问题
I want to get domain name only using javascript. Ex
vn.search.yahoo.com -> yahoo
vn.search.yahoo.com.vn -> yahoo
sub1.sub2.sub3.abcdef.co.uk -> abcdef
Thank you!
Edit: "domain" = domain without extension (ex: .com, .net, .co.uk...) and without sub domain (ex: www, email, cdn, support...)
回答1:
Use location.host and cut off subdomains and the TLD:
var domain = (location.host.match(/([^.]+)\.\w{2,3}(?:\.\w{2})?$/) || [])[1]
update: as @demix pointed out, this fails for 2 and 3-letter domains. It also won't work for domains like aero, jobs and dozens others.
The only way around is to know valid TLDs in advance, so here is a more appropriate function:
// http://data.iana.org/TLD/tlds-alpha-by-domain.txt
var TLDs = ["ac", "ad", "ae", "aero", "af", "ag", "ai", "al", "am", "an", "ao", "aq", "ar", "arpa", "as", "asia", "at", "au", "aw", "ax", "az", "ba", "bb", "bd", "be", "bf", "bg", "bh", "bi", "biz", "bj", "bm", "bn", "bo", "br", "bs", "bt", "bv", "bw", "by", "bz", "ca", "cat", "cc", "cd", "cf", "cg", "ch", "ci", "ck", "cl", "cm", "cn", "co", "com", "coop", "cr", "cu", "cv", "cx", "cy", "cz", "de", "dj", "dk", "dm", "do", "dz", "ec", "edu", "ee", "eg", "er", "es", "et", "eu", "fi", "fj", "fk", "fm", "fo", "fr", "ga", "gb", "gd", "ge", "gf", "gg", "gh", "gi", "gl", "gm", "gn", "gov", "gp", "gq", "gr", "gs", "gt", "gu", "gw", "gy", "hk", "hm", "hn", "hr", "ht", "hu", "id", "ie", "il", "im", "in", "info", "int", "io", "iq", "ir", "is", "it", "je", "jm", "jo", "jobs", "jp", "ke", "kg", "kh", "ki", "km", "kn", "kp", "kr", "kw", "ky", "kz", "la", "lb", "lc", "li", "lk", "lr", "ls", "lt", "lu", "lv", "ly", "ma", "mc", "md", "me", "mg", "mh", "mil", "mk", "ml", "mm", "mn", "mo", "mobi", "mp", "mq", "mr", "ms", "mt", "mu", "museum", "mv", "mw", "mx", "my", "mz", "na", "name", "nc", "ne", "net", "nf", "ng", "ni", "nl", "no", "np", "nr", "nu", "nz", "om", "org", "pa", "pe", "pf", "pg", "ph", "pk", "pl", "pm", "pn", "pr", "pro", "ps", "pt", "pw", "py", "qa", "re", "ro", "rs", "ru", "rw", "sa", "sb", "sc", "sd", "se", "sg", "sh", "si", "sj", "sk", "sl", "sm", "sn", "so", "sr", "st", "su", "sv", "sy", "sz", "tc", "td", "tel", "tf", "tg", "th", "tj", "tk", "tl", "tm", "tn", "to", "tp", "tr", "travel", "tt", "tv", "tw", "tz", "ua", "ug", "uk", "us", "uy", "uz", "va", "vc", "ve", "vg", "vi", "vn", "vu", "wf", "ws", "xn--0zwm56d", "xn--11b5bs3a9aj6g", "xn--3e0b707e", "xn--45brj9c", "xn--80akhbyknj4f", "xn--90a3ac", "xn--9t4b11yi5a", "xn--clchc0ea0b2g2a9gcd", "xn--deba0ad", "xn--fiqs8s", "xn--fiqz9s", "xn--fpcrj9c3d", "xn--fzc2c9e2c", "xn--g6w251d", "xn--gecrj9