Regular expression - replace word except url / uri

Question

I’m writing a globalization module for a web application and I need a regular expression to replace all instances of a word with another word (translation) - with the exception of words found in the URL / URI.

EDIT: I forgot to mention that I'm using Ruby, so I can't use "Lookbehind"

Answer 1

• Split into a regular expression URI; include the URI in the result.
• For each part:
  • if it is a URI, leave it alone
  • otherwise replace words
• Join the parts

the code:

# From RFC 3986 Appendix B, with these modifications:
#   o Spaces disallowed
#   o All groups non-matching, except for added outermost group
#   o Not anchored
#   o Scheme required
#   o Authority required
URI_REGEX = %r"((?:(?:[^ :/?#]+):)(?://(?:[^ /?#]*))(?:[^ ?#]*)(?:\?(?:[^ #]*))?(?:#(?:[^ ]*))?)"

def replace_except_uris(text, old, new)
  text.split(URI_REGEX).collect do |s|
    if s =~ URI_REGEX
      s
    else
      s.gsub(old, new)
    end
  end.join
end

text = <<END
stack http://www.stackoverflow.com stack
stack http://www.somewhere.come/stack?stack=stack#stack stack
END

puts replace_except_uris(text, /stack/, 'LINKED-LIST')

# => LINKED-LIST http://www.stackoverflow.com LINKED-LIST
# => LINKED-LIST http://www.somewhere.come/stack?stack=stack#stack LINKED-LIST

Answer 2

You can probaby use something like

(?<!://[^ ]*)\bfoo\b

, , , , , :// - .

PS Home:\> "foo foobar http://foo_bar/baz?gak=foobar baz foo" -replace '(?<!://[^ ]*)\bfoo\b', 'FOO'
FOO foobar http://foo_bar/baz?gak=foobar baz FOO

Answer 3

Did you try to split the text into words and repeat the words? Then you can check each word, determine if it is a URI, translate it if it is not.