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
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_]*)?$','i'); // fragment locator
return pattern.test(str);
}
其他回答
这里还有另一种方法。
// ***note***: if the incoming value is empty(""), the function returns true var elm; function isValidURL(u){ //A precaution/solution for the problem written in the ***note*** if(u!==""){ if(!elm){ elm = document.createElement('input'); elm.setAttribute('type', 'url'); } elm.value = u; return elm.validity.valid; } else{ return false } } console.log(isValidURL('')); console.log(isValidURL('http://www.google.com/')); console.log(isValidURL('//google.com')); console.log(isValidURL('google.com')); console.log(isValidURL('localhost:8000'));
一个有答案的相关问题
或者来自Devshed的Regexp:
function validURL(str) {
var pattern = new RegExp('^(https?:\\/\\/)?'+ // protocol
'((([a-z\\d]([a-z\\d-]*[a-z\\d])*)\\.)+[a-z]{2,}|'+ // domain name
'((\\d{1,3}\\.){3}\\d{1,3}))'+ // OR ip (v4) address
'(\\:\\d+)?(\\/[-a-z\\d%_.~+]*)*'+ // port and path
'(\\?[;&a-z\\d%_.~+=-]*)?'+ // query string
'(\\#[-a-z\\d_]*)?$','i'); // fragment locator
return !!pattern.test(str);
}
和我一起工作
function isURL(str) {
var regex = /(http|https):\/\/(\w+:{0,1}\w*)?(\S+)(:[0-9]+)?(\/|\/([\w#!:.?+=&%!\-\/]))?/;
var pattern = new RegExp(regex);
return pattern.test(str);
}
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);
}
您可以使用ajax请求来检查字符串是否有效的url和可访问的
(function() { $("input").change(function() { const check = $.ajax({ url : this.value, dataType: "jsonp" }); check.then(function() { console.log("Site is valid and registered"); }); //expected output check.catch(function(reason) { if(reason.status === 200) { return console.log("Site is valid and registered"); } console.log("Not a valid site"); }) }); })() <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <input type="text" placeholder="Please input url to check ? ">