2016-08-21 274 views
-5

我怎么能找到一个字符串的IP地址一样,127.0.0.1Java在String中查找IP地址?

fdgdfgd127.0.0.1fhj 

但当然,这将是一个不同的IP地址

+2

使用[了'Pattern'类(https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html),并把[IPv4的匹配正则表达式] (https://www.safaribooksonline.com/library/view/regular-expressions-cookbook/9780596802837/ch07s16.html)在其中,voilà! – Jezor

回答

1
  String ip="raje127.0.0.1something"; 
      String IPADDRESS_PATTERN = 
      "^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." + 
      "([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." + 
      "([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." + 
      "([01]?\\d\\d?|2[0-4]\\d|25[0-5])$"; 

     public IPAddressValidator(){ 
      pattern = Pattern.compile(IPADDRESS_PATTERN); 
      Matcher matcher = pattern.matcher(ip); 
      return matcher.matches(); 

     } 
0

在这种情况下,你需要找到它包含四个正则表达式数字和三个点。每个点将在两个数字之间。阅读PatternRegex将帮助您解决您的问题。