是否可以检测用户是否安装并使用Google健身应用程序？

Question

我想为我的应用程序添加步骤计数器。 现在我使用Google fit来衡量用户步数。

我面临的问题是，并非所有设备都安装了Google Fit应用程序，即使这样做 - 并非所有用户都注册到应用程序服务（作为客户端）。

所以所需流量是：

1. 检测是否安装Google fit应用程序。
2. 如果已安装，请检查用户是否注册了Google fit应用程序。
3. 如果已安装并注册到Google fit应用程序，请检查用户帐户类型支持是否使用健身服务。 （记录步骤等数据）
4. 如果以上都好，请检查用户是否确认了Google适配器弹出窗口。

基本上我想如果用户使用健身应用（使用上述所有条件）来检查，如果一个失败，则它将使用在设备中的传感器STEPCOUNT（如果存在的话），如果传感器不存在它将使用其他传感器来实现这一目标。

下面是我使用与谷歌飞度API连接代码：

private void buildFitnessClient() { 
    // Create the Google API Client 
    mClient = new GoogleApiClient.Builder(getContext()) 
      .addApi(Fitness.HISTORY_API) 
      .addScope(new Scope(Scopes.FITNESS_ACTIVITY_READ)) 
      .addConnectionCallbacks(
        new GoogleApiClient.ConnectionCallbacks() { 

         @Override 
         public void onConnected(Bundle bundle) { 
          Log.i(TAG, "Connected!!!"); 
          // Now you can make calls to the Fitness APIs. 
          // Put application specific code here. 


          new getFitnessTask().execute(); 
         } 


         @Override 
         public void onConnectionSuspended(int i) { 
          // If your connection to the sensor gets lost at some point, 
          // you'll be able to determine the reason and react to it here. 
          if (i == GoogleApiClient.ConnectionCallbacks.CAUSE_NETWORK_LOST) { 
           Log.i(TAG, "Connection lost. Cause: Network Lost."); 
          } else if (i == GoogleApiClient.ConnectionCallbacks.CAUSE_SERVICE_DISCONNECTED) { 
           Log.i(TAG, "Connection lost. Reason: Service Disconnected"); 
          } 
         } 
        } 
      ) 
      .addOnConnectionFailedListener(
        new GoogleApiClient.OnConnectionFailedListener() { 
         // Called whenever the API client fails to connect. 
         @Override 
         public void onConnectionFailed(ConnectionResult result) { 
          Log.i(TAG, "Connection failed. Cause: " + result.toString()); 
          if (!result.hasResolution()) { 
           // Show the localized error dialog 
           if (getActivity() != null) { 

            GooglePlayServicesUtil.getErrorDialog(result.getErrorCode(), 
              getActivity(), 0).show(); 
           } 
           return; 
          } 
          // The failure has a resolution. Resolve it. 
          // Called typically when the app is not yet authorized, and an 
          // authorization dialog is displayed to the user. 
          if (!authInProgress) { 

           try { 
            Log.i(TAG, "Attempting to resolve failed connection; activity; "+ getActivity()); 
            if (getActivity() != null) { 

             authInProgress = true; 
             result.startResolutionForResult(getActivity(), 
               REQUEST_OAUTH); 
            } 
           } catch (IntentSender.SendIntentException e) { 
            Log.e(TAG, 
              "Exception while starting resolution activity", e); 
           } 
          } 
         } 
        } 
      ) 
      .build(); 
} 

public void onActivityResult(int requestCode, int resultCode, Intent data){ 
    Log.e(TAG, "onActivityResult requestCode=" + requestCode + ", resultCode=" + resultCode); 
    if (requestCode == REQUEST_OAUTH) { 
     authInProgress = false; 
     if (resultCode == Activity.RESULT_OK) { 

      Log.e(TAG, "RESULT_OK"); 
      // Make sure the app is not already connected or attempting to connect 
      if (!mClient.isConnecting() && !mClient.isConnected()) { 
       mClient.connect(); 
      } 
     } else if (resultCode == Activity.RESULT_CANCELED) { 
      Log.e(TAG, "RESULT_CANCELED"); 
     } 
    } 
} 


private class getFitnessTask extends AsyncTask<Void, Void, Integer> { 
    ... 
} 
} 

需要您的帮助球员，

感谢。

Answer 1

您可以拨打PackageManager致电getPackageInfo()。如果它抛出PackageManager.NameNotFoundException这意味着该应用程序未安装。如果它不抛出异常，谷歌适合安装

private static final String PACKAGE_NAME = "com.google.android.apps.fitness"; 

@CheckResult public boolean fitInstalled() { 
    try { 
     context.getPackageManager().getPackageInfo(PACKAGE_NAME, PackageManager.GET_ACTIVITIES); 
     return true; 
    } catch (PackageManager.NameNotFoundException e) { 
     return false; 
    } 
}