Read command line arguments - Unix / Linux Bash Script

Q) How to read the arguments or parameters passed to a shell script from the command line?

In general the command line arguments are passed to the bash or shell script to change the behavior of the script. In this article, I am going to provide a simple shell script to read and print the command line parameters.

Also Read: 201-450: Linux Engineer - 201 (LPIC-2 201)

Take a look at the following unix shell script:

> cat OS_Print.sh
#!/bin/bash
echo "Script execution starts"
echo "$@"
echo "$0"
echo "$1"
echo "$2"
echo "$#"
echo "Script execution ends"

The basic functionality of the above script is to print the values stored in the $ variables. Now we will run the above script by passing some arguments.

> OS_Print.sh unix linux
Script execution starts
unix linux
OS_Print.sh
unix
linux
2
Script execution ends

You can see, the command line arguments passed here are unix and linux. Command line arguments are a list of parameters separated by space delimiters passed to the shell script.

Explanation of $ variables:

◈ $@ : contains all the arguments
◈ $0 : contains script name
◈ $1 : First argument
◈ $2 : Second argument
◈ $n : Nth argument
◈ $# : Count of arguments passed.

Examples:

1. Script to iterate through arguments.

The following script prints the parameters using for loop.

#!/bin/bash

for value in $@
do
  echo $value
done

2. Print only the last argument.

There are many ways to display the last argument. The following script shows the different ways of printing the last argument.

#!/bin/bash

echo "${@: -1}"
echo "${BASH_ARGV[0]}"
echo "${@: $#}"
echo "${!#}"

for value in $@; do :; done
echo $value