"""
Waylan -
The problem here is not parse actions, but with the behavior of scanString.
It is possible to hit some false positives with scanString, especially with
a grammar such as this. scanString works by walking the input string
character by character, trying to match the scan expression.
In your case, scanString predictably finds the first three foo's, but also
finds an unexpected match at the fourth foo. This is because scanString
tries each successive character location, that is, after matching the third
foo ending at locn 23, scanString does:
loc 24: `````foo`` -> no match
loc 25: ````foo`` -> no match
loc 26: ```foo`` -> no match
loc 27: ``foo`` -> match!
Notice the different behavior when using parseString with a grammar of
OneOrMore(Group(foo)). Now there is no successive matching of each
character location in turn - parseString looks for a foo match at only one
location, 24. Now it should also be obvious why `bar` was matched
successfully.
-- Paul
"""
from pyparsing import *
begin = Word("`")
end = matchPreviousExpr(begin)
foo = begin + Literal('foo') + end
a = foo.scanString("``foo`` ```foo``` `foo` `````foo``")
for i in a:
print i
# this only recognizes the first three foo's ("All the Foos down in
Fooville...")
print OneOrMore(Group(foo)).parseString("``foo`` ```foo``` `foo`
`````foo``")
-----Original Message-----
From: Waylan Limberg [mailto:wa...@gm...]
Sent: Thursday, January 25, 2007 12:35 PM
To: Paul McGuire
Subject: Re: [Pyparsing] matching a previous match
On 1/25/07, Paul McGuire <pa...@al...> wrote:
> Version 1.4.5 includes a couple of new helper methods,
> matchPreviousLiteral and matchPreviousExpr, for exactly this
> situation. I think there is some sample code in the HTML docs, if
> not, write back and I'll post some sample code using these methods.
Cool. Exactly what I had in mind. Unfortunately, the only place I've found
any docs on it is using pydoc, but that should give me enough to get going.
Thanks for the pointer.
>
> There is a current known bug if you are using parse actions with these
> two methods, I've got a fix in the works, just need to push out the
> next release if its critical for you.
Hmm, I might have run into it. Here is what i have so far:
>>> begin = Word("`")
>>> end = matchPreviousExpr(begin)
>>> foo = begin + Literal('foo') + end
>>> a = foo.scanString("``foo`` ```foo``` `foo` `````foo``") print [i
>>> for i in a]
[((['``', 'foo', '``'], {}), 0, 7), ((['```', 'foo', '```'], {}), 8, 17),
((['`', 'foo', '`'], {}), 18, 23), ((['``', 'foo', '``'], {}), 27, 34)]
That last one shouldn't match. Which would explain why the following doesn't
work:
>>> match = begin + SkipTo(end) + end
>>> b = match.scanString("`foo` ``foo`bar`` `` `bar` ``") print [i for i
>>> in b]
[((['`', 'foo', '`'], {}), 0, 5), ((['`', 'foo', '`'], {}), 7, 12), ((['``',
'', '``'], {}), 15, 20), ((['`', 'bar', '`'], {}), 21, 26)]
>
> -- Paul
>
>
> -----Original Message-----
> From: pyp...@li...
> [mailto:pyp...@li...] On Behalf Of
> Waylan Limberg
> Sent: Thursday, January 25, 2007 11:00 AM
> To: pyp...@li...
> Subject: [Pyparsing] matching a previous match
>
> I'm trying to match an string enclosed in backticks. Rather than using
> an escape character, the string is simply wrapped in more backticks.
> Here are some examples:
>
> `foo` => foo
> ``foo`bar`` => foo`bar
> `` `bar` `` => `bar` #note the spaces in this one
>
> This is easy in regex:
>
> (?P<backtick>`+)(?P<string>.*?)(?P=backtick)
>
> Of course, the trick is that the ending string of backticks must
> exactly match the opening string of backticks.
>
> Sure I can use Regex() for this, but I'm trying to figure you how to
> do that without regex. How can I refer back to a previous match?
>
> --
> ----
> Waylan Limberg
> wa...@gm...
>
> ----------------------------------------------------------------------
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>
>
--
----
Waylan Limberg
wa...@gm...
|