c", "xn--h2brj9c", "xn--hgbk6aj7f53bba", "xn--hlcj6aya9esc7a", "xn--j6w193g", "xn--jxalpdlp", "xn--kgbechtv", "xn--kprw13d", "xn--kpry57d", "xn--lgbbat1ad8j", "xn--mgbaam7a8h", "xn--mgbayh7gpa", "xn--mgbbh1a71e", "xn--mgbc0a9azcg", "xn--mgberp4a5d4ar", "xn--o3cw4h", "xn--ogbpf8fl", "xn--p1ai", "xn--pgbs0dh", "xn--s9brj9c", "xn--wgbh1c", "xn--wgbl6a", "xn--xkc2al3hye2a", "xn--xkc2dl3a5ee0h", "xn--yfro4i67o", "xn--ygbi2ammx", "xn--zckzah", "xxx", "ye", "yt", "za", "zm", "zw"].join()
function getDomain(url){
var parts = url.split('.');
if (parts[0] === 'www' && parts[1] !== 'com'){
parts.shift()
}
var ln = parts.length
, i = ln
, minLength = parts[parts.length-1].length
, part
// iterate backwards
while(part = parts[--i]){
// stop when we find a non-TLD part
if (i === 0 // 'asia.com' (last remaining must be the SLD)
|| i < ln-2 // TLDs only span 2 levels
|| part.length < minLength // 'www.cn.com' (valid TLD as second-level domain)
|| TLDs.indexOf(part) < 0 // officialy not a TLD
){
return part
}
}
}
getDomain(location.host)
I hope I didn't miss too many corner cases. This should be available in the location object :(
Test cases: http://jsfiddle.net/hqBKd/4/
A list of TLDs can be found here: http://mxr.mozilla.org/mozilla-central/source/netwerk/dns/effective_tld_names.dat?raw=1
回答2:
I was looking for something that would work for the majority of cases, without having to maintain the TLD list (and skip it's size!). It seems to me that you can do this pretty accurately by looking instead at the Second-Level Domain for common ones:
function getDomainName(domain) {
var parts = domain.split('.').reverse();
var cnt = parts.length;
if (cnt >= 3) {
// see if the second level domain is a common SLD.
if (parts[1].match(/^(com|edu|gov|net|mil|org|nom|co|name|info|biz)$/i)) {
return parts[2] + '.' + parts[1] + '.' + parts[0];
}
}
return parts[1]+'.'+parts[0];
}
Fiddle & Tests @ http://jsfiddle.net/mZPaf/2/
Critiques/thoughts welcome.
回答3:
var docdomain = document.domain.split('.');
var dom1 = "";
if (typeof (docdomain[docdomain.length - 2]) != 'undefined') dom1 = docdomain[docdomain.length - 2] + '.';
var domain = dom1 + docdomain[docdomain.length - 1];
console.log(domain);
//without subdomains
回答4:
The only way I can imagine is list all the TLD. Sample code like below.
function getDomainName(){
var domainList = ['com','org','net',...];//all TLD
var tokens = document.domain.split('.');
while(tokens.length){
var token = tokens.pop();
if( domainList.indexOf(token) == -1 ){
return token;
}
}
return null;
}
Array.prototype.indexOf should do some fix in IE.
