Re: backreference in regexp

2006-01-31 Thread Schüle Daniel 
thank you, I completely forgot that + is one of metacharacters

Regards, Daniel

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Re: backreference in regexp

2006-01-31 Thread Fredrik Lundh 
Schüle Daniel wrote:

> Hello @all,
>
> >>> p = re.compile(r"(\d+) = \1 + 0")
> >>> p.search("123 = 123 + 0")
>
> 'search' returns None but I would expect it to
> find 123 in group(1)
>
> Am I using something that is not supported by Python
> RegExp engine or what is the problem with my regexp?

plus matches one or more instances of the previous item.  to make
it match a plug sign, you have to escape it:

p = re.compile(r"(\d+) = \1 \+ 0")





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backreference in regexp

2006-01-31 Thread Schüle Daniel 
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Hello @all,

 >>> p = re.compile(r"(\d+) = \1 + 0")
 >>> p.search("123 = 123 + 0")

'search' returns None but I would expect it to
find 123 in group(1)

Am I using something that is not supported by Python
RegExp engine or what is the problem with my regexp?

Regards, Daniel

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