如何检查给定的字符串是否是有效的URL地址?

我对正则表达式的知识是基本的,不允许我从我已经在网上看到的数百个正则表达式中进行选择。


当前回答

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));

    }

其他回答

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; 
}

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 >

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

检查URL正则表达式将是:

^http(s{0,1})://[a-zA-Z0-9_/\\-\\.]+\\.([A-Za-z/]{2,5})[a-zA-Z0-9_/\\&\\?\\=\\-\\.\\~\\%]*