From: Ross Bamford <rosco@...>
Date: 2005-12-22T22:42:49+09:00
Subject: Re: Regular expression question

On Thu, 22 Dec 2005 09:42:00 -0000, Robert Klemme <bob.news@gmx.net> wrote:

> Ross Bamford wrote:
>> On Wed, 21 Dec 2005 18:03:13 -0000, DeZo <nobody@nowhere.com> wrote:
>>
>>> Why does the following code:
>>>
>>> line = "  rows = 10  cols = 1  occupied cells = 0"
>>> line =~ /.*(\d+).*(\d+).*(\d+)/
>>> print("  scanned rows = ",$1," cols = ",$2," occ = ",$3,"\n")
>>>
>>> print this when it runs:
>>>
>>>   scanned rows = 0 cols = 1 occ = 0
>>>
>>> (notice rows is zero!)
>>>
>>> What have I done wrong?
>>
>> Your problem is that '*' is greedy so it'll match as many 'any
>> characters' as it can. Try
>>
>> /.*?(\d+).*?(\d+).*?(\d+)/
>>
>> Usually I'd tend to use something like:
>>
>> /[^\d]*(\d+)[^\d]*(\d+)[^\d]*(\d+)/
>>
>> instead, to make it explicit I want not digits, followed by digits,
>> etc...
>
> Some other solutions with individual pros and cons:
>
>>> line = "  rows = 10  cols = 1  occupied cells = 0"
> => "  rows = 10  cols = 1  occupied cells = 0"
>>> line.scan(/\d+/)
> => ["10", "1", "0"]
>>> line.scan(/\d+/).map {|s| s.to_i}
> => [10, 1, 0]

>>> line.scan(/\w+\s*=\s*(\d+)/)
> => [["10"], ["1"], ["0"]]
>>> line.scan(/\w+\s*=\s*(\d+)/).map {|m| m[0].to_i}
> => [10, 1, 0]
>

Ahh, much better. Another KISS reminder gets it's own page (again) in my  
notebook..

Thanks :)

-- 
Ross Bamford - rosco@roscopeco.remove.co.uk