blocks|key|414839|text|我写了一个方法(见下文)，它提取一个url的域名，并使用简单的字符串匹配。它实际做的是提取第一个"://"+(如果没有包含"://"，则为索引0+)和第一个后续"/"+(如果没有后续"/"，则为索引String.length()+)之间的位。剩余的、在前的"www(_)*."位被截断。我肯定会有这样做不够好的情况，但在大多数情况下应该足够好！|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|414840|Mike+Samuel在上面的帖子中说java.net.URI类可以做到这一点(并且比java.net.URL类更受欢迎)，但是我遇到了URI类的问题。值得注意的是，如果url不包括该方案，即"http(s)"比特，则URI.getHost()给出空值。|414841|/**
+*+Extracts+the+domain+name+from+{@code+url}
+*+by+means+of+String+manipulation
+*+rather+than+using+the+{@link+URI}+or+{@link+URL}+class.
+*
+*+@param+url+is+non-null.
+*+@return+the+domain+name+within+{@code+url}.
+*/
public+String+getUrlDomainName(String+url)+{
++String+domainName+=+new+String(url);

++int+index+=+domainName.indexOf("://");

++if+(index+!=+-1)+{
++++//+keep+everything+after+the+"://"
++++domainName+=+domainName.substring(index+%2B+3);
++}

++index+=+domainName.indexOf('/');

++if+(index+!=+-1)+{
++++//+keep+everything+before+the+'/'
++++domainName+=+domainName.substring(0,+index);
++}

++//+check+for+and+remove+a+preceding+'www'
++//+followed+by+any+sequence+of+characters+(non-greedy)
++//+followed+by+a+'.'
++//+from+the+beginning+of+the+string
++domainName+=+domainName.replaceFirst("%5Ewww.*?\\.",+"");

++return+domainName;
}|code-block|syntax|javascript|414842|entityMap^0|1C|5|1P|5|1Z|1|28|3|2J|3|2R|F|3K|A|0|J|C|17|C|1W|3|2O|9|31|D|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1E|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|N|$]]

I wrote a method (see below) which extracts a url's domain name and which uses simple String matching. What it actually does is extract the bit between the first <code>"://"</code> (or index <code>0</code> if there's no <code>"://"</code> contained) and the first subsequent <code>"/"</code> (or index <code>String.length()</code> if there's no subsequent <code>"/"</code>). The remaining, preceding <code>"www(_)*."</code> bit is chopped off. I'm sure there'll be cases where this won't be good enough but it should be good enough in most cases!

Mike Samuel's post above says that the <code>java.net.URI</code> class could do this (and was preferred to the <code>java.net.URL</code> class) but I encountered problems with the <code>URI</code> class. Notably, <code>URI.getHost()</code> gives a null value if the url does not include the scheme, i.e. the <code>"http(s)"</code> bit.

<pre><code>/**
 * Extracts the domain name from {@code url}
 * by means of String manipulation
 * rather than using the {@link URI} or {@link URL} class.
 *
 * @param url is non-null.
 * @return the domain name within {@code url}.
 */
public String getUrlDomainName(String url) {
 String domainName = new String(url);

 int index = domainName.indexOf("://");

 if (index != -1) {
 // keep everything after the "://"
 domainName = domainName.substring(index + 3);
 }

 index = domainName.indexOf('/');

 if (index != -1) {
 // keep everything before the '/'
 domainName = domainName.substring(0, index);
 }

 // check for and remove a preceding 'www'
 // followed by any sequence of characters (non-greedy)
 // followed by a '.'
 // from the beginning of the string
 domainName = domainName.replaceFirst("^www.*?\\.", "");

 return domainName;
}
</code></pre>

blocks|key|630008|text|import+java.net.*;
import+java.io.*;

public+class+ParseURL+{
++public+static+void+main(String[]+args)+throws+Exception+{

++++URL+aURL+=+new+URL("http://example.com:80/docs/books/tutorial"
+++++++++++++++++++++++%2B+"/index.html?name=networking#DOWNLOADING");

