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;
}
}