want - Here's 2 stackoverflow answers that go into greater detail: dfeault values and inside a function
    – Brad Parks
    Commented Feb 17, 2022 at 12:47
  • Note also, that to confuse even more (with JSON/python dicts) no space is permitted anywhere, e.g. after the colon and/or after/before the dash... where is the principle of least astonishment here?:)
    – mirekphd
    Commented May 19, 2022 at 8:39
68
if [ ! -z $1 ] 
then 
    : # $1 was given
else
    : # $1 was not given
fi
5
  • 2
    Technically, if you pass in an empty string '' that might count as a parameter, but your check will miss it. In that case $# would say how many parameters were given
    – vmpstr
    Commented Feb 17, 2012 at 17:40
  • 31
    -n is the same as ! -z.
    – l0b0
    Commented Feb 21, 2012 at 15:15
  • 1
    I get different results using -n and ! -z so I would say that is not the case here.
    – Eliezer
    Commented Dec 19, 2019 at 21:31
  • 6
    because you failed to quote the variable, [ -n $1 ] will always be true. If you use bash, [[ -n $1 ]] will behave as you expect, otherwise you must quote [ -n "$1" ] Commented Mar 19, 2020 at 12:33
  • 3
    For ones like me: forget the ":" in the code, it's not required, replace it with your real commands! Commented Nov 12, 2021 at 13:52
35

You can check the number of arguments with $#

#!/bin/bash
if [ $# -ge 1 ]
then
    $1
else
    foo
fi
0
13

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")}
2
  • 2
    Brad's answer above proves that argument variables can also be substituted without intermediate vars.
    – Vadzim
    Commented Mar 28, 2016 at 10:05
  • 2
    +1 for noting the way to use a command like date as the default instead of a fixed value. This is also possible: DAY=${1:-$(date +%F -d "yesterday")}
    – Garren S
    Commented Jan 6, 2017 at 21:07
10

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 
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.

5

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 $DAY
2
while getopts a: flag
do
    case "$flag" in
        a) arg1=${OPTARG};;
    esac
done

echo "Print optional argument: $arg1"

if [[ -z "$arg1" ]]; then
    ARG=DEFAULT_VAL
else
    ARG=$arg1
fi  

#Run using below command (eg: file name : runscript.sh)
bash runscript.sh -a argument_val &  
-1

When you use ${1: } you can catch the first parameter(1) passed to your function or(:) you can catch a blank space like a default value.

For example. To be able to use Laravel artisan, I put this into my .bash_aliases file:

artisan() {
    docker exec -it **container_name** php artisan ${1: } ${2: } ${3: } ${4: } ${5: } ${6: } ${7: }
}  

and now, I can just type in command line:

  • artisan -- and all parameters(${1: } ${2: } ${3: } ${4: } ${5: } ${6: } ${7: }) will be just blank spaces
  • artisan cache:clear -- and the first parameter ( ${1: } ) will be cache:clear and all the others will be just blank spaces

So, in this case I can pass 7 parameters optionally.

I hope it can help someone.

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