function validateURL(textval) {
            var urlregex = new RegExp(
            "^(http|https|ftp)\://[a-zA-Z0-9\-\.]+\.[a-zA-Z]{2,3}(:[a-zA-Z0-9]*)?/?([a-zA-Z0-9\-\._\?\,\'/\\\+&%\$#\=~])*$");
            return urlregex.test(textval);
        }

匹配 http://www.asdah.com/~joe | ftp://ftp.asdah.co.uk:2828/asdah%20asdah.gif | https://asdah.gov/asdh-ah.as

Regardless the broad question asked, I post this for anyone in the future who is looking for something simple... as I think validating a URL has no perfect regular expression that fit all needs, it depends on your requirements, i.e: in my case, I just needed to verify if a URL is in the form of domain.extension and I wanted to allow the www or any other subdomain like blog.domain.extension I don't care about http(s) as in my app I have a field which says "enter the URL" so it's obvious what that entered string is.

这是regEx:

/^(www\.|[a-zA-Z0-9](.*[a-zA-Z0-9])?\.)?((?!www)[a-zA-Z0-9][a-zA-Z0-9-]+[a-zA-Z0-9])\.[a-z]{2,5}(:[0-9]{1,5})?$/i

这个regExp中的第一个块是:

(www \ | [a-zA-Z0-9] (. * [a-zA-Z0-9]) ? \。)?——>我们开始检查URL是否以www开头。[a-zA-Z0-9]或[a-zA-Z0-9] (. *) ?这意味着一个字母或数字+ (anyCharacter(0或多次)+另一个字母或数字)，然后是一个点

注意(.*[a-zA-Z0-9])?\.)?我们翻译由(anyCharacter(0或多次)+另一个字母或数字) 是可选的(可以是或不是)这就是为什么我们将它分组在括号之间，后面跟着问号?

到目前为止我们讨论的整个块也放在括号之间，后面跟着?这意味着WWW或任何其他词(表示子域)都是可选的。

第二部分是:((? ! www) [a-zA-Z0-9] [a-zA-Z0-9 -] + [a-zA-Z0-9]) \。——>表示“域”部分，它可以是任何单词(www除外)，以字母或数字开头+任何其他字母(包括破折号“-”)重复一次或多次，以任何字母或数字结尾，后面跟一个点。

最后一部分是[a-z]{2，}——>，它代表“扩展名”，它可以是任何字母重复2次或更多次，所以它可以是com, net, org, art基本上任何扩展名

Mathias Bynens有一篇关于大量正则表达式的最佳比较的很棒的文章:寻找完美的URL验证正则表达式

最好的一个帖子有点长，但它几乎可以匹配任何你可以扔给它的东西。

JavaScript版本

/^(?:(?:(?:https?|ftp):)?\/\/)(?:\S+(?::\S*)?@)?(?:(?!(?:10|127)(?:\.\d{1,3}){3})(?!(?:169\.254|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-z0-9\u00a1-\uffff][a-z0-9\u00a1-\uffff_-]{0,62})?[a-z0-9\u00a1-\uffff]\.)+(?:[a-z\u00a1-\uffff]{2,}\.?))(?::\d{2,5})?(?:[/?#]\S*)?$/i

PHP版本(使用%符号作为分隔符)

%^(?:(?:(?:https?|ftp):)?\/\/)(?:\S+(?::\S*)?@)?(?:(?!(?:10|127)(?:\.\d{1,3}){3})(?!(?:169\.254|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-z0-9\x{00a1}-\x{ffff}][a-z0-9\x{00a1}-\x{ffff}_-]{0,62})?[a-z0-9\x{00a1}-\x{ffff}]\.)+(?:[a-z\x{00a1}-\x{ffff}]{2,}\.?))(?::\d{2,5})?(?:[/?#]\S*)?$%iuS

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);

经过严格的搜索，我最终确定如下

^[a-zA-Z0-9]+\:\/\/[a-zA-Z0-9]+\.[-a-zA-Z0-9]+\.?[a-zA-Z0-9]+$|^[a-zA-Z0-9]+\.[-a-zA-Z0-9]+\.[a-zA-Z0-9]+$

这个在未来的url中也适用。

我认为这是一个非常简单的方法。效果非常好。

var hasURL = (str) =>{ var url_pattern = new RegExp(”(www。| | | http:// https:// ftp://) \ w *”); 如果(! url_pattern.test (str)) { . getelementbyid(“演示”)。innerHTML = '没有URL'; ｝ 其他的 . getelementbyid(“演示”)。innerHTML = '字符串有一个URL'; }; <p>请输入一个字符串并测试它是否有任何url <input type="text" id="url" placeholder="url" onkeyup="hasURL(document.getElementById("url").value)" > < p id = "演示" > < / p >