Checking for the correct number of arguments

How do i check for the correct number of arguments (one argument). If somebody tries to invoke the script without passing in the correct number of arguments, and checking to make sure the command line argument actually exists and is a directory.

Answers


#!/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


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