Re: Regular expression to parse out "host" part of URL strin

I have that :).

I’m trying to handle the case when I get a “URL” without the http://

Wes

Instead of beating yourself up trying to come up with a regex, just
stick the ‘http://’ in front of it yourself.

unless path.index(‘http://’, 0)
path = ‘http://’ + path
host = URI.parse(path).host
end

It’s not necessarily the most robust solution, but you get the general
idea.

Regards,

Dan

Berger, Daniel wrote:

I have that :).

I’m trying to handle the case when I get a “URL” without the http://

Wes

Instead of beating yourself up trying to come up with a regex, just
stick the ‘http://’ in front of it yourself.

Definitely a solution, but the regexp shouldn’t be that complicated too.

url.scan(%r{(?:.+://)?([^/]+)}).first.first

should do the trick. (?) well i hope thats not like the email-regexp…

cheers

Simon

“Berger, Daniel” [email protected] writes:

unless path.index(‘http://’, 0)
unless path.index(‘http://’) == 0

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