确定是否正在使用仿真器设备 [英] Determine if emulator device is in use

问题描述

有人知道hasClients在android sdk中的作用吗?

Does anyone know what the method hasClients does in the android sdk?

boolean com.android.ddmlib.IDevice.hasClients()

它看起来不像它记录的那样.

It doesn't look like its documented.

我正在尝试找到一种方法来查看是否正在使用模拟器.有什么好办法吗?

I am trying to find a way to see if an emulator is being used. Any good way to do this?

for(int i =0; i < devices.length; i++){

    if(!devices[i].hasClients()){

        monkeyDevice = devices[i];

    }

}

当我说正在使用时，我的意思是说设备上当前是否正在运行一个应用程序，或者该应用程序正在接收来自任何内容的命令.

When I say is being used, I mean if there is currently an application running on the device or if its receiving commands from anything.

我应该提到我想在我的应用程序中外部测试这些条件.我在应用程序外部运行了一个单独的类，该类在可用的模拟器中启动该应用程序.我想让这个猴子类知道现有的模拟器是否已经用于测试.

I should of mentioned that I want to test for these conditions outside from my application. I have a seperate class running outside the application that starts the application within an available emulator. I want this monkey class to know if an existing emulator is already being used for testing.

推荐答案

看看这个问题，弄清楚您是否正在模拟器中运行:
如何检测何时Android应用程序正在模拟器中运行?

Have a look at this question to figure out if you are running in the emulator:
How can I detect when an Android application is running in the emulator?

关于猴子，您可能想看看 Activity.isUserAMonkey()方法(自API级别8，OS 2.2起). Google DeviceAdminSample代码进行了简要说明:

On a monkey-related point, you might want to have a look at Activity.isUserAMonkey() method (since API level 8, OS 2.2). The Google DeviceAdminSample code gives a brief explanation:

/**
 * If the "user" is a monkey, post an alert and notify the caller.  This prevents automated
 * test frameworks from stumbling into annoying or dangerous operations.
 */
private static boolean alertIfMonkey(Context context, int stringId) {
    if (ActivityManager.isUserAMonkey()) {
        AlertDialog.Builder builder = new AlertDialog.Builder(context);
        builder.setMessage(stringId);
        builder.setPositiveButton(R.string.monkey_ok, null);
        builder.show();
        return true;
    } else {
        return false;
    }
}

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