How to parse a URL and extract the required substring

Question

Say I have a string like this: "http://something.example.com/directory/"

What I want to do is to parse this string, and extract the "something" from the string.

The first step, is to obviously check to make sure that the string contains "http://" - otherwise, it should ignore the string.

But, how do I then just extract the "something" in that string? Assume that all the strings that this will be evaluating will have a similar structure (i.e. I am trying to extract the subdomain of the URL - if the string being examined is indeed a valid URL - where valid is starts with "http://").

Thanks.

P.S. I know how to check the first part, i.e. I can just simply split the string at the "http://" but that doesn't solve the full problem because that will produce "http://something.example.com/directory/". All I want is the "something", nothing else.

@durron597: Don't hammer everything with regexen. URL is a well defined object, treated million times both in Ruby standard library and in million other gems. If I was an expert, I would answer. — Boris Stitnicky, Nov 06 '12 at 01:48
What if you have something lik `http://a.b.c.d/directory`, more than 2 dots in the host ? — oldergod, Nov 06 '12 at 02:11
The downvotes are probably because there are a number of questions similar to this one. You can find out how to get the host from a URL, then find the first element of a `'.'` delimited string. — the Tin Man, Nov 06 '12 at 03:09

score 39 · Accepted Answer · edited Jul 05 '20 at 16:54

39

I'd do it this way:

require 'uri'

uri = URI.parse('http://something.example.com/directory/')
uri.host.split('.').first
=> "something"

URI is built into Ruby. It's not the most full-featured but it's plenty capable of doing this task for most URLs. If you have IRIs then look at Addressable::URI.

edited Jul 05 '20 at 16:54

Curious Sam

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answered Nov 06 '12 at 03:13

the Tin Man

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I love this. Nice, elegant and isn't expensive. Thanks! – marcamillion Nov 06 '12 at 03:19
1

And it doesn't require us to maintain any regex. – the Tin Man Nov 06 '12 at 03:27
I think it won't work if the string has colon( : ) eg: URI.parse('abc123') – Ram Kumar Dec 12 '18 at 14:36
@RamKumar First, read the [URI RFC](https://tools.ietf.org/html/rfc3986) to see what it says about the use of colons, then provide a more usable test string in a separate question if you still can't get it to work. – the Tin Man Dec 16 '18 at 23:14

score 9 · Answer 2 · edited May 23 '17 at 12:09

9

You could use URI like

uri = URI.parse("http://something.example.com/directory/")
puts uri.host
# "something.example.com"

and you could then just work on the host.
Or there is a gem domainatrix from Remove subdomain from string in ruby

require 'rubygems'
require 'domainatrix'

url = Domainatrix.parse("http://foo.bar.pauldix.co.uk/asdf.html?q=arg")
url.public_suffix       # => "co.uk"
url.domain              # => "pauldix"
url.subdomain           # => "foo.bar"
url.path                # => "/asdf.html?q=arg"
url.canonical           # => "uk.co.pauldix.bar.foo/asdf.html?q=arg"

and you could just take the subdomain.

edited May 23 '17 at 12:09

Community

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answered Nov 06 '12 at 02:39

oldergod

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I think this is a good overall solution for anyone that wants to parse all sorts of URLs. In my case, though, I just need to parse a specific format of URL - which is `foo.bar/somedir`. – marcamillion Nov 06 '12 at 03:20
@marcamillion That is why I gave you the first alternative as well, which is the same as what the Tin Man answered. – oldergod Nov 06 '12 at 04:03
@oldergod: How well is that dominatrix maintained? Is it mature? – Boris Stitnicky Nov 06 '12 at 05:03
@BorisStitnicky I never used it and just found it out today. I don't know. – oldergod Nov 06 '12 at 07:47

score 2 · Answer 3 · edited Apr 08 '19 at 19:33

2

Well, you can use regular expressions. Something like /http:\/\/([^\.]+)/, that is, the first group of non '.' letters after http.

Check out http://rubular.com/. You can test your regular expressions against a set of tests too, it's great for learning this tool.

edited Apr 08 '19 at 19:33

the Tin Man

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answered Nov 06 '12 at 01:49

resilva87

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I started off doing this, but it got pretty messy pretty quickly. – marcamillion Nov 06 '12 at 03:21

score 2 · Answer 4 · answered Jul 19 '20 at 12:10

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with URI.parse you can get:

require "uri"

uri = URI.parse("http://localhost:3000")
uri.scheme # http
uri.host # localhost
uri.port # 3000

answered Jul 19 '20 at 12:10

Dorian

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How to parse a URL and extract the required substring

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