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;
}
到此这篇关于Android判断是否Root方法介绍的文章就介绍到这了,更多相关Android判断Root内容请搜索易知道(ezd.cc)以前的文章或继续浏览下面的相关文章希望大家以后多多支持易知道(ezd.cc)！
