对已接受答案的改进…

Check for ftp/ftps as protocol Has double escaping for backslashes (\\) Ensures that domains have a dot and an extension (.com .io .xyz) Allows full colon (:) in the path e.g. http://thingiverse.com/download:1894343 Allows ampersand (&) in path e.g http://en.wikipedia.org/wiki/Procter_&_Gamble Allows @ symbol in path e.g. https://medium.com/@techytimo isURL(str) { var pattern = new RegExp('^((ft|htt)ps?:\\/\\/)?'+ // protocol '((([a-z\\d]([a-z\\d-]*[a-z\\d])*)\\.)+[a-z]{2,}|'+ // domain name and extension '((\\d{1,3}\\.){3}\\d{1,3}))'+ // OR ip (v4) address '(\\:\\d+)?'+ // port '(\\/[-a-z\\d%@_.~+&:]*)*'+ // path '(\\?[;&a-z\\d%@_.,~+&:=-]*)?'+ // query string '(\\#[-a-z\\d_]*)?$','i'); // fragment locator return pattern.test(str); }

已经有很多答案了，但这里有另一个贡献: 直接从URL polyfill有效性检查中获取，使用type=" URL "的输入元素来利用浏览器内置的有效性检查:

var inputElement = doc.createElement('input');
inputElement.type = 'url';
inputElement.value = url;

if (!inputElement.checkValidity()) {
    throw new TypeError('Invalid URL');
}

源

使用javascript验证Url如下所示

function ValidURL(str) {
  var regex = /(?:https?):\/\/(\w+:?\w*)?(\S+)(:\d+)?(\/|\/([\w#!:.?+=&%!\-\/]))?/;
  if(!regex .test(str)) {
    alert("Please enter valid URL.");
    return false;
  } else {
    return true;
  }
}

我不能评论最接近#5717133的帖子，但下面是我想出如何让@tom-gullen正则表达式工作的方法。

/^(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

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'));

如果你可以改变输入类型，我认为这个解决方案会更容易:

你可以简单地在输入中使用type="url"，并在js中使用checkValidity()检查它

E.g:

your.html

<input id="foo" type="url">

your.js

// The selector is JQuery, but the function is plain JS
$("#foo").on("keyup", function() {
    if (this.checkValidity()) {
        // The url is valid
    } else {
        // The url is invalid
    }
});