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网上有很多区配IP的正则表达式,自己写了几个测试的小例子,总是出Bug,不能令人满意,所以就想自己写一个出来,研究了下,最终终于成功了,很简单,有测试用例,代码如下:
package com.autonavi.regex;import java.util.regex.Matcher;import java.util.regex.Pattern;public class TestRegex { public static boolean isIP(String ip) { String ipRegex = "(25[0-5]|2[0-4]\\d|1\\d{2}|[1-9][0-9]|[0-9])\\.(25[0-5]|2[0-4]\\d|1\\d{2}|[1-9][0-9]|[0-9])\\.(25[0-5]|2[0-4]\\d|1\\d{2}|[1-9][0-9]|[0-9])\\.(25[0-5]|2[0-4]\\d|1\\d{2}|[1-9][0-9]|[0-9])"; Pattern pattern = Pattern.compile(ipRegex); Matcher matcher = pattern.matcher(ip); return matcher.matches(); } public static void main(String[] args) { //true,正常,应该true String ip1 = "192.168.1.1"; if(isIP(ip1)) { System.out.println("IP1正确!"); } else { System.out.println("IP1无效!"); } //true,正常,应该true String ip2 = "255.255.255.255"; if(isIP(ip2)) { System.out.println("IP2正确!"); } else { System.out.println("IP2无效!"); } //false,正常,应该false String ip3 = "256.168.1.1"; if(isIP(ip3)) { System.out.println("IP3正确!"); } else { System.out.println("IP3无效!"); } //true,正常,应该true String ip4 = "0.168.1.1"; if(isIP(ip4)) { System.out.println("IP4正确!"); } else { System.out.println("IP4无效!"); } //true,异常,应该false String ip5 = "00.168.1.1"; if(isIP(ip5)) { System.out.println("IP5正确!"); } else { System.out.println("IP5无效!"); } //false,正常,应该false String ip6 = "000.168.1.1"; if(isIP(ip6)) { System.out.println("IP6正确!"); } else { System.out.println("IP6无效!"); } //true,异常,应该false String ip7 = "01.100.100.100"; if(isIP(ip7)) { System.out.println("IP7正确!"); } else { System.out.println("IP7无效!"); } //false,正常,,应该false String ip8 = "001.100.100.100"; if(isIP(ip8)) { System.out.println("IP8正确!"); } else { System.out.println("IP8无效!"); } //false,正常,应该false String ip9 = "010.100.100.100"; if(isIP(ip9)) { System.out.println("IP9正确!"); } else { System.out.println("IP9无效!"); } //false,正常,应该false String ip10 = "-10.100.100.100"; if(isIP(ip10)) { System.out.println("IP10正确!"); } else { System.out.println("IP10无效!"); } }}最终返回结果:
IP1正确!IP2正确!IP3无效!IP4正确!IP5无效!IP6无效!IP7无效!IP8无效!IP9无效!IP10无效!
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