回答5:
Without having a complete list of TLD's (which would get very long). If you just need the domain name from the current page you can use my technique (using cookies to find the root domain)
Javascript - Get Domain Name Excluding Subdomain
To remove the extension you can then use the first element from a str.split('.')[0]
回答6:
i needed to do this and whipped up something simple that accounted for my use case
function stripSubDomainAndTLD (domain) {
return domain.replace(/^(?:[a-z0-9\-\.]+\.)??([a-z0-9\-]+)(?:\.com|\.net|\.org|\.biz|\.ws|\.in|\.me|\.co\.uk|\.co|\.org\.uk|\.ltd\.uk|\.plc\.uk|\.me\.uk|\.edu|\.mil|\.br\.com|\.cn\.com|\.eu\.com|\.hu\.com|\.no\.com|\.qc\.com|\.sa\.com|\.se\.com|\.se\.net|\.us\.com|\.uy\.com|\.ac|\.co\.ac|\.gv\.ac|\.or\.ac|\.ac\.ac|\.af|\.am|\.as|\.at|\.ac\.at|\.co\.at|\.gv\.at|\.or\.at|\.asn\.au|\.com\.au|\.edu\.au|\.org\.au|\.net\.au|\.id\.au|\.be|\.ac\.be|\.adm\.br|\.adv\.br|\.am\.br|\.arq\.br|\.art\.br|\.bio\.br|\.cng\.br|\.cnt\.br|\.com\.br|\.ecn\.br|\.eng\.br|\.esp\.br|\.etc\.br|\.eti\.br|\.fm\.br|\.fot\.br|\.fst\.br|\.g12\.br|\.gov\.br|\.ind\.br|\.inf\.br|\.jor\.br|\.lel\.br|\.med\.br|\.mil\.br|\.net\.br|\.nom\.br|\.ntr\.br|\.odo\.br|\.org\.br|\.ppg\.br|\.pro\.br|\.psc\.br|\.psi\.br|\.rec\.br|\.slg\.br|\.tmp\.br|\.tur\.br|\.tv\.br|\.vet\.br|\.zlg\.br|\.br|\.ab\.ca|\.bc\.ca|\.mb\.ca|\.nb\.ca|\.nf\.ca|\.ns\.ca|\.nt\.ca|\.on\.ca|\.pe\.ca|\.qc\.ca|\.sk\.ca|\.yk\.ca|\.ca|\.cc|\.ac\.cn|\.com\.cn|\.edu\.cn|\.gov\.cn|\.org\.cn|\.bj\.cn|\.sh\.cn|\.tj\.cn|\.cq\.cn|\.he\.cn|\.nm\.cn|\.ln\.cn|\.jl\.cn|\.hl\.cn|\.js\.cn|\.zj\.cn|\.ah\.cn|\.gd\.cn|\.gx\.cn|\.hi\.cn|\.sc\.cn|\.gz\.cn|\.yn\.cn|\.xz\.cn|\.sn\.cn|\.gs\.cn|\.qh\.cn|\.nx\.cn|\.xj\.cn|\.tw\.cn|\.hk\.cn|\.mo\.cn|\.cn|\.cx|\.cz|\.de|\.dk|\.fo|\.com\.ec|\.tm\.fr|\.com\.fr|\.asso\.fr|\.presse\.fr|\.fr|\.gf|\.gs|\.co\.il|\.net\.il|\.ac\.il|\.k12\.il|\.gov\.il|\.muni\.il|\.ac\.in|\.co\.in|\.org\.in|\.ernet\.in|\.gov\.in|\.net\.in|\.res\.in|\.is|\.it|\.ac\.jp|\.co\.jp|\.go\.jp|\.or\.jp|\.ne\.jp|\.ac\.kr|\.co\.kr|\.go\.kr|\.ne\.kr|\.nm\.kr|\.or\.kr|\.li|\.lt|\.lu|\.asso\.mc|\.tm\.mc|\.com\.mm|\.org\.mm|\.net\.mm|\.edu\.mm|\.gov\.mm|\.ms|\.nl|\.no|\.nu|\.pl|\.ro|\.org\.ro|\.store\.ro|\.tm\.ro|\.firm\.ro|\.www\.ro|\.arts\.ro|\.rec\.ro|\.info\.ro|\.nom\.ro|\.nt\.ro|\.se|\.si|\.com\.sg|\.org\.sg|\.net\.sg|\.gov\.sg|\.sk|\.st|\.tf|\.ac\.th|\.co\.th|\.go\.th|\.mi\.th|\.net\.th|\.or\.th|\.tm|\.to|\.com\.tr|\.edu\.tr|\.gov\.tr|\.k12\.tr|\.net\.tr|\.org\.tr|\.com\.tw|\.org\.tw|\.net\.tw|\.ac\.uk|\.uk\.com|\.uk\.net|\.gb\.com|\.gb\.net|\.vg|\.sh|\.kz|\.ch|\.info|\.ua|\.gov|\.name|\.pro|\.ie|\.hk|\.com\.hk|\.org\.hk|\.net\.hk|\.edu\.hk|\.us|\.tk|\.cd|\.by|\.ad|\.lv|\.eu\.lv|\.bz|\.es|\.jp|\.cl|\.ag|\.mobi|\.eu|\.co\.nz|\.org\.nz|\.net\.nz|\.maori\.nz|\.iwi\.nz|\.io|\.la|\.md|\.sc|\.sg|\.vc|\.tw|\.travel|\.my|\.se|\.tv|\.pt|\.com\.pt|\.edu\.pt|\.asia|\.fi|\.com\.ve|\.net\.ve|\.fi|\.org\.ve|\.web\.ve|\.info\.ve|\.co\.ve|\.tel|\.im|\.gr|\.ru|\.net\.ru|\.org\.ru|\.hr|\.com\.hr)$/, '$1');
}
mainly i just wanted to remove all subdomains, unfortunately this isn't 100% for some of the new TLD's but it works pretty well, and you can always add to the regex.
