Question

我有两个observable：LoadLocal和LoadServer。 LoadLocal从本地源加载并返回一个元素，LoadServer从服务器获取它。我想将它们组合成另一个可观察的：加载。我希望Load从LoadLocal获取元素，如果它为null，我想从LoadServer返回该元素。关于如何做到这一点的任何想法？

由于

真实场景的详细信息：

// loadLocal(id) gives me an observable that returns an asset from a local source
Func<Guid, IObservable<IAsset>> loadLocal = Observable.ToAsync<Guid, IAsset>(id => GetLocalAsset(id));

var svcClient = new ServiceClient<IDataService>();
var svc = Observable.FromAsyncPattern<Request, Response>(svcClient.BeginInvoke, svcClient.EndInvoke);

// calling loadServer(id) gives me an observable that returns an asset from a server
var loadServer = id => from response in svc(new Request(id)) select response.Asset;

// so at this point i can call loadServer(id).Subscribe() to get an asset with specified id from the server, or I can call loadLocal(id).Subscribe() to get it from local source.
// however I want another observable that combines the two so I can do: load(id).Subscribe() that gets the asset from loadLocal(id) and if it is null it gets it from loadServer(id)
var load = ???

以下几乎给了我想要的结果，但是loadLocal（id）和loadServer（id）都会运行。如果loadLocal（id）返回一个元素，我不希望loadServer（id）运行。

var load = id => loadLocal(id).Zip(loadServer(id), (local, server) => local ?? server);

Answer 1

这个怎么样：

loadLocal(id)
    .Where(x => x != null)
    .Concat(Observable.Defer(() => loadServer(id)))
    .Take(1);

Answer 2

也许是这样的东西？

var load = LoadLocal.Zip(LoadServer, (local, server) => local ?? server);

结合可观察量

2 个答案: