Check existence of input argument in a Bash shell script

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.

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.

Try:

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

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.

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