From: Robert Klemme <bob.news@...>
Date: 2005-12-22T18:42:49+09:00
Subject: Re: Regular expression question

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]

And explicitely matching the pattern:

>> /rows\s*=\s*(\d+)\s*cols\s*=\s*(\d+)\s*occupied cells\s*=\s*(\d+)/ =~
line and [$1, $2, $3]
=> ["10", "1", "0"]
>> /rows\s*=\s*(\d+)\s*cols\s*=\s*(\d+)\s*occupied cells\s*=\s*(\d+)/ =~
line and [$1.to_i, $2.to_i, $3.to_i]
=> [10, 1, 0]

Kind regards

    robert