http://jsfiddle.net/icodeforlove/TzjJE/2/
回答7:
With the help of other friend's code examples given above I created a function which will return only domain name and if it is not valid domain for example TLD is missing then it will attach ".com" as per my requirement.
function getDomain(url){
var TLDs = ["ac", "ad", "ae", "aero", "af", "ag", "ai", "al", "am", "an", "ao", "aq", "ar", "arpa", "as", "asia", "at", "au", "aw", "ax", "az", "ba", "bb", "bd", "be", "bf", "bg", "bh", "bi", "biz", "bj", "bm", "bn", "bo", "br", "bs", "bt", "bv", "bw", "by", "bz", "ca", "cat", "cc", "cd", "cf", "cg", "ch", "ci", "ck", "cl", "cm", "cn", "co", "com", "coop", "cr", "cu", "cv", "cx", "cy", "cz", "de", "dj", "dk", "dm", "do", "dz", "ec", "edu", "ee", "eg", "er", "es", "et", "eu", "fi", "fj", "fk", "fm", "fo", "fr", "ga", "gb", "gd", "ge", "gf", "gg", "gh", "gi", "gl", "gm", "gn", "gov", "gp", "gq", "gr", "gs", "gt", "gu", "gw", "gy", "hk", "hm", "hn", "hr", "ht", "hu", "id", "ie", "il", "im", "in", "info", "int", "io", "iq", "ir", "is", "it", "je", "jm", "jo", "jobs", "jp", "ke", "kg", "kh", "ki", "km", "kn", "kp", "kr", "kw", "ky", "kz", "la", "lb", "lc", "li", "lk", "lr", "ls", "lt", "lu", "lv", "ly", "ma", "mc", "md", "me", "mg", "mh", "mil", "mk", "ml", "mm", "mn", "mo", "mobi", "mp", "mq", "mr", "ms", "mt", "mu", "museum", "mv", "mw", "mx", "my", "mz", "na", "name", "nc", "ne", "net", "nf", "ng", "ni", "nl", "no", "np", "nr", "nu", "nz", "om", "org", "pa", "pe", "pf", "pg", "ph", "pk", "pl", "pm", "pn", "pr", "pro", "ps", "pt", "pw", "py", "qa", "re", "ro", "rs", "ru", "rw", "sa", "sb", "sc", "sd", "se", "sg", "sh", "si", "sj", "sk", "sl", "sm", "sn", "so", "sr", "st", "su", "sv", "sy", "sz", "tc", "td", "tel", "tf", "tg", "th", "tj", "tk", "tl", "tm", "tn", "to", "tp", "tr", "travel", "tt", "tv", "tw", "tz", "ua", "ug", "uk", "us", "uy", "uz", "va", "vc", "ve", "vg", "vi", "vn", "vu", "wf", "ws", "xn--0zwm56d", "xn--11b5bs3a9aj6g", "xn--3e0b707e", "xn--45brj9c", "xn--80akhbyknj4f", "xn--90a3ac", "xn--9t4b11yi5a", "xn--clchc0ea0b2g2a9gcd", "xn--deba0ad", "xn--fiqs8s", "xn--fiqz9s", "xn--fpcrj9c3d", "xn--fzc2c9e2c", "xn--g6w251d", "xn--gecrj9c", "xn--h2brj9c", "xn--hgbk6aj7f53bba", "xn--hlcj6aya9esc7a", "xn--j6w193g", "xn--jxalpdlp", "xn--kgbechtv", "xn--kprw13d", "xn--kpry57d", "xn--lgbbat1ad8j", "xn--mgbaam7a8h", "xn--mgbayh7gpa", "xn--mgbbh1a71e", "xn--mgbc0a9azcg", "xn--mgberp4a5d4ar", "xn--o3cw4h", "xn--ogbpf8fl", "xn--p1ai", "xn--pgbs0dh", "xn--s9brj9c", "xn--wgbh1c", "xn--wgbl6a", "xn--xkc2al3hye2a", "xn--xkc2dl3a5ee0h", "xn--yfro4i67o", "xn--ygbi2ammx", "xn--zckzah", "xxx", "ye", "yt", "za", "zm", "zw"].join()
