This stuff:
$node = ("$ARGV[1]"); $interface = ("$ARGV[3]");
note that $ARGV[1] is the second command line argument to the script and $ARGV[3] is the fourth. Since you don't mention anything about how you're supplying the arguments the problem is probably there somewhere.

Also note that you usually don't want to quote array lookups, it will work here since the values are strings.

In my opinion

$node = $ARGV[1]; $interface = $ARGV[3];
is cleaner.

Update: if you're wondering what's going on adding something like this just behind the argument assignment might help:

warn "Supplied arguments: ",join(",",map { "'$_'" } @ARGV); warn "Node is '$node', interface is '$interface'.";

In reply to Re: Problem with adding variable from array by Joost
in thread Problem with adding variable from array by Bennco99

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