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Bash Script - How to use Command Line Arguments

Last Updated : 07 Feb, 2022

In this article, let us see about Command Line Arguments usage in Bash script. Arguments are inputs that are necessary to process the flow. Instead of getting input from a shell program or assigning it to the program, the arguments are passed in the execution part.

Command-line arguments are passed in the positional way i.e. in the same way how they are given in the program execution.  Let us see with an example. 

Create a shell program that can display the command line arguments in a positional way. "Nano" editor is used to create the shell program"

The program can be executed by using the "sh" command.

sh <script filename> arg1 arg2 arg3 ......

So, our code of execution will become

sh displayPositionalArgument.sh Welcome To GeeksForGeeks

Diagrammatic representation of the above code:

Always the first argument starts after the <script filename>. <script filename> will be in the 0th location, We need to take positional arguments after the <script filename>. Hence in the above example

$1-> "Welcome"
$2-> "To"
$3-> "GeeksForFeeks"

Note: We can pass n number of arguments and they are identified by means of their position.

The output of the above code :

Some special variables are also to be noted while handling command-line arguments.

Arguments can be passed along with the flags. The arguments can be identified with a single letter having - before that. A single letter can be meaningful and here let us take -1, -2, and -3.

We need to use getopts function to read the flags in the input, and OPTARG refers to the corresponding values:

The above program can be executed as

sh usingFlags.sh -1 'GeeksForGeeks' -2 'JavaProgramming' -3 100

Output

Though positional parameters help to send as command-line arguments, if there are more arguments, we cannot count and use them. Instead by using $@, we can achieve that. It is nothing but the array of all the parameters passed. Iterating over a for loop, we can get that.

Output :

Another way is by using the Shift operator instead of $@ - Shift operator:

The $# variable is used to return the input size. By  using that and with the shift operator we can achieve instead of $@

Output:

General Use case scenario using Shift operator:

Assume that in the command arguments, we may pass the "-d" argument which is nothing but to know it is a directory. 

• If there is -d is given, then we need to pick the next argument and check is it a directory like that.
• If there is no -d, then check whether is it a file or not

Then display accordingly as to whether the argument is a directory or file

Let the name of the script be "shiftArgument.sh"

sh  shiftArgument.sh -d /home shiftArgument.sh 

Output:

The above code will check "/home" as a directory as we have "-d" as the first positional argument and hence shift operator is done and the next argument is checked against to see is it a directory. If there is no "-d" argument as the prior one, it just checks "whether that argument is a file or not"

We can pass command-line arguments at our convenience in a positional way or with flags. Moreover useful concepts like $@ (array of all the parameters), $# (by calculating input size)  are available and with that, we can iterate in the  "for" or "while" loops, we can access the arguments.

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