I want to replace all multiple slashes in URL, apart from those in protocol definition ('http[s]://', 'ftp://' and etc). How do I do this?
This code replaces without any exceptions:
url.gsub(/\/\/+/, '/')
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3 Answers
You just need to exclude any match which is preceeded by :
url.gsub(/([^:])\/\//, '\1/')
5 Comments
krn
Thank you, but how could I now prevent it from selecting the closest character from the left? rubular.com/r/PhVk4JSxcx
Phrogz
Use a negative lookbehind:
%r{(?<!:)//} (not available in Ruby 1.8)
Stuart
If you don't have negative lookbehind you can use a capture
url.gsub(/([^:])\/\//, '\1/')
ocodo
@Stuart, thanks I've updated my answer to use the back reference, since it's more widely compatible.
masimplo
how about protocol-less urls like //www.domain.com/ used for http/https swaping?
I tried using URI:
require "uri"
url = "http://host.com//foo//bar"
components = URI.split(url)
components[-4].gsub!(/\/+/, "/")
fixed_url = [components[0], "://", components[2], components[-4]].join
But that seemed hardly better than using a regex.
Comments
gsub can take a block:
url = 'http://host.com//foo/bar'
puts url.gsub(%r{.//}) { |s| (s == '://') ? s : s[0] + '/' }
>> http://host.com/foo/bar
Or, as @Phrogz so kindly reminded us:
puts url.gsub(%r{(?<!:)//}, '/')
>> http://host.com/foo/bar
