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IP 正则表达式有关问题

发布时间: 2012-01-19 00:22:28 作者: rapoo

IP 正则表达式问题
String pattern = "\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}";
String value = "192.168.4.35"

Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(value);

这样能match,


String pattern = "\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}";
String value = "asdad 192.168.4.35 asdada"

Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(value);

不match?

[解决办法]
var sIPAddress = form.appIp.value;
var exp=/^(\d{1,2}|1\d\d|2[0-4]\d|25[0-5])\.(\d{1,2}|1\d\d|2[0-4]\d|25[0-5])\.(\d{1,2}|1\d\d|2[0-4]\d|25[0-5])\.(\d{1,2}|1\d\d|2[0-4]\d|25[0-5])$/;
var reg = sIPAddress.match(exp);
var ErrMsg="你输入的是一个非法的IP地址段!\nIP段为::xxx.xxx.xxx.xxx(xxx为0-255)!";
[解决办法]

Java code
String pattern = "\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}"; String value = "asdad 192.168.4.35 asdada";Pattern p = Pattern.compile(pattern); Matcher m = p.matcher(value);if(m.find()) {   System.out.println("success.");}
[解决办法]
Java code
(".*?\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}.*?"))
[解决办法]
楼上说的是不是用这个。
Java code
Pattern = Pattern.compile(".*?((25[0-5]|2[0-4]\\d|1\\d{2}|[1-9]\\d|\\d)\\.){3}(25[0-5]|2[0-4]\\d|1\\d{2}|[1-9]\\d|\\d).*?"); 

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