++++System.out.println("protocol+=+"+%2B+aURL.getProtocol());+//http
++++System.out.println("authority+=+"+%2B+aURL.getAuthority());+//example.com:80
++++System.out.println("host+=+"+%2B+aURL.getHost());+//example.com
++++System.out.println("port+=+"+%2B+aURL.getPort());+//80
++++System.out.println("path+=+"+%2B+aURL.getPath());+//++/docs/books/tutorial/index.html
++++System.out.println("query+=+"+%2B+aURL.getQuery());+//name=networking
++++System.out.println("filename+=+"+%2B+aURL.getFile());+///docs/books/tutorial/index.html?name=networking
++++System.out.println("ref+=+"+%2B+aURL.getRef());+//DOWNLOADING
++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|630009|Read+more|unstyled|offset|length|630010|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/tutorial/networking/urls/urlInfo.html^0|0|0|9|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code>import java.net.*;
import java.io.*;

public class ParseURL {
 public static void main(String[] args) throws Exception {

 URL aURL = new URL("http://example.com:80/docs/books/tutorial"
 + "/index.html?name=networking#DOWNLOADING");

 System.out.println("protocol = " + aURL.getProtocol()); //http
 System.out.println("authority = " + aURL.getAuthority()); //example.com:80
 System.out.println("host = " + aURL.getHost()); //example.com
 System.out.println("port = " + aURL.getPort()); //80
 System.out.println("path = " + aURL.getPath()); // /docs/books/tutorial/index.html
 System.out.println("query = " + aURL.getQuery()); //name=networking
 System.out.println("filename = " + aURL.getFile()); ///docs/books/tutorial/index.html?name=networking
 System.out.println("ref = " + aURL.getRef()); //DOWNLOADING
 }
}
</code></pre>

<a href="http://docs.oracle.com/javase/tutorial/networking/urls/urlInfo.html">Read more</a>

blocks|key|630041|text|在创建URI对象之后，我做了一个小处理|type|unstyled|depth|inlineStyleRanges|entityRanges|data|630042|+if+(url.startsWith("http:/"))+{
++++++++if+(!url.contains("http://"))+{
++++++++++++url+=+url.replaceAll("http:/",+"http://");
++++++++}
++++}+else+{
++++++++url+=+"http://"+%2B+url;
++++}
++++URI+uri+=+new+URI(url);
++++String+domain+=+uri.getHost();
++++return+domain.startsWith("www.")+?+domain.substring(4)+:+domain;|code-block|syntax|javascript|630043|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I made a small treatment after the URI object creation

<pre><code> if (url.startsWith("http:/")) {
 if (!url.contains("http://")) {
 url = url.replaceAll("http:/", "http://");
 }
 } else {
 url = "http://" + url;
 }
 URI uri = new URI(url);
 String domain = uri.getHost();
 return domain.startsWith("www.") ? domain.substring(4) : domain;
</code></pre>

blocks|key|414945|text|如果输入url是用户输入。此方法提供最合适的主机名。如果没有找到，返回输入的url。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|414946|private+String+getHostName(String+urlInput)+{
++++++++urlInput+=+urlInput.toLowerCase();
++++++++String+hostName=urlInput;
++++++++if(!urlInput.equals("")){
++++++++++++if(urlInput.startsWith("http")+%7C%7C+urlInput.startsWith("https")){
++++++++++++++++try{
++++++++++++++++++++URL+netUrl+=+new+URL(urlInput);
++++++++++++++++++++String+host=+netUrl.getHost();
++++++++++++++++++++if(host.startsWith("www")){
++++++++++++++++++++++++hostName+=+host.substring("www".length()%2B1);
++++++++++++++++++++}else{
++++++++++++++++++++++++hostName=host;
++++++++++++++++++++}
++++++++++++++++}catch+(MalformedURLException+e){
++++++++++++++++++++hostName=urlInput;
++++++++++++++++}
++++++++++++}else+if(urlInput.startsWith("www")){
++++++++++++++++hostName=urlInput.substring("www".length()%2B1);
++++++++++++}
++++++++++++return++hostName;
++++++++}else{
++++++++++++return++"";
++++++++}
++++}|code-block|syntax|javascript|414947|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If the input url is user input. this method gives the most appropriate host name. if not found gives back the input url.

