Re: LPeg help...

[Philipp Janda <siffiejoe@...>]

Am 23.03.2013 00:55 schröbte dan young:
> Hello all,

Hi!

> FIrst of all Lua is awesome!  I'm a fairly newbie @ Lua and a total newbie
> @ LPeg.
>
> I'm trying to figure out how to do the following. FOr example, I have the
> following strings, and I only want to match strings a and e.
>
> a = 'disease_show/881/////'
> b = 'disease_show/266/tests///'
> c = 'disease_show/191/description////'
> d = 'disease_show//description////'
> e = 'disease_show/881/'
>
>
> I don't know how to exclude anything that might have letters after the
> digits and the slash (i.e... b and c)
>
> disease_match = (P'disease_show/'*R'09'^1*P'/'^0)
>
> > = disease_match:match(a)
> 22
> > return  disease_match:match(b)
> 18
> > return  disease_match:match(c)
> 18
> > return  disease_match:match(d)
> nil
> > return  disease_match:match(e)
> 18
> >

If you add a capture to your pattern, you can see exactly which parts of 
the strings are matched:

disease_match = (C(P'disease_show/'*R'09'^1*P'/'^0))

disease_show/881/////
disease_show/266/
disease_show/191/
nil
disease_show/881/

So it appears in this case you need something that works like '$' for 
regexes (anchor the pattern at the end of the string), and thats P(-1) 
or -P(1) in LPeg[1][2]:

disease_match = (P'disease_show/'*R'09'^1*P'/'^0*P(-1))

22
nil
nil
nil
18

  [1]: http://www.inf.puc-rio.br/~roberto/lpeg/lpeg.html#op-p
  [2]: http://www.inf.puc-rio.br/~roberto/lpeg/lpeg.html#op-unm

> Regards,
>
> Dano

HTH,
Philipp