我写了一个很棒的版本，你可以运行

它匹配以下url(这对我来说已经足够好了)

public static void main(args) {
    String url = "go to http://www.m.abut.ly/abc its awesome"
    url = url.replaceAll(/https?:\/\/w{0,3}\w*?\.(\w*?\.)?\w{2,3}\S*|www\.(\w*?\.)?\w*?\.\w{2,3}\S*|(\w*?\.)?\w*?\.\w{2,3}[\/\?]\S*/ , { it ->
        "woof${it}woof"
    })
    println url 
}

http://google.com
http://google.com/help.php
http://google.com/help.php?a=5

http://www.google.com
http://www.google.com/help.php
http://www.google.com?a=5

google.com?a=5
google.com/help.php
google.com/help.php?a=5

http://www.m.google.com/help.php?a=5 (and all its permutations)
www.m.google.com/help.php?a=5 (and all its permutations)
m.google.com/help.php?a=5 (and all its permutations)

对于任何不以http或www开头的url，重要的是它们必须包含/或?

我打赌这可以稍作调整，但它的工作非常好，因为它是如此简短和紧凑……因为你可以把它分成三份:

找到任何以http开头的内容:

https?:\/\/w{0,3}\w*?\.\w{2,3}\S*

找到任何以www开头的东西:

www\.\w*?\.\w{2,3}\S*

或者找到任何必须有一个文本，然后一个点，然后至少两个字母，然后一个?或/:

\w*?\.\w{2,3}[\/\?]\S*

这个怎么样:

^(https?:\/\/(?:www\.|(?!www))[a-zA-Z0-9][a-zA-Z0-9-]+[a-zA-Z0-9]\.[^\s]{2,}|www\.[a-zA-Z0-9][a-zA-Z0-9-]+[a-zA-Z0-9]\.[^\s]{2,}|https?:\/\/(?:www\.|(?!www))[a-zA-Z0-9]\.[^\s]{2,}|www\.[a-zA-Z0-9]\.[^\s]{2,})$

这些是测试用例:

你可以在这里试试:https://regex101.com/r/mS9gD7/41

一个简单的URL检查是

^(ftp|http|https):\/\/[^ "]+$

Javascript现在有一个名为new URL()的URL构造函数。它允许您完全跳过REGEX。

/** * * The URL() constructor returns a newly created URL object representing * the URL defined by the parameters. * * https://developer.mozilla.org/en-US/docs/Web/API/URL/URL * */ let requestUrl = new URL('https://username:password@developer.mozilla.org:8080/en-US/docs/search.html?par1=abc&par2=123&par3=true#Recent'); let urlParts = { origin: requestUrl.origin, href: requestUrl.href, protocol: requestUrl.protocol, username: requestUrl.username, password: requestUrl.password, host: requestUrl.host, hostname: requestUrl.hostname, port: requestUrl.port, pathname: requestUrl.pathname, search: requestUrl.search, searchParams: { par1: String(requestUrl.searchParams.get('par1')), par2: Number(requestUrl.searchParams.get('par2')), par3: Boolean(requestUrl.searchParams.get('par3')), }, hash: requestUrl.hash }; console.log(urlParts);

用这个吧，它对我有用

function validUrl(Url) {
    var myRegExp  =/^(?:(?:https?|ftp):\/\/)(?:\S+(?::\S*)?@)?(?:(?!10(?:\.\d{1,3}){3})(?!127(?:\.\d{1,3}){3})(?!169\.254(?:\.\d{1,3}){2})(?!192\.168(?:\.\d{1,3}){2})(?!172\.(?:1[6-9]|2\d|3[0-1])(?:\.\d{1,3}){2})(?:[1-9]\d?|1\d\d|2[01]\d|22[0-3])(?:\.(?:1?\d{1,2}|2[0-4]\d|25[0-5])){2}(?:\.(?:[1-9]\d?|1\d\d|2[0-4]\d|25[0-4]))|(?:(?:[a-z\u00a1-\uffff0-9]+-?)*[a-z\u00a1-\uffff0-9]+)(?:\.(?:[a-z\u00a1-\uffff0-9]+-?)*[a-z\u00a1-\uffff0-9]+)*(?:\.(?:[a-z\u00a1-\uffff]{2,})))(?::\d{2,5})?(?:\/[^\s]*)?$/i;

    if (!RegExp.test(Url.value)) {
        $("#urlErrorLbl").removeClass('highlightNew');
        return false;
    } 

    $("#urlErrorLbl").addClass('highlightNew'); 
    return true; 
}

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]*))?)/