我有一个未知大小的用户列表。我想要的是查询前30并更新UI。然后,我想用步骤100的偏移量来查询所有其他用户,直到我得到最后一包用户--我应该在这里使用takeUntil吗?)当我得到-我通过添加剩余的用户来更新UI (结合reduce(),我相信)。
这是我的密码:
final int INITIAL_OFFSET = 0;
final int INITIAL_LIMIT = 30;
// Loading first 30 users to immediately update UI (better UX)
getServerApi().getAllFriends(userId, "photo_50", INITIAL_OFFSET, INITIAL_LIMIT)
// Loading remaining users 100 by 100 and updating UI after all users been loaded
.flatMap(users -> {
AtomicInteger newOffset = new AtomicInteger(INITIAL_LIMIT);
return Observable.just(users)
.flatMap(users1 -> getServerApi().getAllFriends(userId, "photo_50", newOffset.get(), Config.DEFAULT_FRIEND_REQUEST_COUNT))
.subscribeOn(Schedulers.io())
.observeOn(Schedulers.io())
.collect(() -> new ArrayList<User>(), (b, s) -> {
b.addAll(s);
newOffset.set(newOffset.get() + Config.DEFAULT_FRIEND_REQUEST_COUNT);
})
.repeat()
.takeUntil(friends -> friends.size() == 0);
})
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.subscribe(users -> getView().appendAllFriends(users),
throwable -> getView().setError(processFail(throwable, ServerApi.Action.GET_ALL_FRIENDS), false));但是似乎我做错了什么,因为每次进行更新调用时都会调用onNext。
发布于 2015-07-12 21:28:12
回答我自己的问题。阿黛尔的回答很好,但我需要一个单一的订阅(我使用的是核MVP文库),我想使用collect()和takeUntil()而不是while循环(后者需要阻塞修改接口方法)。
花了几个小时终于得到了它:
final int INITIAL_LIMIT = 30;
// Loading first 30 users to immediately update UI (better UX)
getServerApi().getAllFriends(userId, "photo_50", null, INITIAL_LIMIT)
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
// Updating UI 1st time or show error
.doOnNext(users -> getView().appendAllFriends(users))
.doOnError(throwable -> getView().setError(processFail(throwable, ServerApi.Action.GET_ALL_FRIENDS), false))
// Loading remaining users 100 by 100 and updating UI after all users been loaded
.flatMap(users -> {
AtomicInteger newOffset = new AtomicInteger(INITIAL_LIMIT);
ArrayList<User> remainingUsers = new ArrayList<>();
AtomicBoolean hasMore = new AtomicBoolean(true);
return Observable.just(users)
.flatMap(users1 -> getServerApi().getAllFriends(userId, "photo_50", newOffset.get(), Config.DEFAULT_FRIEND_REQUEST_COUNT))
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.collect(() -> remainingUsers, (b, s) -> {
// Needed for takeUntil
hasMore.set(b.addAll(s));
newOffset.set(newOffset.get() + Config.DEFAULT_FRIEND_REQUEST_COUNT);
})
.repeat()
.takeUntil(friends -> !hasMore.get())
// Grab all items emitted by collect()
.last()
// Updating UI last time
.doOnNext(users2 -> getView().appendAllFriends(users2));
})
.subscribe();也许这对使用核的其他人是有用的。
发布于 2015-07-12 19:08:27
// cache() will ensure that we load the first pack only once
Observable<Users> firstPack = firstPack().cache();
// this subscription is for updating the UI on the first batch
firstPack
.observeOn(AndroidSchedulers.mainThread())
.subscribe(x -> draw(x), e -> whoops(e));
// this subscription is for collecting all the stuff
// do whatever tricks you need to do with your backend API to get the full list of stuff
firstPack
.flatMap(fp -> rest(fp))
.observeOn(AndroidSchedulers.mainThread())
.subscribe(x -> allUsers(x), e -> whoops(e));
// I would do this in a simple while loop
Observable<List<User>> rest(List<User> firstPack) {
return Observable.create(sub -> {
final List<User> total = firstPack;
try {
while (!sub.isUnsubscribed()) {
final List<User> friends = api.getFriendsBlocking(total.size());
if (friends.isEmpty()) {
sub.onNext(total);
sub.onCompleted();
} else {
total.addAll(friends);
}
}
} catch(IOException e) {
sub.onError(e);
}
})
}https://stackoverflow.com/questions/31366699
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