Question

完全披露，我还在学习RxJava，如果可用的许多教程都不是新手友好的话，有点难以理解这个想法。

此错误发生在API 16上，在API 23＆amp;以上（已在下面测试过）。

正如您将看到的，我尝试用RxJava替换异步任务。

这是我的代码：

private void getGps() {
    TrackGPS gps = new TrackGPS(this);
    Single.fromCallable(() -> {
        if (gps.canGetLocation()) {
            mMainVariables.setLongitude(gps.getLongitude());
            mMainVariables.setLatitude(gps.getLatitude());

            if (mMainVariables.getLongitude() != 0.0) {
                Geocoder geocoder;
                List<Address> addresses = null;

                geocoder = new Geocoder(this, Locale.getDefault());
                addresses = geocoder.getFromLocation(mMainVariables.getLatitude(), mMainVariables.getLongitude(), 1); // Here 1 represent max location result to returned, by documents it recommended 1 to 5
                Log.e("Address:", addresses.get(0).getAddressLine(0));
                mMainVariables.setAddress(addresses.get(0).getAddressLine(0));
                mMainVariables.setCity(addresses.get(0).getLocality());
                mMainVariables.setState(addresses.get(0).getAdminArea());
                mMainVariables.setCountry(addresses.get(0).getCountryName());
                mMainVariables.setPostalCode(addresses.get(0).getPostalCode());
                mMainVariables.setKnownName(addresses.get(0).getFeatureName());
                Log.d("Lat Long:", "Lat: " + Double.toString(mMainVariables.getLatitude()) + " Long: " + Double.toString(mMainVariables.getLongitude()));
                return addresses.get(0).getAddressLine(0);
            } else {
                gps.showSettingsAlert();
            }

        } else {
            Toasty.error(this, "Can't locate GPS", Toast.LENGTH_SHORT, true).show();
        }
        return "";
    }).subscribeOn(Schedulers.io())
            .observeOn(AndroidSchedulers.mainThread())
            .subscribe((result) -> {
                mTxtResult.setText(result);
            });
}

编辑：堆叠如下：

05-23 20:26:58.936 3420-3420/com.example.ga.realm3 E/AndroidRuntime: FATAL EXCEPTION: main
                                                                 io.reactivex.exceptions.OnErrorNotImplementedException: Can't create handler inside thread that has not called Looper.prepare()
                                                                     at io.reactivex.internal.functions.Functions$OnErrorMissingConsumer.accept(Functions.java:704)
                                                                     at io.reactivex.internal.functions.Functions$OnErrorMissingConsumer.accept(Functions.java:701)
                                                                     at io.reactivex.internal.observers.ConsumerSingleObserver.onError(ConsumerSingleObserver.java:45)
                                                                     at io.reactivex.internal.operators.single.SingleObserveOn$ObserveOnSingleObserver.run(SingleObserveOn.java:79)
                                                                     at io.reactivex.android.schedulers.HandlerScheduler$ScheduledRunnable.run(HandlerScheduler.java:109)
                                                                     at android.os.Handler.handleCallback(Handler.java:730)

Answer 1

您无法在Scheduler主题中显示与UI相关的内容。您正试图展示Toast，并且我还假设您的showSettingsAlert()也在尝试展示对话框。

这违反了线程政策。与

非常相似

Can't create handler inside thread that has not called Looper.prepare() inside AsyncTask for ProgressDialog

RxJava - 无法在未调用Looper.prepare（）的线程内创建处理程序 - API 16

1 个答案: