对我来说，URL的最佳正则表达式是:

"(([\w]+:)?//)?(([\d\w]|%[a-fA-F\d]{2,2})+(:([\d\w]|%[a-fA-f\d]{2,2})+)?@)?([\d\w][-\d\w]{0,253}[\d\w]\.)+[\w]{2,4}(:[\d]+)?(/([-+_~.\d\w]|%[a-fA-f\d]{2,2})*)*(\?(&?([-+_~.\d\w]|%[a-fA-f\d]{2,2})=?)*)?(#([-+_~.\d\w]|%[a-fA-f\d]{2,2})*)?"

我希望这对你有帮助…

^(http|https):\/\/+[\www\d]+\.[\w]+(\/[\w\d]+)?

我想我找到了一个更通用的regexp来验证url，特别是网站

​(https?:\/\/)?(www\.)[-a-zA-Z0-9@:%._\+~#=]{2,256}\.[a-z]{2,4}\b([-a-zA-Z0-9@:%_\+.~#?&//=]*)|(https?:\/\/)?(www\.)?(?!ww)[-a-zA-Z0-9@:%._\+~#=]{2,256}\.[a-z]{2,4}\b([-a-zA-Z0-9@:%_\+.~#?&//=]*)

它不允许例如www.something或http://www或http://www.something

点击这里查看:http://regexr.com/3e4a2

我试着制定我的url版本。我的需求是在一个字符串中捕获实例，其中可能的url可以是cse.uom.ac.mu -注意它的前面没有http或www

String regularExpression = "((((ht{2}ps?://)?)((w{3}\\.)?))?)[^.&&[a-zA-Z0-9]][a-zA-Z0-9.-]+[^.&&[a-zA-Z0-9]](\\.[a-zA-Z]{2,3})";

assertTrue("www.google.com".matches(regularExpression));
assertTrue("www.google.co.uk".matches(regularExpression));
assertTrue("http://www.google.com".matches(regularExpression));
assertTrue("http://www.google.co.uk".matches(regularExpression));
assertTrue("https://www.google.com".matches(regularExpression));
assertTrue("https://www.google.co.uk".matches(regularExpression));
assertTrue("google.com".matches(regularExpression));
assertTrue("google.co.uk".matches(regularExpression));
assertTrue("google.mu".matches(regularExpression));
assertTrue("mes.intnet.mu".matches(regularExpression));
assertTrue("cse.uom.ac.mu".matches(regularExpression));

//cannot contain 2 '.' after www
assertFalse("www..dr.google".matches(regularExpression));

//cannot contain 2 '.' just before com
assertFalse("www.dr.google..com".matches(regularExpression));

// to test case where url www must be followed with a '.'
assertFalse("www:google.com".matches(regularExpression));

// to test case where url www must be followed with a '.'
//assertFalse("http://wwwe.google.com".matches(regularExpression));

// to test case where www must be preceded with a '.'
assertFalse("https://www@.google.com".matches(regularExpression));

我刚刚写了一篇博客文章，介绍了一个很好的解决方案，可以识别大多数常用格式的url，比如:

www.google.com http://www.google.com mailto: somebody@google.com somebody@google.com url www.url-with-querystring.com/ ? = has-querystring

使用的正则表达式是:

/((([A-Za-z]{3,9}:(?:\/\/)?)(?:[-;:&=\+\$,\w]+@)?[A-Za-z0-9.-]+|(?:www.|[-;:&=\+\$,\w]+@)[A-Za-z0-9.-]+)((?:\/[\+~%\/.\w-_]*)?\??(?:[-\+=&;%@.\w_]*)#?(?:[\w]*))?)/

如果你想采用更严格的规则，以下是我提出的:

isValidUrl(input) {
    var regex = /^(((H|h)(T|t)(T|t)(P|p)(S|s)?):\/\/)?[-a-zA-Z0-9@:%._\+~#=]{2,100}\.[a-zA-Z]{2,10}(\/([-a-zA-Z0-9@:%_\+.~#?&//=]*))?/
    return regex.test(input)
}