Java获取URL中的顶级域名domain的工具类

方式一：

 1 import java.net.MalformedURLException;
 2 import java.net.URL;
 3 import java.util.Arrays;
 4 import java.util.HashSet;
 5 import java.util.Set;
 6 import java.util.regex.Pattern;
 7 
 8 public class URLUtil {
 9 
10     private final static Set<String> PublicSuffixSet = new HashSet<String>(
11             Arrays.asList(new String(
12                     "com|org|net|gov|edu|co|tv|mobi|info|asia|xxx|onion|cn|com.cn|edu.cn|gov.cn|net.cn|org.cn|jp|kr|tw|com.hk|hk|com.hk|org.hk|se|com.se|org.se")
13                     .split("\\|")));
14 
15     private static Pattern IP_PATTERN = Pattern.compile("(\\d{1,3}\\.){3}(\\d{1,3})");
16 
17     /**
18      * 获取url的顶级域名
19      * @param url
20      * @return
21      */
22     public static String getDomainName(URL url) {
23         String host = url.getHost();
24         if (host.endsWith(".")){
25             host = host.substring(0, host.length() - 1);
26         }
27         if (IP_PATTERN.matcher(host).matches()){
28             return host;
29         }
30 
31         int index = 0;
32         String candidate = host;
33         for (; index >= 0;) {
34             index = candidate.indexOf('.');
35             String subCandidate = candidate.substring(index + 1);
36             if (PublicSuffixSet.contains(subCandidate)) {
37                 return candidate;
38             }
39             candidate = subCandidate;
40         }
41         return candidate;
42     }
43 
44     /**
45      * 获取url的顶级域名
46      * @param url
47      * @return
48      * @throws MalformedURLException
49      */
50     public static String getDomainName(String url) throws MalformedURLException {
51         return getDomainName(new URL(url));
52     }
53 
54     /**
55      * 判断两个url顶级域名是否相等
56      * @param url1
57      * @param url2
58      * @return
59      */
60     public static boolean isSameDomainName(URL url1, URL url2) {
61         return getDomainName(url1).equalsIgnoreCase(getDomainName(url2));
62     }
63 
64     /**
65      * 判断两个url顶级域名是否相等
66      * @param url1
67      * @param url2
68      * @return
69      * @throws MalformedURLException
70      */
71     public static boolean isSameDomainName(String url1, String url2)
72             throws MalformedURLException {
73         return isSameDomainName(new URL(url1), new URL(url2));
74     }
75     
76     public static void main(String[] args) throws Exception {
77         String urlStr = "http://news.hexun.com/2017-09-23/190978248.html";
78         getDomainName(urlStr);
79         getDomainName(new URL(urlStr));
80     }
81 
82 }

方式二：

 1 import java.net.MalformedURLException;
 2 import java.net.URL;
 3 import java.util.regex.Matcher;
 4 import java.util.regex.Pattern;
 5 
 6 public class DomainUtils {
 7     /**
 8      * 获取url的顶级域名
 9      * @param
10      * @return
11      */
12     public static String getTopDomain(String url){
13         try{
14             //获取值转换为小写
15             String host = new URL(url).getHost().toLowerCase();//news.hexun.com
16             Pattern pattern = Pattern.compile("[^\\.]+(\\.com\\.cn|\\.net\\.cn|\\.org\\.cn|\\.gov\\.cn|\\.com|\\.net|\\.cn|\\.org|\\.cc|\\.me|\\.tel|\\.mobi|\\.asia|\\.biz|\\.info|\\.name|\\.tv|\\.hk|\\.公司|\\.中国|\\.网络)");
17             Matcher matcher = pattern.matcher(host);
18             while(matcher.find()){
19                 return matcher.group();
20             }
21         }catch(MalformedURLException e){
22             e.printStackTrace();
23         }
24         return null;
25     }
26     public static void main(String[] args) {
27         System.out.println(getTopDomain("http://news.hexun.com/2017-09-23/190978248.html"));//hexun.com
28 
29     }
30 }

 

posted @ 2017-09-25 22:25  SummerChill  阅读(6454)  评论(0)  编辑  收藏  举报