java如何处理SyncOnSubscribe RxJava中generateState中的错误?
我试图掌握创建SyncOnSubscribe的诀窍,但不太确定如果generateState
方法在状态初始化期间失败,该怎么办
return Observable.create(new SyncOnSubscribe<MyState, String>() {
@Override
protected MyState generateState() {
return new MyState(); // <---- what if this fails?
}
@Override
protected MyState next(MyState state, Observer<? super String> observer) {
// do something with state
}
});
我可以想出几个方法来处理这个问题:
- 如果引发运行时异常,库是否会自动
打电话给
o.onError
?(参见编辑)李> - 我可以将
MyState
包装在另一个存储错误的变量中,并且我可以在第一次调用next
时自己调用o.onError
李>
我只是好奇有没有一个建议的做法
谢谢
编辑:我尝试在generateState
方法中抛出一个运行时异常,我认为它指向我用一个错误包装MyState
,我将在next
方法中检查这个错误。如果你有更好的建议,请评论/回答
public static Observable<String> getEventsOnSubscribe1() {
return Observable.create((s) -> {
throw new UnsupportedOperationException("getEvents3");
});
};
/** Like 'getEventsOnSubscribe1' but wrap exception and call onError manually. */
public static Observable<String> getEventsOnSubscribe2() {
return Observable.create((s) -> {
try {
throw new UnsupportedOperationException("getEvents3");
} catch (Exception ex) {
s.onError(ex);
}
});
};
public static Observable<String> getEventsSyncOnSubscribe() {
return Observable.create(new SyncOnSubscribe<Channel, String>() {
@Override
protected Channel generateState() {
System.out.println("SyncOnSubscribe.generateState");
throw new UnsupportedOperationException("Not supported yet.");
}
@Override
protected Channel next(Channel state, Observer<? super String> observer) {
System.out.println("SyncOnSubscribe.next");
throw new UnsupportedOperationException("Not supported yet.");
}
@Override
protected void onUnsubscribe(Channel state) {
System.out.println("SyncOnSubscribe.onUnsubscribe");
throw new UnsupportedOperationException("Not supported yet.");
}
});
}
public static void main(String[] args) throws IOException, TimeoutException {
getEventsOnSubscribe1()
//getEventsOnSubscribe2()
//getEventsOnSyncSubscribe()
.toBlocking()
.subscribe(new Subscriber<String>() {
@Override
public void onCompleted() {
System.out.println("onCompleted");
}
@Override
public void onError(Throwable e) {
System.out.println("onError: " + e.getLocalizedMessage());
}
@Override
public void onNext(String t) {
System.out.println("onNext: " + t);
}
});
}
使用上面的main
函数getEventsOnSubscribe1
和getEventsSyncOnSubscribe
都会调用订户onError
,但它们会让程序挂起。使用getEventsOnSubscribe2
包装异常并手动调用s.onError
,程序可以退出
共 (0) 个答案