Check existence of input argument in a Bash shell script

Asked 2023-09-21 08:07:54 View 747,128

I need to check the existence of an input argument. I have the following script

if [ "$1" -gt "-1" ]
  then echo hi
fi

I get

[: : integer expression expected

How do I check the input argument1 first to see if it exists?

  • You did not tell us how you invoked this script. most likely, the first parameter was not an integral number. - anyone
  • How do I check the input argument1 first to see if it exists? : Do you mean "how do I check whether the argument 1 is non-empty?" or do you mean "how do I check whether the number of arguments is greater than zero?". Please clarify this in your question. - anyone

Answers

It is:

if [ $# -eq 0 ]
  then
    echo "No arguments supplied"
fi

The $# variable will tell you the number of input arguments the script was passed.

Or you can check if an argument is an empty string or not like:

if [ -z "$1" ]
  then
    echo "No argument supplied"
fi

The -z switch will test if the expansion of "$1" is a null string or not. If it is a null string then the body is executed.

Answered   2023-09-21 08:07:54

  • I like to do it this way, in terse syntax and still POSIX acceptable. [ -z "$1" ] && echo "No argument supplied" I prefer one-liners, as they are easier for me; and it's also faster to check exit value, compared to using if - anyone
  • You probably want to add an exit 1 at the end of your echos inside the if block when the argument is required for the script to function. Obvious, but worth noting for completeness. - anyone
  • It is possible, though rarely useful, for the first argument to be initialized but empty; programname "" secondarg third. The $# check unambiguously checks the number of arguments. - anyone
  • For a noob, especially someone who comes from a non-scripting background, it is also important to mention some peculiarities about these things. You could have also mentioned that we need a space after the opening and the closing brace. Otherwise things do not work. I am myself a scripting noob (I come from C background) and found it the hard way. It was only when I decided to copy the entire thing "as is" that things worked for me. It was then I realized I had to leave a space after the opening brace and before the closing one. - anyone
  • and for optional args if [ ! -z "$1" ]; then ... - anyone

It is better to demonstrate this way

if [[ $# -eq 0 ]] ; then
    echo 'some message'
    exit 1
fi

You normally need to exit if you have too few arguments.

Answered   2023-09-21 08:07:54

  • To know more about the difference between [ ] and [[ ]] see stackoverflow.com/questions/3427872/… - anyone
  • This answer is better than the accepted answer because 1. exiting with an error code is better than continuing with invalid data, and 2. [[ is normally more reasonable than [. - anyone

In some cases you need to check whether the user passed an argument to the script and if not, fall back to a default value. Like in the script below:

scale=${2:-1}
emulator @$1 -scale $scale

Here if the user hasn't passed scale as a 2nd parameter, I launch Android emulator with -scale 1 by default. ${varname:-word} is an expansion operator. There are other expansion operators as well:

  • ${varname:=word} which sets the undefined varname instead of returning the word value;
  • ${varname:?message} which either returns varname if it's defined and is not null or prints the message and aborts the script (like the first example);
  • ${varname:+word} which returns word only if varname is defined and is not null; returns null otherwise.

Answered   2023-09-21 08:07:54

  • The example above seems to use ${varname?message}. Is the extra : a typo, or does it change behavior? - anyone
  • Eki, the ":" is a builtin command and shorthand for /bin/true in this example. It represents a do-nothing command that basically ignores the arguments it is provided. It is essential in this test in order to keep the interpreter from trying to execute the contents of "$varname" (which you certainly do NOT want to happen). Also worth noting; you can test as many variables with this method as you wish. And all with specific error messages. i.e. : ${1?"First argument is null"} ${2?"Please provide more than 1 argument"} - anyone
  • I have bash scrip file name sitelog the show me nginx log file, I want to pass to it argument like sitelog -c to clear log file. - anyone

Try:

 #!/bin/bash
 if [ "$#" -eq  "0" ]
   then
     echo "No arguments supplied"
 else
     echo "Hello world"
 fi

Answered   2023-09-21 08:07:54

  • Why do you need double-quotes for $# and 0? - anyone
  • No problem if we use without double-quotes as like $# and 0 - anyone
  • on windows, mingw this is the only way to go. - anyone
  • This answer provides great starting point for a script I just made. Thanks for showing the else, too. - anyone
  • @user13107 double quoted variables in bash prevent globbing (i.e. expanding filenames like foo*) and word splitting (i.e. splitting the contents if the value contains whitespace). In this case it's not necessary to quote $# because both of those cases do not apply. Quoting the 0 is also not necessary, but some people prefer to quote values since they are really strings and that makes it more explicit. - anyone

Only because there's a more base point to point out I'll add that you can simply test your string is null:

if [ "$1" ]; then
  echo yes
else
  echo no
fi

Likewise if you're expecting arg count just test your last:

if [ "$3" ]; then
  echo has args correct or not
else
  echo fixme
fi

and so on with any arg or var

Answered   2023-09-21 08:07:54

Another way to detect if arguments were passed to the script:

((!$#)) && echo No arguments supplied!

Note that (( expr )) causes the expression to be evaluated as per rules of Shell Arithmetic.

