版本升级比较常用的接口，字符串解析，不是很难，但没必须重复造轮子，保存一份网上搜到的实现：

/*** 比较版本号的大小,前者大则返回一个正数,后者大返回一个负数,相等则返回0** @param version1* @param version2* @return*/public static int compareVersion(String version1, String version2) throws Exception {if (version1 == null || version2 == null) {throw new Exception("compareVersion error:illegal params.");}String[] versionArray1 = version1.split("\\.");//注意此处为正则匹配，不能用"."；String[] versionArray2 = version2.split("\\.");int idx = 0;int minLength = Math.min(versionArray1.length, versionArray2.length);//取最小长度值int diff = 0;while (idx < minLength&& (diff = versionArray1[idx].length() - versionArray2[idx].length()) == 0//先比较长度&& (diff = versionArray1[idx].compareTo(versionArray2[idx])) == 0) {//再比较字符++idx;}//如果已经分出大小，则直接返回，如果未分出大小，则再比较位数，有子版本的为大；diff = (diff != 0) ? diff : versionArray1.length - versionArray2.length;return diff;}

判断一个字符串是否符合mac地址规则：”78:DA:07:11:14:75”

 private boolean stringIsMac(String val) {String trueMacAddress = "([A-Fa-f0-9]{2}:){5}[A-Fa-f0-9]{2}";if (val.matches(trueMacAddress)) {return true;} else {return false;}}