检查JavaScript字符串是否是URL

在JavaScript中是否有一种方法来检查字符串是否是URL?

regex被排除在外，因为URL很可能写成stackoverflow;也就是说，它可能没有。com, WWW或http。

当前回答

我已经修改了所有的评论，注释和备注是这个主题，并做了一个新的正则表达式:

^((javascript:[\w-_]+(\([\w-_\s,.]*\))?)|(mailto:([\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+\.)*[\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+@([\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+\.)*[\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+)|(\w+:\/\/(([\w\u00C0-\u1FFF\u2C00-\uD7FF-]+\.)*([\w\u00C0-\u1FFF\u2C00-\uD7FF-]*\.?))(:\d+)?(((\/[^\s#$%^&*?]+)+|\/)(\?[\w\u00C0-\u1FFF\u2C00-\uD7FF:;&%_,.~+=-]+)?)?(#[\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+)?))$

你可以在这里测试和改进它https://regexr.com/668mt。

我检查了下一个值的表达式:

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&docid=nIv5rk2GyP3hXM&tbnid=isiOkMe3nCtexM:&ved=0CAUQjRw&url=http%3A%2F%2Fanimalcrossing.wikia.com%2Fwiki%2FLion&ei=ygZXU_2fGKbMsQTf4YLgAQ&bvm=bv.65177938,d.aWc&psig=AFQjCNEpBfKnal9kU7Zu4n7RnEt2nerN4g&ust=1398298682009707
http://192.168.0.4:55/
https://web.archive.org/web/20170817095211/https://github.com/Microsoft/vscode/issues/32405
http://www.example.com
javascript:void()
http://.
https://example.
https://en.m.wikipedia.org/wiki/C_Sharp_(programming_language)
http://zh.wikipedia.org/wiki/Wikipedia:关于中文维基百科/en?a#a
https://medium.com/@User_name/
https://test-test-test-test-test-test-test-test-test.web.app/
http://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&docid=nIv5rk2GyP3hXM&tbnid=isiOkMe3nCtexM:&ved=0CAUQjRw&url=http%3A%2F%2Fanimalcrossing.wikia.com%2Fwiki%2FLion&ei=ygZXU_2fGKbMsQTf4YLgAQ&bvm=bv.65177938,d.aWc&psig=AFQjCNEpBfKnal9kU7Zu4n7RnEt2nerN4g&ust=1398298682009707
https://sdfasdp.ppppppppppp
mailto:sadf@gmail.com
https://тест.юа

2021-09-24 12:42:11

其他回答

2020年更新。为了扩展@iamnewton和@ fernando Chavez Herrera的精彩回答，我已经开始看到@被用于url的路径。

所以更新后的正则表达式是:

RegExp('(https?:\\/\\/)?((([a-z\\d]([a-z\\d-]*[a-z\\d])*)\\.)+[a-z]{2,}|((\\d{1,3}\\.){3}\\d{1,3}))(\\:\\d+)?(\\/[-a-z\\d%_.~+@]*)*(\\?[;&a-z\\d%_.~+=-]*)?(\\#[-a-z\\d_]*)?$', 'i');

如果你想在查询字符串和哈希中允许它，使用:

RegExp('(https?:\\/\\/)?((([a-z\\d]([a-z\\d-]*[a-z\\d])*)\\.)+[a-z]{2,}|((\\d{1,3}\\.){3}\\d{1,3}))(\\:\\d+)?(\\/[-a-z\\d%_.~+@]*)*(\\?[;&a-z\\d%_.~+=-@]*)?(\\#[-a-z\\d_@]*)?$', 'i');

话虽如此，我不确定是否有白皮书规则禁止在查询字符串或哈希中使用@。

2020-05-18 18:26:32

使用纯正则表达式很难做到这一点，因为url有很多“不方便”的地方。

For example domain names have complicated restrictions on hyphens: a. It is allowed to have many consecutive hyphens in the middle. b. but the first character and last character of the domain name cannot be a hyphen c. The 3rd and 4th character cannot be both hyphen Similarly port number can only be in the range 1-65535. This is easy to check if you extract the port part and convert to int but quite difficult to check with a regular expression. There is also no easy way to check valid domain extensions. Some countries have second-level domains(such as 'co.uk'), or the extension can be a long word such as '.international'. And new TLDs are added regularly. This type of things can only be checked against a hard-coded list. (see https://en.wikipedia.org/wiki/Top-level_domain) Then there are magnet urls, ftp addresses etc. These all have different requirements.

