在JavaScript中有没有办法检查一个字符串是否是URL?
RegExes被排除在外,因为它很可能写成stackoverflow
;也就是说,它可能没有.com
、www
或http
。
发布于 2011-04-19 21:29:35
一个有答案的相关问题:
或者来自Devshed的这个Regexp
function validURL(str) {
var pattern = new RegExp('^(https?:\\/\\/)?'+ // protocol
'((([a-z\\d]([a-z\\d-]*[a-z\\d])*)\\.)+[a-z]{2,}|'+ // domain name
'((\\d{1,3}\\.){3}\\d{1,3}))'+ // OR ip (v4) address
'(\\:\\d+)?(\\/[-a-z\\d%_.~+]*)*'+ // port and path
'(\\?[;&a-z\\d%_.~+=-]*)?'+ // query string
'(\\#[-a-z\\d_]*)?$','i'); // fragment locator
return !!pattern.test(str);
}
发布于 2013-01-29 19:50:37
function isURL(str) {
var pattern = new RegExp('^(https?:\\/\\/)?'+ // protocol
'((([a-z\\d]([a-z\\d-]*[a-z\\d])*)\\.?)+[a-z]{2,}|'+ // domain name
'((\\d{1,3}\\.){3}\\d{1,3}))'+ // OR ip (v4) address
'(\\:\\d+)?(\\/[-a-z\\d%_.~+]*)*'+ // port and path
'(\\?[;&a-z\\d%_.~+=-]*)?'+ // query string
'(\\#[-a-z\\d_]*)?$','i'); // fragment locator
return pattern.test(str);
}
发布于 2018-04-16 12:04:28
我正在使用下面的函数来验证带或不带http/https
的网址
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
https://stackoverflow.com/questions/5717093
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