Bashscripts.org

I want to write a script which runs with command line arguments. The script can be passed with several arguments (the number may vary). I want to assign those arguments to a variable and do some modifications. Here is part of my code



Here I get the output as 0 to the number of arguments. That's because "$i" returns the value of the variable "i". I want to print the argument number "i".

How can I do this ?

usually the easiest way to handle this type of script is to use the shift function.

lets say you ran the script with 3 arguments.. test1 test2 test3


this would create an array with the following content:
$argv[0] = test1
$argv[1] = test2
$argv[2] = test3

This happens because shift moves the entire argument list on step in. $1 will disappear and $2 will become $1, $3 will become $2.

You don't need to use an array in the while loop, you can do a case or anything you wish, the shift statement will still do what it supposed to

ps. while [ ! -z $1 ] means while $1 is not empty keep on going ds.

Best regards
Fredrik Eriksson

I suggest you also check out the 'getopts' function in the bash man page

getopts rules for things like this! Here's a quick example from my "functions" file:



Hope this helps!