javascript - 正则表达式匹配字符串直到空格 Javascript

Question

我希望能够匹配下面的例子:

www.example.com
http://example.com
https://example.com

我有以下正则表达式，它不匹配 www. 但会匹配 http:// https://。我需要匹配上面示例中的任何前缀，直到下一个空格，即整个 URL。

var regx = ((\s))(http?:\/\/)|(https?:\/\/)|(www\.)(?=\s{1});

假设我有一个如下所示的字符串:

我从 www.stackoverflow.com 和那里的人那里找到了很多帮助!

我想在那个字符串上运行匹配并得到

www.stackoverflow.com

谢谢!

Answer 1

你可以试试

(?:www|https?)[^\s]+

这里是 online demo

示例代码:

var str="I have found a lot of help off www.stackoverflow.com and the people on there!";
var found=str.match(/(?:www|https?)[^\s]+/gi);
alert(found);

图案解释:

  (?:                      group, but do not capture:
    www                      'www'
   |                        OR
    http                     'http'
    s?                       's' (optional)
  )                        end of grouping
  [^\s]+                   any character except: whitespace 
                            (\n, \r, \t, \f, and " ") (1 or more times)