有没有办法可以在下面的代码中从userObservable返回用户列表。基本上如何读取Observable的对象值。
public List<User> getUserList(){
Observable<Map<String, Object>> userDetailsObservable = getUserDetails(...);
Observable<Map<String, Object>> userLikesObservable = getUserLikes(...);
Observable<List<User>> userObservable = Observable.zip(userDetailsObservable,
userLikesObservable, new Func2<Map<String, Object>, Map<String, Object>, List<User>>() {
public List<User> call(Map<String, Object> value1, Map<String, Object> value2) {
List<User> userList = new ArrayList<User>();
//Iterating both maps and composing user list here.
return userList;
}
});
userObservable.subscribe(new Subscriber<List<User>>() {
@Override
public void onNext(List<User> userList) {
System.out.println("Merged values " + userList);
}
@Override
public void onError(Throwable error) {
System.err.println("Error: " + error.getMessage());
}
@Override
public void onCompleted() {
System.out.println("Sequence complete.");
}
});
// Throws ClassCastException which is expected... How do I get the user list from userObservable
List<User> userList = List<User> userObservable;
return userList; //Fetch the user list from userObservable and return
}
答案 0 :(得分:11)
您需要订阅userObservable
。然后,您将获得数据可用时发出的用户列表。
userObservable
.subscribe(users -> {
for (User user : users) {
// Do what you want with user.
}
});
编辑:要返回列表,您需要使用toBlocking()
来电。然后,您可以使用其中一个Blocking Observable Operators来获得所需的效果。最有可能first()
或single()
适合您。
return userObservable
.toBlocking()
.first();