Question

我正在处理项目，我希望迭代List并在返回结果之前将它们合并。基本上我正在尝试将private void btnStart_Click(object sender, RoutedEventArgs e) { string path = ""; OpenFileDialog ofd = new OpenFileDialog(); ofd.Filter = "Properties | *.properties"; if (ofd.ShowDialog() == true) { path = ofd.FileName; } List<ServerProperties> serverProperties = new List<ServerProperties>(); using (StreamReader sr = new StreamReader(path)) { string line; while ((line = sr.ReadLine()) != null) { if (!line.StartsWith("#")) { string[] lines = line.Split('='); string property = lines[0]; string value = lines[1]; serverProperties.Add(new ServerProperties { Property = property, Value = value }); Debug.Print($"Property: {property} Value: {value}"); } } } dgItems.ItemsSource = serverProperties; }对象转换为import java.util.*; public class xy { public static void main(String[] args) { System.out.println("Enter a number with the format 'xy'. x is the row number and y the column number."); Scanner scanner = new Scanner(System.in); String[] xy = scanner.nextLine().split(charAt(1)); // not sure what to put here so I tried charAt but it doesn't work String x = xy [0]; String y = xy [1]; } }。

一些相关代码。

Post.java

PostResponse

PostResponse.java

Post

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public class Post {
    @NonNull private User user;
    @NonNull private String title;
    @NonNull private String body;
    @NonNull private Integer id;
}

上面提到的变换函数。

public final class PostResponse{
    private Integer userId;
    private Integer id;
    private String title;
    private String body;
}

但我没有收到订阅@Override public Observable<List<Post>> posts() { return postDataStore.getAllPosts() .flatMap(postResponses -> Observable.fromIterable(postResponses) .flatMap(postResponse -> postDataMapper.transform(postResponse)) // Receiving output here .doOnNext(post -> Log.i(TAG, "posts: " + post)) .toList() .toObservable() // Not Receiving output here .doOnNext(posts -> Log.i(TAG, "posts: " + posts))); }的{{1}}的任何数据。不确定是什么问题。

如果我犯了一个非常愚蠢的错误，我道歉。

Answer 1

快速解决方法是添加：

emitter.onComplete();

在emitter.onNext(post);

之后

但我建议重新考虑设计和逻辑。特别是您的转换不需要创建新的Observable。我会用这样的方法重新设计你的tranform方法：

Observable<Post> transform(PostResponse postResponse) {
    return userRepository.user(postResponse.getUserId())
        .map(user -> parseResponse(parseResponse, user));
}


Post parseResponse(PostResponse response, User user) {
    final Post post = new Post(postResponse.getId());
    post.setTitle(postResponse.getTitle());
    post.setBody(postResponse.getBody());   
    post.setUser(user);

    return post;
}

通过这种方式，您可以避免创建新的Observable，避免内部订阅，并且您有一个结果的控制点，即最终结果Observable的订阅。 / p>

RxJava2在toList之后没有收到doOnNext

1 个答案: