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。
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
这显然不是最有效的方法,但它是可读的,并且很容易形成您需要的任何形式。从这里添加regex/complexity更容易。这是一个非常实用的方法
const validFirstBits = ["ftp://", "http://", "https://", "www."];
const invalidPatterns = [" ", "//.", ".."];
export function isUrl(word) {
// less than www.1.dk
if (!word || word.length < 8) return false;
// Let's check and see, if our candidate starts with some of our valid first bits
const firstBitIsValid = validFirstBits.some(bit => word.indexOf(bit) === 0);
if (!firstBitIsValid) return false;
const hasInvalidPatterns = invalidPatterns.some(
pattern => word.indexOf(pattern) !== -1,
);
if (hasInvalidPatterns) return false;
const dotSplit = word.split(".");
if (dotSplit.length > 1) {
const lastBit = dotSplit.pop(); // string or undefined
if (!lastBit) return false;
const length = lastBit.length;
const lastBitIsValid =
length > 1 || (length === 1 && !isNaN(parseInt(lastBit)));
return !!lastBitIsValid;
}
return false;
}
测试:
import { isUrl } from "./foo";
describe("Foo", () => {
test("should validate correct urls correctly", function() {
const validUrls = [
"http://example.com",
"http://example.com/blah",
"http://127.0.0.1",
"http://127.0.0.1/wow",
"https://example.com",
"https://example.com/blah",
"https://127.0.0.1:1234",
"ftp://example.com",
"ftp://example.com/blah",
"ftp://127.0.0.1",
"www.example.com",
"www.example.com/blah",
];
validUrls.forEach(url => {
expect(isUrl(url) && url).toEqual(url);
});
});
test("should validate invalid urls correctly", function() {
const inValidUrls = [
"http:// foo.com",
"http:/foo.com",
"http://.foo.com",
"http://foo..com",
"http://.com",
"http://foo",
"http://foo.c",
];
inValidUrls.forEach(url => {
expect(!isUrl(url) && url).toEqual(url);
});
});
});
其他回答
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;
}
}
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_]*)?$','i'); // fragment locator return pattern.test(str); }
这似乎是CS中最难的问题之一;)
这是另一个不完整的解决方案,它对我来说足够好,比我在这里看到的其他解决方案更好。为了支持IE11,我使用了一个输入[type=url],否则使用window会简单得多。URL来执行验证:
const ipv4Regex = /^(\d{1,3}\.){3}\d{1,3}$/; function isValidIpv4(ip) { if (!ipv4Regex.test(ip)) return false; return !ip.split('.').find(n => n > 255); } const domainRegex = /(?:[a-z0-9-]{1,63}\.){1,125}[a-z]{2,63}$/i; function isValidDomain(domain) { return isValidIpv4(domain) || domainRegex.test(domain); } let input; function validateUrl(url) { if (! /^https?:\/\//.test(url)) url = `http://${url}`; // assuming Babel is used // to support IE11 we'll resort to input[type=url] instead of window.URL: // try { return isValidDomain(new URL(url).host) && url; } catch(e) { return false; } if (!input) { input = document.createElement('input'); input.type = 'url'; } input.value = url; if (! input.validity.valid) return false; const domain = url.split(/^https?:\/\//)[1].split('/')[0].split('@').pop(); return isValidDomain(domain) && url; } console.log(validateUrl('google'), // false validateUrl('user:pw@mydomain.com'), validateUrl('https://google.com'), validateUrl('100.100.100.100/abc'), validateUrl('100.100.100.256/abc')); // false
为了接受不完整的输入,例如“www.mydomain.com”,它还将使其有效,假设在这些情况下协议是“http”,如果地址有效,则返回有效的URL。无效时返回false。
它还支持IPv4域,但不支持IPv6域。
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'));
我使用下面的函数来验证URL是否有http/https:
function isValidURL(string) { var res = string.match(/(http(s)?:\/\/.)?(www\.)?[-a-zA-Z0-9@:%._\+~#=]{2,256}\.[a-z]{2,6}\b([-a-zA-Z0-9@:%_\+.~#?&//=]*)/g); return (res !== null) }; var testCase1 = "http://en.wikipedia.org/wiki/Procter_&_Gamble"; console.log(isValidURL(testCase1)); // return true var testCase2 = "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"; console.log(isValidURL(testCase2)); // return true var testCase3 = "https://sdfasd"; console.log(isValidURL(testCase3)); // return false var testCase4 = "dfdsfdsfdfdsfsdfs"; console.log(isValidURL(testCase4)); // return false var testCase5 = "magnet:?xt=urn:btih:123"; console.log(isValidURL(testCase5)); // return false var testCase6 = "https://stackoverflow.com/"; console.log(isValidURL(testCase6)); // return true var testCase7 = "https://w"; console.log(isValidURL(testCase7)); // return false var testCase8 = "https://sdfasdp.ppppppppppp"; console.log(isValidURL(testCase8)); // return false