Isn't re.findall supposed to find all?

[Rick Lee]

Well, try   'XYZ([^L]|(L(?!MN)))*PQR'

- Rick Lee

June Kim wrote:

> "June Kim" <junaftnoon at nospamplzyahoo.com> wrote in message
> news:8ui0i5$nfc$1 at news.nuri.net...
> >
> > Well, I forgot the old rule "Longest match possible."
> > It worked perfectly with this:
> > >>> z
> > '[abcdefrdofhd]kdeioslkdfj[sdkfj]'
> > >>> p=re.compile('\[[^\[]*\]')
> > >>> re.findall(p,z)
> > ['[abcdefrdofhd]', '[sdkfj]']
> >
> > But another problem emerged. What if the brackets were
> > longer than one character? for example:
> > z='this is yet another start of a sting end but never start an end'
> >
> > opening bracket : "start"
> > closing bracket : "end"
> >
> > What I want to get is,
> >
> > re.findall(p,z) == ['[start of a string end]','[start an end]']
> >
> > How should I render the regular expression?
> >
> > Thanks in advance
> >
> > ---------
> > June
> >
>
> I found out one way to do it! (bravo)
>
> p=re.compile('start.*?end')
> re.findall(p,z)
>
> But my question still remains; I want to find a substring
> that starts with a specific string, say XYZ and ends with
> another specific string, say QPR, but hasn't got a string,
> LMN in between the opener and the closer.
>
> e.g. XYZ....(any length of letters but no LMN here).....QPR
>
> Could any one enlighten me kindly?
>
> =====
> June