我用这个:/ ((https ?: \ / \ / | ftp: \ / \ / | www \) \ S + \[^。\ n ]+((?:\([^)]*\))|[^.,;:?!"'\ n \)\]<* ])+)/

它很短，但它处理了一些边缘情况，比如某些以括号结尾的维基百科链接(https://en.wikipedia.org/wiki/Sally_(name)，这里投票最多的答案似乎没有涵盖这一点。

        function validateURL(textval) {
            var urlregex = new RegExp(
            "^(http|https|ftp)\://([a-zA-Z0-9\.\-]+(\:[a-zA-Z0-9\.&%\$\-]+)*@)*((25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9])\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[0-9])|localhost|([a-zA-Z0-9\-]+\.)*[a-zA-Z0-9\-]+\.(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-zA-Z]{2}))(\:[0-9]+)*(/($|[a-zA-Z0-9\.\,\?\'\\\+&%\$#\=~_\-]+))*$");
            return urlregex.test(textval);
        }

匹配 http://site.com/dir/file.php?var=moo | ftp://user:pass@site.com:21/file/dir

Non-Matches site。com | http://site.com/dir//

我无法找到我正在寻找的正则表达式，所以我修改了一个正则表达式来满足我的要求，显然现在它似乎工作得很好。我的要求是:

匹配带有协议的url (www.gooogle.com) 使用查询参数和路径匹配url (http://subdomain.web-site.com/cgi-bin/perl.cgi?key1=value1&key2=value2e) 不要匹配有不可接受字符的url(例如。' '£)，例如:(www.google.com/somthing"/somethingmore)

以下是我的想法，任何建议都很感激:

@Test
    public void testWebsiteUrl(){
        String regularExpression = "((http|ftp|https):\\/\\/)?[\\w\\-_]+(\\.[\\w\\-_]+)+([\\w\\-\\.,@?^=%&:/~\\+#]*[\\w\\-\\@?^=%&/~\\+#])?";

        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));

        assertTrue("http://www.google.com/path".matches(regularExpression));
        assertTrue("http://subdomain.web-site.com/cgi-bin/perl.cgi?key1=value1&key2=value2e".matches(regularExpression));
        assertTrue("http://www.google.com/?queryparam=123".matches(regularExpression));
        assertTrue("http://www.google.com/path?queryparam=123".matches(regularExpression));

        assertFalse("www..dr.google".matches(regularExpression));

        assertFalse("www:google.com".matches(regularExpression));

        assertFalse("https://www@.google.com".matches(regularExpression));

        assertFalse("https://www.google.com\"".matches(regularExpression));
        assertFalse("https://www.google.com'".matches(regularExpression));

        assertFalse("http://www.google.com/path'".matches(regularExpression));
        assertFalse("http://subdomain.web-site.com/cgi-bin/perl.cgi?key1=value1&key2=value2e'".matches(regularExpression));
        assertFalse("http://www.google.com/?queryparam=123'".matches(regularExpression));
        assertFalse("http://www.google.com/path?queryparam=12'3".matches(regularExpression));

    }

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)
}

我发现下面的正则表达式的url，测试成功与500+ url:

/\b(?:(?: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\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)(?:\.(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)*(?:\.(?:[a-z\x{00a1}-\x{ffff}]{2,})))(?::\d{2,5})?(?:\/[^\s]*)?\b/gi

我知道这看起来很丑，但好处是它很有效。：）

regex101上581个随机url的解释和演示。

来源:在搜索完美的URL验证正则表达式

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 >