Question

我正在使用文件共享应用程序。我需要通过调用WifiManager.startLocalOnlyHotspot（）以编程方式打开设备的仅本地热点。

根据此页面上的android文档-https://developer.android.com/reference/android/net/wifi/WifiManager#startLocalOnlyHotspot(android.net.wifi.WifiManager.LocalOnlyHotspotCallback,%2520android.os.Handler)，

应用程序需要具有以下权限才能启动LocalOnlyHotspot：Manifest.permission.CHANGE_WIFI_STATE和ACCESS_FINE_LOCATION。没有权限的调用者将触发SecurityException。

我已经在清单中添加了这些权限，还在运行时授予了ACCESS_FINE_LOCATION（因为这是运行时权限（危险权限））。

但是在某些设备中调用startLocalOnlyHotspot（）仍会引发SecurityException。
引发SecurityException的设备：Samsung Galaxy J7 Max（j7maxlte），Android 8.1 运行正常且不会引发异常的设备：Redmi Note 7 Pro，Android 9 PKQ1.181203.001

我想念什么？

Answer 1

您需要使用GPS启用位置。

首先，添加gms的依赖项。

implementation 'com.google.android.gms:play-services-location:16.0.0'

然后，设置位置服务

  private void setupLocationServices() {
    LocationRequest mLocationRequest = new LocationRequest();
    mLocationRequest.setInterval(10);
    mLocationRequest.setSmallestDisplacement(10);
    mLocationRequest.setFastestInterval(10);
    mLocationRequest.setPriority(LocationRequest.PRIORITY_HIGH_ACCURACY);
    LocationSettingsRequest.Builder builder = new
        LocationSettingsRequest.Builder();
    builder.addLocationRequest(mLocationRequest);

    task = LocationServices.getSettingsClient(this).checkLocationSettings(builder.build());

    locationSettingsResponseBuilder();
  }

LocationSettingsResponseBuilder：

  private void locationSettingsResponseBuilder() {
    task.addOnCompleteListener(task -> {
      try {
        LocationSettingsResponse response = task.getResult(ApiException.class);
        // All location settings are satisfied. The client can initialize location
        // requests here.

        //Everything is good. You can turn on hotspot here.

        //}
      } catch (ApiException exception) {
        switch (exception.getStatusCode()) {
          case LocationSettingsStatusCodes.RESOLUTION_REQUIRED:
            // Location settings are not satisfied. But could be fixed by showing the
            // user a dialog.
            try {
              // Cast to a resolvable exception.
              ResolvableApiException resolvable = (ResolvableApiException) exception;
              // Show the dialog by calling startResolutionForResult(),
              // and check the result in onActivityResult().
              resolvable.startResolutionForResult(
                  MainActivity.this,
                  101);
            } catch (IntentSender.SendIntentException e) {
              // Ignore the error.
            } catch (ClassCastException e) {
              // Ignore, should be an impossible error.
            }
            break;
          case LocationSettingsStatusCodes.SETTINGS_CHANGE_UNAVAILABLE:
            // Location settings are not satisfied. However, we have no way to fix the
            // settings so we won't show the dialog.
            break;
        }
      }
    });
  }

将此案例添加到 onActivityResult 方法。

      case 101:
        final LocationSettingsStates states = LocationSettingsStates.fromIntent(data);
        switch (resultCode) {
          case Activity.RESULT_OK:
            // All required changes were successfully made
            Log.v("case 101", states.isLocationPresent() + "");

            //Everything is good. You can turn on hotspot here.

            break;
          case Activity.RESULT_CANCELED:
            // The user was asked to change settings, but chose not to
            Log.v("case 101", "Canceled");
            break;
          default:
            break;
        }
        break;

我最近创建了一个名为Spotserve的演示应用程序。这会为所有API> = 15的设备打开wifi热点，并在该热点上托管一个演示服务器。您可以检查更多细节。希望这会有所帮助！

甚至在授予某些设备上的所有必需权限后，调用WifiManager.startLocalOnlyHotspot（）也会引发SecurityException

1 个答案: