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how do I pattern match a string within a string and then extract it into a variable

Answer 1

A:

if rhyme =~ /(\d\.\d)/
    version = $1
end

The regexp matches a digit, followed by a period, followed by another digit. The parenthesis captures its contents. Since it is the first pair of parenthesis, it is mapped to $1.

Magnar 2010-09-20 09:55:17

looks like Perl.

Mark Thomas 2010-09-20 10:58:08

Answer 2

+2 A:

First you need to decide what the pattern for a version number should be. One possibility would be \d+(\.\d+)*$ (a number followed by zero or more (dot followed by a number) at the end of the string).

Then you can use String#[] to get the substring that matches the pattern:

rhyme[ /\d+(\.\d+)*$/ ] #=> "6.0"

sepp2k 2010-09-20 09:57:29

Your regex misses the 6. It should be /(\d+\.\d+)/.

Gerhard 2010-09-20 10:44:10

@Gerhard: Nope. My regex works as is. If you run my code in irb, you'll see that it does indeed return "6.0" like I said. Also unlike your proposed regex, mine would also match "6" or "6.0.0".

sepp2k 2010-09-20 10:48:18

@Gerhard: To make it work with `scan`, you need to make the group non-capturing (and remove the anchor of course). Like this: `"1.0 2.0.0 3".scan(/\d+(?:\.\d+)*/) #=> ["1.0", "2.0.0", "3"]`

sepp2k 2010-09-20 10:52:43

@Gerhard: That being said, I anchored the regex specifically because I assumed that the OP would not want stray numbers appearing in the name to affect the result, so I wouldn't recommend using scan. I.e. I think for "Sepp2k's base 2 to base 16 converter Version 42" it should just return 42 and ignore the 2 and the 16.

sepp2k 2010-09-20 10:56:44

@sepp2K: I checked it in irb and rubular and it works. But what is the purpose of the brackets. Would `/\d+\.\d+/` not be sufficient?

Gerhard 2010-09-20 11:16:48

@Gerhard: As I said, my regex also matches "1.0.0" or "1", while `/\d+\.\d+/` would only match "1.0".

sepp2k 2010-09-20 11:18:48

Ok. () is not just capture group but also groupings. Your Regex Fu exceeds mine!

Gerhard 2010-09-20 11:27:16

got it to work with this one rhyme[ /\d+(\.\d+)*$/ ] #=> "6.0" - nice simple and easy! Thanks!

2010-09-21 06:18:13

Answer 3

+1 A:

You need to use regular expressions. I would use rhyme.scan(/(\d+\.\d+)/) since it can return an array if multiple matches occur. It can also take a block so that you can add range checks or other checks to ensure the right one is captured.

version = "0.0"
rhyme = "three blind mice Version 6.0"
rhyme.scan(/(\d+\.\d+)/){|x| version = x[0] if x[0].to_f < 99}
p version

If the input can be trusted to yield only one match or if you always are going to use the first match you can just use the solution in this answer.

Edit: So after our discussion just go with that answer.

Gerhard 2010-09-20 10:30:59

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