ruby regex: how to get group value

Question

Here is my regex:

s = /(?<head>http|https):\/\/(?<host>[^\/]+)/.match("http://www.myhost.com")

How do I get the head and host groups?

Why don't you reuse an invented wheel, like Ruby's built-in URI or Addressable::URI? — the Tin Man
– the Tin Man, Commented Nov 28, 2012 at 1:40

Philip Hallstrom · Accepted Answer · 2012-11-27 23:59:01Z

5

s['head'] => "http"
s['host'] => "www.myhost.com"

You could also use URI...

1.9.3p327 > require 'uri'
 => true 
1.9.3p327 > u = URI.parse("http://www.myhost.com")
 => #<URI::HTTP:0x007f8bca2239b0 URL:http://www.myhost.com> 
1.9.3p327 > u.scheme
 => "http" 
1.9.3p327 > u.host
 => "www.myhost.com"

answered Nov 27, 2012 at 23:59

Philip Hallstrom

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Comments

user1858213 · Accepted Answer · 2012-11-28 00:01:17Z

2

Use captures >>

string = ...
one, two, three = string.match(/pattern/).captures

answered Nov 28, 2012 at 0:01

user1858213

2 Comments

Just Jake Over a year ago

How do you do this in context, ie, with his regex?

user1858213 Over a year ago

@JustJake - With your "current" regex pattern (with some minor optimization) it would be head, host = "http://www.myhost.com".match(/(https?):\/\/([^\/]+)/).captures. As you can see, the regex pattern has two groups (...), and their matches will be assigned into head, host.

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

You should probably use the uri library for this purpose as suggested above, but whenever you match a string to a regex, you can grab captured values using the special variable:

"foo bar baz" =~ /(bar)\s(baz)/

$1

=> 'bar'

$2

=> 'baz'

and so on...

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered Nov 28, 2012 at 4:22

James Martin

964 bronze badges

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