AngularJS ng模式网站模式

Question

/^[_a-z0-9]+(\.[_a-z0-9]+)*[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,7})$/

That's the pattern. It says "ok" when it's correct, and "not ok" when it's incorrect. So it says "ok" on www.google.com, but "not ok" when I type http://www.google.com

What I'd like is this pattern to allow http:// too, but it should never be a requirement.

Answer 1

You can use the following regular expression, which I lifted off borrowed from the documentation of the Perl module URI on CPAN (escaping of slashes mine).

/(?:([^:\/?#]+):)?(?:\/\/([^\/?#]*))?([^?#]*)(?:\?([^#]*))?(?:#(.*))?/

It will give you all the different parts of the URI in capture groups.

Those parts are:

• scheme ( http: )
• authority (not applicable here)
• path ( www.google.com )
• query ( q=querystring )
• fragment ( #anker )

See https://regex101.com/r/vS5qO1/1 to try it out.

Also note that this will parse all types of URIs, not only http(s). So stuff like ftp://anonymous@example.org will also work.

Answer 2

If you're looking to only allow http/https URL schemes (when the scheme is provided), the following modification to your regular expression will do the trick:

/^(http:\\/\\/|https:\\/\\/)?[_a-z0-9]+(\\.[_a-z0-9]+)*[a-z0-9-]+(\\.[a-z0-9-]+)*(\\.[az]{2,7})$/

You can use my answer as an example of how to add the url scheme group to your existing regex, but credit @simbabque, as he has a much more complete answer.