<pre><code>private String getHostName(String urlInput) {
 urlInput = urlInput.toLowerCase();
 String hostName=urlInput;
 if(!urlInput.equals("")){
 if(urlInput.startsWith("http") || urlInput.startsWith("https")){
 try{
 URL netUrl = new URL(urlInput);
 String host= netUrl.getHost();
 if(host.startsWith("www")){
 hostName = host.substring("www".length()+1);
 }else{
 hostName=host;
 }
 }catch (MalformedURLException e){
 hostName=urlInput;
 }
 }else if(urlInput.startsWith("www")){
 hostName=urlInput.substring("www".length()+1);
 }
 return hostName;
 }else{
 return "";
 }
 }
</code></pre>

blocks|key|630138|text|试试这个:+java.net.URL；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|630139|JOptionPane.showMessageDialog(null，getDomainName(新的URL(“https://en.wikipedia.org/wiki/List_of_Internet_top-level_domains”)；|offset|length|630140|public+String+getDomainName(URL+url){
String+strDomain;
String[]+strhost+=+url.getHost().split(Pattern.quote("."));
String[]+strTLD+=+{"com","org","net","int","edu","gov","mil","arpa"};

if(Arrays.asList(strTLD).indexOf(strhost[strhost.length-1])>=0)
++++strDomain+=+strhost[strhost.length-2]%2B"."%2Bstrhost[strhost.length-1];
else+if(strhost.length>2)
++++strDomain+=+strhost[strhost.length-3]%2B"."%2Bstrhost[strhost.length-2]%2B"."%2Bstrhost[strhost.length-1];
else
++++strDomain+=+strhost[strhost.length-2]%2B"."%2Bstrhost[strhost.length-1];
return+strDomain;}|code-block|syntax|javascript|630141|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/List_of_Internet_top-level_domains^0|0|1K|1S|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@$D|U|E|V|1|W]]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

try this one : java.net.URL; 
JOptionPane.showMessageDialog(null, getDomainName(new URL("<a href="https://en.wikipedia.org/wiki/List_of_Internet_top-level_domains" rel="nofollow">https://en.wikipedia.org/wiki/List_of_Internet_top-level_domains</a>")));

<pre><code>public String getDomainName(URL url){
String strDomain;
String[] strhost = url.getHost().split(Pattern.quote("."));
String[] strTLD = {"com","org","net","int","edu","gov","mil","arpa"};

if(Arrays.asList(strTLD).indexOf(strhost[strhost.length-1])&gt;=0)
 strDomain = strhost[strhost.length-2]+"."+strhost[strhost.length-1];
else if(strhost.length&gt;2)
 strDomain = strhost[strhost.length-3]+"."+strhost[strhost.length-2]+"."+strhost[strhost.length-1];
else
 strDomain = strhost[strhost.length-2]+"."+strhost[strhost.length-1];
return strDomain;}
</code></pre>

blocks|key|415013|text|Extract+main+domain+name+from+a+given+url也有类似的问题。如果你看一下这个answer+，你会发现它非常简单。您只需要使用java.net.URL和String实用程序-+Split|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|415014|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/7217271/extract-main-domain-name-from-a-given-url/|1|https://stackoverflow.com/a/34466483/1808829^0|29|C|2M|6|2Y|5|0|15|0|1L|7|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@$9|W|A|X|1|Y]|$9|Z|A|10|1|11]]|E|$]]|$1|F|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]|N|$5|I|J|K|E|$L|O]]]]