然而，这里有一个函数可以处理几乎所有的事情，除了:

案例1。c 接受任何1-5位数的端口号接受任何扩展2-13个字符不接受ftp，磁铁等…

2019-02-10 19:51:48

我已经修改了所有的评论，注释和备注是这个主题，并做了一个新的正则表达式:

^((javascript:[\w-_]+(\([\w-_\s,.]*\))?)|(mailto:([\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+\.)*[\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+@([\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+\.)*[\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+)|(\w+:\/\/(([\w\u00C0-\u1FFF\u2C00-\uD7FF-]+\.)*([\w\u00C0-\u1FFF\u2C00-\uD7FF-]*\.?))(:\d+)?(((\/[^\s#$%^&*?]+)+|\/)(\?[\w\u00C0-\u1FFF\u2C00-\uD7FF:;&%_,.~+=-]+)?)?(#[\w\u00C0-\u1FFF\u2C00-\uD7FF-_]+)?))$

你可以在这里测试和改进它https://regexr.com/668mt。

我检查了下一个值的表达式:

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&docid=nIv5rk2GyP3hXM&tbnid=isiOkMe3nCtexM:&ved=0CAUQjRw&url=http%3A%2F%2Fanimalcrossing.wikia.com%2Fwiki%2FLion&ei=ygZXU_2fGKbMsQTf4YLgAQ&bvm=bv.65177938,d.aWc&psig=AFQjCNEpBfKnal9kU7Zu4n7RnEt2nerN4g&ust=1398298682009707
http://192.168.0.4:55/
https://web.archive.org/web/20170817095211/https://github.com/Microsoft/vscode/issues/32405
http://www.example.com
javascript:void()
http://.
https://example.
https://en.m.wikipedia.org/wiki/C_Sharp_(programming_language)
http://zh.wikipedia.org/wiki/Wikipedia:关于中文维基百科/en?a#a
https://medium.com/@User_name/
https://test-test-test-test-test-test-test-test-test.web.app/
http://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&docid=nIv5rk2GyP3hXM&tbnid=isiOkMe3nCtexM:&ved=0CAUQjRw&url=http%3A%2F%2Fanimalcrossing.wikia.com%2Fwiki%2FLion&ei=ygZXU_2fGKbMsQTf4YLgAQ&bvm=bv.65177938,d.aWc&psig=AFQjCNEpBfKnal9kU7Zu4n7RnEt2nerN4g&ust=1398298682009707
https://sdfasdp.ppppppppppp
mailto:sadf@gmail.com
https://тест.юа

2021-09-24 12:42:11

我建议使用锚元素，而不是使用正则表达式。

当你设置一个锚的href属性时，其他各种属性也会被设置。

var parser = document.createElement('a');
parser.href = "http://example.com:3000/pathname/?search=test#hash";

parser.protocol; // => "http:"
parser.hostname; // => "example.com"
parser.port;     // => "3000"
parser.pathname; // => "/pathname/"
parser.search;   // => "?search=test"
parser.hash;     // => "#hash"
parser.host;     // => "example.com:3000"

源

但是，如果href绑定的值不是一个有效的url，那么这些辅助属性的值将是空字符串。

编辑:正如评论中指出的:如果使用了无效的url，则可以替换当前url的属性。

所以，只要你没有传递当前页面的URL，你可以这样做:

function isValidURL(str) {
   var a  = document.createElement('a');
   a.href = str;
   return (a.host && a.host != window.location.host);
}

2016-01-09 15:00:57

这里还有另一种方法。

// ***note***: if the incoming value is empty(""), the function returns true var elm; function isValidURL(u){ //A precaution/solution for the problem written in the ***note*** if(u!==""){ if(!elm){ elm = document.createElement('input'); elm.setAttribute('type', 'url'); } elm.value = u; return elm.validity.valid; } else{ return false } } console.log(isValidURL('')); console.log(isValidURL('http://www.google.com/')); console.log(isValidURL('//google.com')); console.log(isValidURL('google.com')); console.log(isValidURL('localhost:8000'));

2018-04-18 22:24:17

检查JavaScript字符串是否是URL

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