I created a parametrized shell script. As per my understanding, $0 is the command name, and $1, $2, etc. are all the parameters that follow. So in the statement
ls -lhtr
This would be the corresponding variable count:
ls $0
-lhtr $1
I have also learned, that if you don't want to do the following:
chmod +x myScript.sh
./myScipt.sh param1 # in this case, "./myScript.sh" is $0 and "param1" is $1
You can just run
sh myScript.sh param1
However, in the second example, would sh be $0? Or would myScript.sh be $0 (according to the source code in myScript.sh)?
Also, what if I aliased
alias myScript="sh myScript.sh"
?
Would myScript, sh, or myScript.sh be $0?
These variables don't change for every command in the script. When you execute
sh scriptname foo bar baz
The shell receives the parameters from the operating system. In C, its argv array will look like:
argv[0] = "sh"
argv[1] = "scriptname"
argv[2] = "foo"
argv[3] = "bar"
argv[4] = "baz"
It then creates shell variables from this. argv[0] isn't put into a shell variable, the rest are put into $0, $1, $2, and so on, so you get:
$0 = scriptname
$1 = foo
$2 = bar
$3 = baz
The script can update the parameters starting from $1 using the set built-in, and can use the shift built-in to remove parameters from the beginning of the argument list (this is often used in loops that process arguments from left to right).
When you write a command like
ls "$1"
it refers to the arguments to the script, not the arguments on that command line.