如何正确处理subcribe()方法？

Question

我在代码中遇到了subscribe()方法的问题(调试控制台下面的消息)

io.reactivex.exceptions.OnErrorNotImplementedException: The exception was not handled due to missing onError handler in the subscribe() method call. Further reading: https://github.com/ReactiveX/RxJava/wiki/Error-Handling | Expected a string but was BEGIN_OBJECT at line 1 column 2 path $

我不知道如何使它正确，这是我的部分代码开始的地方

private fun startSearch(query: String) {
        disposables.addAll(IMyService.searchCourse(query)
            .subscribeOn(Schedulers.io())
            .observeOn(AndroidSchedulers.mainThread())
            .subscribe ({ courses ->
                adapter = CourseAdapter(baseContext, courses)
                recycler_search.adapter = adapter
            }, {
                Toast.makeText(this, "Not found", Toast.LENGTH_LONG).show()
            }))
    }

    private fun getAllCourses() {
        disposables.addAll(IMyService.coursesList
            .subscribeOn(Schedulers.io())
            .observeOn(AndroidSchedulers.mainThread())
            .subscribe ({ courses ->
                adapter = CourseAdapter(baseContext, courses)
                recycler_search.adapter = adapter
            }, {
                Toast.makeText(this, "Not found", Toast.LENGTH_LONG).show()
            }))
    }

还有完整的代码

参数

Answer 1

在反应性编程中，将订阅者传递给可观察的用户应包括如何处理以下三种情况：

1. onSuccess
2. onError
3. onFailure

但是，如果您只想传递一个订阅服务器，您知道它肯定不会有任何错误或任何故障，并且确定它总是成功的话，那么就试试@EpicPandaForce中提到的onSuccess或onFailure吧。然而，一个好的做法是始终实现这三种情况，因为你永远不知道。