url = url.replace(/.*?:\/\//g, "");
url = url.replace(/www./g, "");
var parts = url.split('/');
url = parts[0];
var parts = url.split('.');
if (parts[0] === 'www' && parts[1] !== 'com'){
parts.shift()
}
var ln = parts.length
, i = ln
, minLength = parts[parts.length-1].length
, part
// iterate backwards
while(part = parts[--i]){
// stop when we find a non-TLD part
if (i === 0 // 'asia.com' (last remaining must be the SLD)
|| i < ln-2 // TLDs only span 2 levels
|| part.length < minLength // 'www.cn.com' (valid TLD as second-level domain)
|| TLDs.indexOf(part) < 0 // officialy not a TLD
){
var actual_domain = part;
break;
//return part
}
}
//console.log(actual_domain);
var tid ;
if(typeof parts[ln-1] != 'undefined' && TLDs.indexOf(parts[ln-1]) >= 0)
{
tid = '.'+parts[ln-1];
}
if(typeof parts[ln-2] != 'undefined' && TLDs.indexOf(parts[ln-2]) >= 0)
{
tid = '.'+parts[ln-2]+tid;
}
if(typeof tid != 'undefined')
actual_domain = actual_domain+tid;
else
actual_domain = actual_domain+'.com';
return actual_domain;
}
回答8:
It's simple:
var tokens = document.domain.split('.');
var domain = tokens[tokens.length - 2];
回答9:
function getDomainName( hostname ) {
var TLDs = new RegExp(/\.(com|net|org|biz|ltd|plc|edu|mil|asn|adm|adv|arq|art|bio|cng|cnt|ecn|eng|esp|etc|eti|fot|fst|g12|ind|inf|jor|lel|med|nom|ntr|odo|ppg|pro|psc|psi|rec|slg|tmp|tur|vet|zlg|asso|presse|k12|gov|muni|ernet|res|store|firm|arts|info|mobi|maori|iwi|travel|asia|web|tel)(\.[a-z]{2,3})?$|(\.[^\.]{2,3})(\.[^\.]{2,3})$|(\.[^\.]{2})$/);
return hostname.replace(TLDs, '').split('.').pop();
}
/* TEST */
var domains = [
'domain.com',
'subdomain.domain.com',
'www.subdomain.domain.com',
'www.subdomain.domain.info',
'www.subdomain.domain.info.xx',
'mail.subdomain.domain.co.uk',
'mail.subdomain.domain.xxx.yy',
'mail.subdomain.domain.xx.yyy',
'mail.subdomain.domain.xx',
'domain.xx'
];
var result = [];
for (var i = 0; i < domains.length; i++) {
result.push( getDomainName( domains[i] ) );
}
alert ( result.join(' | ') );
// result: domain | domain | domain | domain | domain | domain | domain | domain | domain | domain
回答10:
What about this?
function getDomain(){
if(document.domain.length){
var parts = document.domain.replace(/^(www\.)/,"").split('.');
//is there a subdomain?
while(parts.length > 2){
//removing it from our array
var subdomain = parts.shift();
}
//getting the remaining 2 elements
var domain = parts.join('.');
return domain.replace(/(^\.*)|(\.*$)/g, "");
}
return '';
}
回答11:
RegEx I use to get domain name only: ([^.]*[.]){0,}([^.]*)(\.[^.]*)
The domain can be found in the second part.
来源：https://stackoverflow.com/questions/8253136/how-to-get-domain-name-only-using-javascript