In order to exit in the absence of any arguments, one can say:

((!$#)) && echo No arguments supplied! && exit 1

Another (analogous) way to say the above would be:

let $# || echo No arguments supplied

let $# || { echo No arguments supplied; exit 1; }  # Exit if no arguments!

help let says:

let: let arg [arg ...]

  Evaluate arithmetic expressions.

  ...

  Exit Status:
  If the last ARG evaluates to 0, let returns 1; let returns 0 otherwise.

Answered   2023-09-21 08:07:54

  • -1 this might be the worst method if validating existence of an argument.. plus it can trigger history substitution and potentially do bad things. - anyone
  • instead of exit which kills my zsh process, I use return which does not kill it - anyone
  • Why would ((!$#)) trigger history substitution? - anyone
  • What's about 3rd input argument , e.g I have ffmpeg script to downscale videos, but sometimes I want to cut part of it using -ss parameter as 3rd argument on my script, if that arg not presented the clip won't be cut. if [ -z "$3" ]; then - anyone

I often use this snippet for simple scripts:

#!/bin/bash

if [ -z "$1" ]; then
    echo -e "\nPlease call '$0 <argument>' to run this command!\n"
    exit 1
fi

Answered   2023-09-21 08:07:54

  • So, this is to be used in you need only one argument? - anyone
  • @Danijel No, this is testing if there is an argument in the first position. You could have a $2 or a $3 argument ($0 is the script name being run). This simply ignores any other arguments passed. - anyone

More modern

#!/usr/bin/env bash

if [[ $# -gt 0 ]]
  then echo Arguments were provided.
  else echo No arguments were provided.
fi

Answered   2023-09-21 08:07:54

If you'd like to check if the argument exists, you can check if the # of arguments is greater than or equal to your target argument number.

The following script demonstrates how this works

test.sh

#!/usr/bin/env bash

if [ $# -ge 3 ]
then
  echo script has at least 3 arguments
fi

produces the following output

$ ./test.sh
~
$ ./test.sh 1
~
$ ./test.sh 1 2
~
$ ./test.sh 1 2 3
script has at least 3 arguments
$ ./test.sh 1 2 3 4
script has at least 3 arguments

Answered   2023-09-21 08:07:54

As a small reminder, the numeric test operators in Bash only work on integers (-eq, -lt, -ge, etc.)

I like to ensure my $vars are ints by

var=$(( var + 0 ))

before I test them, just to defend against the "[: integer arg required" error.

Answered   2023-09-21 08:07:54

  • Neat trick, but please note: due to bash's inability to handle floats in arithmetic, this method can cause a syntax error and return non-zero which would be a hindrance where errexit is enabled. var=$(printf "%.0f" "$var") can handle floats but suffers from the non-zero exit when given a string. If you don't mind an awk, this method I use seems to be the most robust for enforcing an integer: var=$(<<<"$var" awk '{printf "%.0f", $0}'). If var is unset, it defaults to "0". If var is a float, it is rounded to the nearest integer. Negative values are also fine to use. - anyone

one liner bash function validation

myFunction() {

    : ${1?"forgot to supply an argument"}
    if [ "$1" -gt "-1" ]; then
        echo hi
    fi

}

add function name and usage

myFunction() {

    : ${1?"forgot to supply an argument ${FUNCNAME[0]}() Usage:  ${FUNCNAME[0]} some_integer"}
    if [ "$1" -gt "-1" ]; then
        echo hi
    fi

}

add validation to check if integer

to add additional validation, for example to check to see if the argument passed is an integer, modify the validation one liner to call a validation function:

: ${1?"forgot to supply an argument ${FUNCNAME[0]}() Usage:  ${FUNCNAME[0]} some_integer"} && validateIntegers $1 || die "Must supply an integer!"

then, construct a validation function that validates the argument, returning 0 on success, 1 on failure and a die function that aborts script on failure

validateIntegers() {

    if ! [[ "$1" =~ ^[0-9]+$ ]]; then
        return 1 # failure
    fi
    return 0 #success

}

die() { echo "$*" 1>&2 ; exit 1; }

Even simpler - just use set -u

set -u makes sure that every referenced variable is set when its used, so just set it and forget it

myFunction() {
    set -u
    if [ "$1" -gt "-1" ]; then
        echo hi
    fi

}

Answered   2023-09-21 08:07:54

In my case (with 7 arguments) the only working solution is to check if the last argument exists:

if [[ "$7" == '' ]] ; then
  echo "error"
  exit
fi

Answered   2023-09-21 08:07:54

  • This is not true. $7 is the 7th argument (8th if you count $0 which is the script name), so this does not check if the last argument exists, it checks if the 7th argument exists. - anyone
  • I do agree that this is not a solution to the question, and a sub-optimal solution to a different (and probably avoidable) problem. Seven positional arguments seems heavy. In addition, exit without exit status will return the exit status of echo "error", which I expect to be zero. Recommend shellcheck and set -euo pipefail. I'm going to stop now... - anyone
  • While not a unique answer, it is similar to other accepted answers that have several upvotes. It seems that the author may not be a native English speaker, and likely meant that in their case of 7 arguments, this was a working solution. I've edited the answer to reflect that. Suggestions from @JackWasey should definitely be taken into consideration. - anyone