One For The Gods Of Linux [ Bash Scripting postional parameters / arguments ]

ally_uk

Good evening

I am in the process of trying to get my head around BASH scripting I am not a programmer and am having trouble with understanding Positional parameters and arguments.

I understand how to declare variables but am failing to understand the concept of arguments and positional parameters.

$0, $1, $2

From my understanding $0 holds the script name?

Could somebody write a noob script example breaking it down and explaining arguments it would really help me out i'm getting nothing but headaches lol

Many Thanks

Appreciate it

NightShade03

Here is a quick low down on this:

$0 is the script name as you mentioned
$1 is the first argument
$2 is the second argument

So on and so forth...

What you also need are these two:

a special variable that holds then entire list
$# the total numerical value of all arguments

For example a BASH script called test:

#!/bin/bash



echo $#
echo $0
echo $1
echo $2


On the command line:

# ./test hello
hello
1
./test
hello
<- this is a blank line

# ./test hello world
hello world
2
./test
hello
world

Obviously you don't want to guess how many arguments a script will have when the user executes it so you leverage a while loop to filter through each parameter. See this Stackoverflow link for an example (sry to much code to post here):

linux - How to get arguments with flags in bash script - Stack Overflow

Notice how the author checks to ensure that the $# is great than (gt) 0 and if that is true...loop through all parameters.

Hope this makes more sense.

ally_uk

Many thanks for the explanation I will fire up a terminal and get busy

NightShade03

No problem...let me know if you get stuck