Question

I want to pass many arguments to a shell script which I don’t know how many arguments they are going to be and I want to handle them. I did the following code:

int=$1
src=$2

r=$3
string=$4

duration=$5

./start.sh $int $r $src "$string"
sleep $duration

shift; shift; shift; shift; shift

while [ $# -gt 2 ]
do
  r=$1
  string=$2
  duration=$3
  ./change.sh $int $r "$string"
  sleep $duration
  shift; shift; shift
done

That works but only for one time, now I want this script to run all the time, I mean using while 1 but that won’t work this way because the argument list is empty at the end of that code!

Is there any way to do something like the following “pseudo code” in shell scripts:

for( i=0 ; i<arguments.count; i++ ){
   //do something with arguments[i]
}

or copying the arguments array into another array so that I can use it later the way I want.
I mean can I copy $* or $@ or arg into another array?

Any help is highly appreciated.

Answer 1

The number of arguments is stored in the parameter #, which can be accessed with $#.

A simple loop over all arguments can be written as follows:

for arg
do
    # do something with the argument "arg"
done