On Tue, 15 Aug 2000, Doug MacEachern wrote:

> what do you mean?  args in or out?  $r->args($string)
> just does a blind copy.  @args = $r->args just splits on
> &,=, it's up to the caller to deal with multiple args of
> the same name.

hm. what about %args = $r->args? won't that give me only the
last arg value? i suppose i should just do @args = $r->args.
thanks for the reminder.

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