How can I detect when an Android application is ru

Question

I would like to have my code run slightly differently when running on the emulator than when running on a device. (For example, using 10.0.2.2 instead of a public URL to run against a development server automatically.) What is the best way to detect when an Android application is running in the emulator?

Answer 1

One common one sems to be Build.FINGERPRINT.contains("generic")

Answer 2

Here is my solution (it works only if you run a web server on your debug machine): I have created a background task that starts when the application starts. It looks for http://10.0.2.2 and if it exists it changes a global parameter (IsDebug) to true. It is a silent way to find out where you are running.

public class CheckDebugModeTask extends AsyncTask<String, Void, String> {
public static boolean IsDebug = false;

public CheckDebugModeTask()
{

}

@Override
protected String doInBackground(String... params) {     
  try {
    HttpParams httpParameters = new BasicHttpParams();
    int timeoutConnection = 1000;
    HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);
    int timeoutSocket = 2000;
    HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);

    String url2 = "http://10.0.2.2";        
          HttpGet httpGet = new HttpGet(url2);
    DefaultHttpClient client = new DefaultHttpClient(httpParameters);

    HttpResponse response2 = client.execute(httpGet);
    if (response2 == null || response2.getEntity() == null || response2.getEntity().getContent() == null)
    return "";

    return "Debug";

} catch (Exception e) {
    return "";
}
}

@Override
protected void onPostExecute (String result)
{       
if (result == "Debug")
{
    CheckDebugModeTask.IsDebug = true;
}
}

from the main activity onCreate:

CheckDebugModeTask checkDebugMode = new CheckDebugModeTask();
checkDebugMode.execute("");

Answer 3

if (Build.BRAND.equalsIgnoreCase("generic")) {
    // Is the emulator
}

All BUILD references are build.prop values, so you have to consider that if you are going to put this into release code, you may have some users with root that have modified theirs for whatever reason. There are virtually no modifications that require using generic as the brand unless specifically trying to emulate the emulator.

Fingerprint is the build compile and kernel compile signature. There are builds that use generic, usually directly sourced from Google.

On a device that has been modified, the IMEI has a possibility of being zeroed out as well, so that is unreliable unless you are blocking modified devices altogether.

Goldfish is the base android build that all other devices are extended from. EVERY Android device has an init.goldfish.rc unless hacked and removed for unknown reasons.

Answer 4

Both the following are set to "google_sdk":

Build.PRODUCT
Build.MODEL

So it should be enough to use either one of the following lines.

"google_sdk".equals(Build.MODEL)

or

"google_sdk".equals(Build.PRODUCT)

Answer 5

Checking the answers, none of them worked when using LeapDroid, Droid4x or Andy emulators,

What does work for all cases is the following:

 private static String getSystemProperty(String name) throws Exception {
    Class systemPropertyClazz = Class.forName("android.os.SystemProperties");
    return (String) systemPropertyClazz.getMethod("get", new Class[]{String.class}).invoke(systemPropertyClazz, new Object[]{name});
}

public boolean isEmulator() {
    boolean goldfish = getSystemProperty("ro.hardware").contains("goldfish");
    boolean emu = getSystemProperty("ro.kernel.qemu").length() > 0;
    boolean sdk = getSystemProperty("ro.product.model").equals("sdk");
    return goldfish || emu || sdk;
}

Answer 6

Well Android id does not work for me, I'm currently using:

"google_sdk".equals( Build.PRODUCT );