为了照顾那些着急的同学，先直接给出结论：
private static final String[] rootRelatedDirs = new String[]{
"/su", "/su/bin/su", "/sbin/su",
"/data/local/xbin/su", "/data/local/bin/su", "/data/local/su",
"/system/xbin/su",
"/system/bin/su", "/system/sd/xbin/su", "/system/bin/failsafe/su",
"/system/bin/cufsdosck", "/system/xbin/cufsdosck", "/system/bin/cufsmgr",
"/system/xbin/cufsmgr", "/system/bin/cufaevdd", "/system/xbin/cufaevdd",
"/system/bin/conbb", "/system/xbin/conbb"};
public static boolean hasRootPrivilege() {
boolean hasRootDir = false;
String[] rootDirs;
int dirCount = (rootDirs = rootRelatedDirs).length;
for (int i = 0; i < dirCount; ++i) {
String dir = rootDirs[i];
if ((new File(dir)).exists()) {
hasRootDir = true;
break;
}
}
return Build.TAGS != null && Build.TAGS.contains("test-keys") || hasRootDir;
}好，接下来我们来看看到底是如何得到上述的解决方案的。首先，这是既有的判断root权限的方案，即判定两个root权限相关文件夹是否存在，以及当前账户是否具备访问其内容的权限，如果都成立，那么就认为当前账号具备root权限。然而，这种root方案在一些情况下不能很好地发挥作用。
/**
* 判断Android设备是否拥有Root权限
*/
public class RootCheck {
private final static String TAG = "RootUtil";
public static boolean isRoot() {
String binPath = "/system/bin/su";
String xBinPath = "/system/xbin/su";
if (new File(binPath).exists() && isExecutable(binPath))
return true;
if (new File(xBinPath).exists() && isExecutable(xBinPath))
return true;
return false;
}
private static boolean isExecutable(String filePath) {
Process p = null;
try {
p = Runtime.getRuntime().exec("ls -l " + filePath);
// 获取返回内容
BufferedReader in = new BufferedReader(new InputStreamReader(p.getInputStream()));
String str = in.readLine();
Log.i(TAG, str);
if (str != null && str.length() >= 4) {
char flag = str.charAt(3);
if (flag == 's' || flag == 'x')
return true;
}
} catch (IOException e) {
e.printStackTrace();
} finally {
if (p != null) {
p.destroy();
}
}
return false;
}
}然后我就找到了如下方案，该方案号称是腾讯bugly的root权限判断方案：
private static final String[] a = new String[]{"/su", "/su/bin/su", "/sbin/su",
"/data/local/xbin/su", "/data/local/bin/su", "/data/local/su",
"/system/xbin/su", "/system/bin/su", "/system/sd/xbin/su",
"/system/bin/failsafe/su", "/system/bin/cufsdosck",
"/system/xbin/cufsdosck", "/system/bin/cufsmgr",
"/system/xbin/cufsmgr", "/system/bin/cufaevdd",
"/system/xbin/cufaevdd", "/system/bin/conbb",
"/system/xbin/conbb"};
public static boolean p() {
boolean var0 = false;
String[] var1 = a;
int var2 = a.length;
for(int var3 = 0; var3 < var2; ++var3) {
String var4 = var1[var3];
if ((new File(var4)).exists()) {
var0 = true;
break;
}
}
return Build.TAGS != null && Build.TAGS.contains("test-keys") || var0;
}当然，本人生性多疑，偶像是曹操曹丞相，所以自然不能人云亦云，还是实际确认一下bugly实际上是否是这样实现的，以及确保bugly在新的版本中有没有对该方案有进一步的改进。
然后我就到bugly官网，下载了其最新发布的jar包，笔者下载时最新的版本为4.4.4，然后直接解压，然后在解压的目录中搜索“test-keys”内容。
grep -r test-keys "D:\迅雷下载\Bugly_v3.4.4
最后找到了对应的文件位置和对应方法：com\tencent\bugly\crashreport\common\info\b.class
private static final String[] a = new String[]{"/su", "/su/bin/su",
"/sbin/su", "/data/local/xbin/su", "/data/local/bin/su",
"/data/local/su", "/system/xbin/su", "/system/bin/su",
"/system/sd/xbin/su", "/system/bin/failsafe/su",
"/system/bin/cufsdosck", "/system/xbin/cufsdosck",
"/system/bin/cufsmgr", "/system/xbin/cufsmgr",
"/system/bin/cufaevdd", "/system/xbin/cufaevdd",
"/system/bin/conbb", "/system/xbin/conbb"};
public static boolean l() {
boolean var0 = false;
String[] var1;
int var2 = (var1 = a).length;
for(int var3 = 0; var3 < var2; ++var3) {
String var4 = var1[var3];
if ((new File(var4)).exists()) {
var0 = true;
break;
}
}
return Build.TAGS != null && Build.TAGS.contains("test-keys") || var0;
}然后分析一下对应变量的意思，我们就能还原出腾讯判断Root的代码，即我们开头所贴出的解决方案：
private static final String[] rootRelatedDirs = new String[]{
"/su", "/su/bin/su", "/sbin/su",
"/data/local/xbin/su", "/data/local/bin/su", "/data/local/su",
"/system/xbin/su",
"/system/bin/su", "/system/sd/xbin/su", "/system/bin/failsafe/su",
"/system/bin/cufsdosck", "/system/xbin/cufsdosck", "/system/bin/cufsmgr",
"/system/xbin/cufsmgr", "/system/bin/cufaevdd", "/system/xbin/cufaevdd",
"/system/bin/conbb", "/system/xbin/conbb"};
public static boolean hasRootPrivilege() {
boolean hasRootDir = false;
String[] rootDirs;
int dirCount = (rootDirs = rootRelatedDirs).length;
for (int i = 0; i < dirCount; ++i) {
String dir = rootDirs[i];
if ((new File(dir)).exists()) {
hasRootDir = true;
break;
}
}
return Build.TAGS != null && Build.TAGS.contains("test-keys") || hasRootDir;
}
