Checking for the correct number of arguments

Question

#!/bin/sh
if [ "$#" -ne 1 ] || ! [ -d "$1" ]; then
  echo "Usage: $0 DIRECTORY" >&2
  exit 1
fi

Translation: If number of arguments is not (numerically) equal to 1 or the first argument is not a directory, output usage to stderr and exit with a failure status code.

More friendly error reporting:

#!/bin/sh
if [ "$#" -ne 1 ]; then
  echo "Usage: $0 DIRECTORY" >&2
  exit 1
fi
if ! [ -e "$1" ]; then
  echo "$1 not found" >&2
  exit 1
fi
if ! [ -d "$1" ]; then
  echo "$1 not a directory" >&2
  exit 1
fi

cat script.sh

    var1=$1
    var2=$2
    if [ "$#" -eq 2 ]
    then
            if [ -d $var1 ]
            then
            echo directory ${var1} exist
            else
            echo Directory ${var1} Does not exists
            fi
            if [ -d $var2 ]
            then
            echo directory ${var2} exist
            else
            echo Directory ${var2} Does not exists
            fi
    else
    echo "Arguments are not equals to 2"
    exit 1
    fi

execute it like below -

./script.sh directory1 directory2

Output will be like -

directory1 exit
directory2 Does not exists

You can check the total number of arguments which are passed in command line with "$#" Say for Example my shell script name is hello.sh

sh hello.sh hello-world
# I am passing hello-world as argument in command line which will b considered as 1 argument 
if [ $# -eq 1 ] 
then
    echo $1
else
    echo "invalid argument please pass only one argument "
fi

Output will be hello-world