android – 用room执行删除(rxjava)

在室内@Delete注释不会发出任何东西.这就是dao的样子

@Dao
public interface UserDao {
    @Delete
    void deleteUser(User user);
    //We can't use Maybe or Single or anything here
}

这使得在做类似事情时出现问题

userRepository.deleteUser().subscribeOn,因为我们没有排放到dao.我使用以下代码在后台线程上调用deleteUser.

Observable.just(appDatabase).
            subscribeOn(SchedulerProvider.getInstance().computation()).
            subscribe(db -> {
                userRepository.logoutUser(loggedUser.getLoggedInUser());
                loggedUser.setLoggedInUser(null);
            },this::handleError);

这很好用.但是,在订阅方法中,我现在需要访问Android UI以显示宣布成功删除的Toast.当然,我得到这个例外(因为链中缺少observeOn)

java.lang.RuntimeException: Can't create handler inside thread that has not called Looper.prepare()

但是当我把observeOn放在这样的时候

Observable.just(appDatabase).
        subscribeOn(SchedulerProvider.getInstance().computation()).
        observeOn(SchedulerProvider.getInstance().ui()).
        subscribe(db -> {
            userRepository.logoutUser(loggedUser.getLoggedInUser());
            loggedUser.setLoggedInUser(null);
            Message message = new Message(R.string.user_logged_out_msg);
            message.setMessageType(Message.MessageType.SUCCESS_MESSAGE);
            view.showMessages(Arrays.asList(message)); //this leads to a taost
        },this::handleError);

我奇怪地得到这个例外：

cannot access database on the main thread since it may potentially lock the UI for a long period of time.

最佳答案

我知道这已经很晚了.但我遇到了同样的问题,并能够通过以下方式解决它

在道类

@Query("DELETE FROM users")
    fun deleteAllUser()

像这样称呼它.这样您就可以订阅它,它将在后台运行.

Single.fromCallable {
        user.deleteAllUser() //User Dao fun
    }
            .subscribeOn(Schedulers.io())
            .observeOn(AndroidSchedulers.mainThread())
            .subscribe(..do stuff..)

android – 用room执行删除(rxjava)

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