There is a similar question <a href="https://stackoverflow.com/questions/7217271/extract-main-domain-name-from-a-given-url/">Extract main domain name from a given url</a>. If you take a look at this <a href="https://stackoverflow.com/a/34466483/1808829">answer </a>, you will see that it is very easy. You just need to use <code>java.net.URL</code> and <code>String</code> utility - <code>Split</code>

blocks|key|630242|text|在我的例子中，我只需要主域，而不需要子域(不需要"www“或任何子域)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|630243|public+static+String+getUrlDomain(String+url)+throws+URISyntaxException+{
++++URI+uri+=+new+URI(url);
++++String+domain+=+uri.getHost();
++++String[]+domainArray+=+domain.split("\\.");
++++if+(domainArray.length+==+1)+{
++++++++return+domainArray[0];
++++}
++++return+domainArray[domainArray.length+-+2]+%2B+"."+%2B+domainArray[domainArray.length+-+1];
}|code-block|syntax|javascript|630244|使用此方法，url+"https://rest.webtoapp.io/llSlider?lg=en&t=8“将具有域"webtoapp.io”的url。|offset|length|630245|entityMap|0|LINK|mutability|MUTABLE|url|https://rest.webtoapp.io/llSlider?lg=en&t=8^0|0|0|B|17|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

In my case i only needed the main domain and not the subdomain (no "www" or whatever the subdomain is) :

<pre><code>public static String getUrlDomain(String url) throws URISyntaxException {
 URI uri = new URI(url);
 String domain = uri.getHost();
 String[] domainArray = domain.split("\\.");
 if (domainArray.length == 1) {
 return domainArray[0];
 }
 return domainArray[domainArray.length - 2] + "." + domainArray[domainArray.length - 1];
}
</code></pre>

With this method the url "<a href="https://rest.webtoapp.io/llSlider?lg=en&amp;t=8" rel="nofollow noreferrer">https://rest.webtoapp.io/llSlider?lg=en&amp;t=8</a>" will have for domain "webtoapp.io".

blocks|key|630314|text|所有这些都是好的。这对我来说似乎真的很简单，很容易理解。请原谅我的引用。我是在一个名为DataCenter的类中为Groovy编写的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|630315|static+String+extractDomainName(String+url)+{
++++int+start+=+url.indexOf('://')
++++if+(start+<+0)+{
++++++++start+=+0
++++}+else+{
++++++++start+%2B=+3
++++}
++++int+end+=+url.indexOf('/',+start)
++++if+(end+<+0)+{
++++++++end+=+url.length()
++++}
++++String+domainName+=+url.substring(start,+end)

++++int+port+=+domainName.indexOf(':')
++++if+(port+>=+0)+{
++++++++domainName+=+domainName.substring(0,+port)
++++}
++++domainName
}|code-block|syntax|javascript|630316|下面是一些junit4测试：|630317|@Test
void+shouldFindDomainName()+{
++++assert+DataCenter.extractDomainName('http://example.com/path/')+==+'example.com'
++++assert+DataCenter.extractDomainName('http://subpart.example.com/path/')+==+'subpart.example.com'
++++assert+DataCenter.extractDomainName('http://example.com')+==+'example.com'
++++assert+DataCenter.extractDomainName('http://example.com:18445/path/')+==+'example.com'
++++assert+DataCenter.extractDomainName('example.com/path/')+==+'example.com'
++++assert+DataCenter.extractDomainName('example.com')+==+'example.com'
}|630318|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

All the above are good. This one seems really simple to me and easy to understand. Excuse the quotes. I wrote it for Groovy inside a class called DataCenter.

<pre><code>static String extractDomainName(String url) {
 int start = url.indexOf('://')
 if (start &lt; 0) {
 start = 0
 } else {
 start += 3
 }
 int end = url.indexOf('/', start)
 if (end &lt; 0) {
 end = url.length()
 }
 String domainName = url.substring(start, end)

 int port = domainName.indexOf(':')
 if (port &gt;= 0) {
 domainName = domainName.substring(0, port)
 }
 domainName
}
</code></pre>

And here are some junit4 tests:

<pre><code>@Test
void shouldFindDomainName() {
 assert DataCenter.extractDomainName('http://example.com/path/') == 'example.com'
 assert DataCenter.extractDomainName('http://subpart.example.com/path/') == 'subpart.example.com'
 assert DataCenter.extractDomainName('http://example.com') == 'example.com'
 assert DataCenter.extractDomainName('http://example.com:18445/path/') == 'example.com'
 assert DataCenter.extractDomainName('example.com/path/') == 'example.com'
 assert DataCenter.extractDomainName('example.com') == 'example.com'
}
</code></pre>

blocks|key|415249|text|我使用Guava库和regex的方法之一就是组合使用Guava+Library和regex。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|415250|public+static+String+getDomainNameWithGuava(String+url)+throws+MalformedURLException,+
++URISyntaxException+{
++++String+host+=new+URL(url).getHost();
++++String+domainName="";
++++try{
++++++++domainName+=+InternetDomainName.from(host).topPrivateDomain().toString();
++++}catch+(IllegalStateException+%7C+IllegalArgumentException+e){
++++++++domainName=+getDomain(url,true);
++++}
++++return+domainName;
}|code-block|syntax|javascript|415251|getDomain()可以是正则表达式的任何常见方法。|415252|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

One of the way I did and worked for all of the cases is using Guava Library and regex in combination.

<pre><code>public static String getDomainNameWithGuava(String url) throws MalformedURLException, 
 URISyntaxException {
 String host =new URL(url).getHost();
 String domainName="";
 try{
 domainName = InternetDomainName.from(host).topPrivateDomain().toString();
 }catch (IllegalStateException | IllegalArgumentException e){
 domainName= getDomain(url,true);
 }
 return domainName;
}
</code></pre>

getDomain() can be any common method with regex.

blocks|key|415335|text|为了获得实际的域名，不包括子域，我使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|415336|private+String+getDomainName(String+url)+throws+URISyntaxException+{
++++String+hostName+=+new+URI(url).getHost();
++++if+(!hostName.contains("."))+{
++++++++return+hostName;
++++}
++++String[]+host+=+hostName.split("\\.");
++++return+host[host.length+-+2];
}|code-block|syntax|javascript|415337|请注意，这不适用于二级域(如.co.uk)。|415338|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

To get the actual domain name, without the subdomain, I use:

<pre><code>private String getDomainName(String url) throws URISyntaxException {
 String hostName = new URI(url).getHost();
 if (!hostName.contains(".")) {
 return hostName;
 }
 String[] host = hostName.split("\\.");
 return host[host.length - 2];
}
</code></pre>

Note that this won't work with second-level domains (like .co.uk).

blocks|key|630484|text|val+host+=+url.split("/")[2]|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|630485|entityMap^0|0|S|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>val host = url.split(&quot;/&quot;)[2]</code>

blocks|key|630494|text|//+groovy
String+hostname+={url+->+url[(url.indexOf('://')%2B+3)..-1]​.split('/')[0]​+}

hostname('http://hello.world.com/something')+//+return+'hello.world.com'
hostname('docker://quay.io/skopeo/stable')+//+return+'quay.io'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|630495|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>// groovy
String hostname ={url -&gt; url[(url.indexOf('://')+ 3)..-1]​.split('/')[0]​ }

hostname('http://hello.world.com/something') // return 'hello.world.com'
hostname('docker://quay.io/skopeo/stable') // return 'quay.io'
</code></pre>

Given a URL, I want to extract domain name(It should not include 'www' part). Url can contain http/https. Here is the java code that I wrote. Though It seems to work fine, is there any better approach or are there some edge cases, that could fail.

<pre><code>public static String getDomainName(String url) throws MalformedURLException{
 if(!url.startsWith("http") &amp;&amp; !url.startsWith("https")){
 url = "http://" + url;
 } 
 URL netUrl = new URL(url);
 String host = netUrl.getHost();
 if(host.startsWith("www")){
 host = host.substring("www".length()+1);
 }
 return host;
}
</code></pre>

Input: <a href="http://google.com/blah" rel="noreferrer">http://google.com/blah</a>

Output: google.com

Get domain name from given url

Docker

给定一个URL，我想提取域名(它不应该包含'www‘部分)。Url可以包含http/https。这是我写的java代码。尽管它似乎工作得很好，但有没有更好的方法，或者有一些边缘情况，这可能会失败。public static String getDomainName(String url) throws Malforme...

问从给定的url获取域名
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