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.

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

More modern

#!/usr/bin/env bash

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

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

}

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