Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
234 views
in Technique[技术] by (71.8m points)

Determining if an Android device is rooted programmatically?


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Rooting detection is a cat and mouse game and it is hard to make rooting detection that will work on all devices for all cases.

See Android Root Beer https://github.com/scottyab/rootbeer for advanced root detection which also uses JNI and native CPP code compiled into .so native library.

If you need some simple and basic rooting detection check the code below:

  /**
   * Checks if the device is rooted.
   *
   * @return <code>true</code> if the device is rooted, <code>false</code> otherwise.
   */
  public static boolean isRooted() {

    // get from build info
    String buildTags = android.os.Build.TAGS;
    if (buildTags != null && buildTags.contains("test-keys")) {
      return true;
    }

    // check if /system/app/Superuser.apk is present
    try {
      File file = new File("/system/app/Superuser.apk");
      if (file.exists()) {
        return true;
      }
    } catch (Exception e1) {
      // ignore
    }

    // try executing commands
    return canExecuteCommand("/system/xbin/which su")
        || canExecuteCommand("/system/bin/which su") || canExecuteCommand("which su");
  }

  // executes a command on the system
  private static boolean canExecuteCommand(String command) {
    boolean executedSuccesfully;
    try {
      Runtime.getRuntime().exec(command);
      executedSuccesfully = true;
    } catch (Exception e) {
      executedSuccesfully = false;
    }

    return executedSuccesfully;
  }

Probably not always correct. Tested on ~10 devices in 2014.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...