Shell Programming and Scripting

Hi All,
I am new to scripting. Could you please assist me .

Here is my requirement. I have written a script that has 2 option flags defined.

-l) calls some function with the arguments passed in front of -l
-r) calls second function with the arguments passed in front of -r
*) calls the third function with arguments specified.

For this I have written-

Code:

#!/usr/bin/bash
 
while [ $# -gt 0 ]
do
 
case "$1" in

-l) echo "$2";
    list_files "$2"
    shift;;
-r)usage;;
*)if [ "$1" == "." ];then 
 echo "Not allowed "
 else
 check_path "$1" 
 fi 
 shift ;;
esac
done
######################################

While executing-

Scenario 1- ./test -l file1 file2 file3
I am only able to capture the argument file1 with -l . The arguments file2 and file3 are passed to the *) option.

Could you please let me know how to pass all the arguments to the function called from -l) option.

Thanks a lot

Code:

-l) echo "$2";
    list_files "$2"
    shift;;

Since you use $2 here, should you not shift 2 for a start?
...

I tried shift 2 but it wouldn't work. -

Since i need to call a function check_path if no option (-l etc) is passed, it is becoming increasingly difficult for me to capture multiple arguments passed to an option. .

what about a loop using for:

Code:

for i in $*
do
  case $i in
...

Enclose the arguments for -l with double quotes and parse them one by one like vbe stated with a for loop for example.

Code:

$ ./test -l "file1 file2 file3"

@zaxxon - It worked partially. I was able to list the arguments passed to -l option.
The only problem was that, these arguments were also passed to the *) option.

* would normaly be used for other "unwanted" cases, so you are to define precisely what you expect as argument for your third case...