Remove full URL from text

Should be an easy pattern match and开发者_开发知识库 replace, but I want to be able to remove a full URL from text.

So:

'You gotta love it! http://www.youtube.com/watch?v=0i_bkLbf3EI check it out!!!'

Becomes:

'You gotta love it! check it out!!!'

Any ideas?

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Step 1: Find a regular expression that matches URLs

http://mathiasbynens.be/demo/url-regex

Seems like the last one (@diegoperini) is the best, but it weighs in at 502 characters.

Step 2: Replace any matches of that regex with the empty string

$output = preg_replace($regex, '', $input);

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$string = preg_replace('/\b(https?):\/\/[-A-Z0-9+&@#\/%?=~_|$!:,.;]*[A-Z0-9+&@#\/%=~_|$]/i', '', $string);

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See Daring Fireball's An Improved Liberal, Accurate Regex Pattern for Matching URLs

Excerpt:

(?xi)
\b
(                           # Capture 1: entire matched URL
  (?:
    [a-z][\w-]+:                # URL protocol and colon
    (?:
      /{1,3}                        # 1-3 slashes
      |                             #   or
      [a-z0-9%]                     # Single letter or digit or '%'
                                    # (Trying not to match e.g. "URI::Escape")
    )
    |                           #   or
    www\d{0,3}[.]               # "www.", "www1.", "www2." … "www999."
    |                           #   or
    [a-z0-9.\-]+[.][a-z]{2,4}/  # looks like domain name followed by a slash
  )
  (?:                           # One or more:
    [^\s()<>]+                      # Run of non-space, non-()<>
    |                               #   or
    \(([^\s()<>]+|(\([^\s()<>]+\)))*\)  # balanced parens, up to 2 levels
  )+
  (?:                           # End with:
    \(([^\s()<>]+|(\([^\s()<>]+\)))*\)  # balanced parens, up to 2 levels
    |                                   #   or
    [^\s`!()\[\]{};:'".,<>?«»“”‘’]        # not a space or one of these punct chars
  )
)

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Try this:

/http:\/\/[a-zA-Z0-9\.\/\?\=\_]+/