I do care and definitely prefer '|' over '**'. Furthermore, it would make
life more pleasant for parsers (at least for my parser) if word boundaries
were taken into account. Normally, space (or whitespace) is used as word
boundary. So I prefer 'Gone | | wind' over 'Gone||the wind'. It is a pity
there is no masking symbol for one or more characters, because that could be
used in combination with word separators to mask words.
> -----Original Message-----
> From: Z39.50 Next-Generation Initiative [mailto:[log in to unmask]]On Behalf Of
> Ray Denenberg
> Sent: 04 September 2002 15:58
> To: [log in to unmask]
> Subject: Re: which masking character for words?
> Based on this discussion I suggest we simply declare that we're
> not going to
> attempt to define what's meant by any concatenation of masking
> characters where
> one masks characters and the other masks words.
> Time's getting short, because we need to have an approved 105
> definition before
> we freeze the SRWU spec. So I'm going to post a proposal to the
> ZIG and if you
> have more to say on this please continue the discussion there.
> I'm going to suggest ** to mask words, an arbitrary choice as
> opposed to |. If
> you prefer | post your preference to the ZIG (though I assume
> nobody cares other
> than Rob and I, since nobody has spoken up).
> Robert Sanderson wrote:
> > > > Also, is '***foobar':
> > > > 1. '(**)(*)foobar' which would match 'fred bertfoobar'
> > > > 2. '(*)(**)foobar' which would match 'fredbert foobar' but
> not the above
> > > > I'm sticking by '|' personally.
> > > Well even though I don't buy either example (the "?" inference isn't
> > > necessary, and '(*)(**)foobar' would indeed match 'fred bertfoobar' --
> > > the first * would match the string 'fred bert') rather than
> debate this
> > Hmmm.
> > This applies to | as well. I would have said that '*|foo' would
> match any
> > number of characters, followed by any number of words, and then
> the _word_
> > foo.
> > So '?|foo' would match 'a foo' 'z bar foo' 'foo' but not '3foo' as foo
> > isn't a word.
> > If this isn't the case, then there's no issue with ** as you say.
> > Rob
> > --
> > ,'/:. Rob Sanderson ([log in to unmask])
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