Java中使用Stream根据ID合并电影与分类列表的实现方法

需求说明 #

需要合并两组数据：电影数据（List<Movie>）和分类数据（List<Genres>），仅当两者的id相同时进行关联。其中：

• Movie类包含id（主键）、name字段
• Genres类包含id（外键）、name字段

示例输入数据 #

List<Movie> movies = new ArrayList<>();
movies.add(new Movie(1, "Titanic"));
movies.add(new Movie(2, "Batman"));
movies.add(new Movie(3, "Silence of the Lambs"));
movies.add(new Movie(4, "1917"));
movies.add(new Movie(5, "Fight Club"));
movies.add(new Movie(6, "Inception"));
movies.add(new Movie(7, "The Godfather"));
movies.add(new Movie(8, "Peral Harbor"));

List<Genres> genres = new ArrayList<>();
genres.add(new Genres(1, "Drama"));
genres.add(new Genres(4, "Drama"));
genres.add(new Genres(8, "Drama"));
genres.add(new Genres(7, "Drama"));
genres.add(new Genres(2, "Action"));
genres.add(new Genres(5, "Action"));
genres.add(new Genres(6, "SF"));
genres.add(new Genres(3, "Thriller"));
genres.add(new Genres(5, "Thriller"));
genres.add(new Genres(5, "Crime"));
genres.add(new Genres(16, "Comedy"));
genres.add(new Genres(3, "Horror"));
genres.add(new Genres(1, "Disaster"));
genres.add(new Genres(7, "Noir"));
genres.add(new Genres(1, "Romance"));

期望输出结果 #

[
    {"id": 1, "name": "Titanic", "genres": ["Drama", "Disaster", "Romance"]},
    {"id": 2, "name": "Batman", "genres": ["Action"]},
    {"id": 3, "name": "Silence of the Lambs", "genres": ["Thriller", "Horror"]},
    {"id": 4, "name": "1917", "genres": ["Drama"]},
    {"id": 5, "name": "Fight Club", "genres": ["Action", "Thriller", "Crime"]},
    {"id": 6, "name": "Inception", "genres": ["SF"]},
    {"id": 7, "name": "The Godfather", "genres": ["Drama", "Noir"]},
    {"id": 8, "name": "Peral Harbor", "genres": ["Drama"]}
]

Stream实现方案 #

步骤说明 #

1. 预分组分类数据：先将Genres列表按id分组，把同一id对应的所有分类名称收集到列表中，得到Map<Integer, List<String>>结构，避免后续多次遍历分类列表，提升效率。
2. 关联电影与分类：遍历Movie列表，对每个电影，从分组后的Map中获取对应的分类列表（若没有匹配的id则返回空列表），最后组装成包含分类列表的目标对象。

完整代码实现 #

首先定义基础实体类和结果DTO类：

// 电影实体类
class Movie {
    private Integer id;
    private String name;

    public Movie(Integer id, String name) {
        this.id = id;
        this.name = name;
    }

    public Integer getId() { return id; }
    public String getName() { return name; }
}

// 分类实体类
class Genres {
    private Integer id;
    private String name;

    public Genres(Integer id, String name) {
        this.id = id;
        this.name = name;
    }

    public Integer getId() { return id; }
    public String getName() { return name; }
}

// 结果DTO类，存储关联后的电影与分类信息
class MovieWithGenres {
    private Integer id;
    private String name;
    private List<String> genres;

    public MovieWithGenres(Integer id, String name, List<String> genres) {
        this.id = id;
        this.name = name;
        this.genres = genres;
    }

    // 重写toString方法，让输出匹配JSON格式
    @Override
    public String toString() {
        return String.format("{\"id\": %d, \"name\": \"%s\", \"genres\": %s}",
                id, name, genres.toString().replaceAll("\\[", "\\[\"").replaceAll("\\]", "\"\\]").replaceAll(", ", "\", \""));
    }
}

然后是Stream处理逻辑：

public class MovieGenreMerge {
    public static void main(String[] args) {
        // 初始化示例数据
        List<Movie> movies = new ArrayList<>();
        movies.add(new Movie(1, "Titanic"));
        movies.add(new Movie(2, "Batman"));
        movies.add(new Movie(3, "Silence of the Lambs"));
        movies.add(new Movie(4, "1917"));
        movies.add(new Movie(5, "Fight Club"));
        movies.add(new Movie(6, "Inception"));
        movies.add(new Movie(7, "The Godfather"));
        movies.add(new Movie(8, "Peral Harbor"));

        List<Genres> genres = new ArrayList<>();
        genres.add(new Genres(1, "Drama"));
        genres.add(new Genres(4, "Drama"));
        genres.add(new Genres(8, "Drama"));
        genres.add(new Genres(7, "Drama"));
        genres.add(new Genres(2, "Action"));
        genres.add(new Genres(5, "Action"));
        genres.add(new Genres(6, "SF"));
        genres.add(new Genres(3, "Thriller"));
        genres.add(new Genres(5, "Thriller"));
        genres.add(new Genres(5, "Crime"));
        genres.add(new Genres(16, "Comedy"));
        genres.add(new Genres(3, "Horror"));
        genres.add(new Genres(1, "Disaster"));
        genres.add(new Genres(7, "Noir"));
        genres.add(new Genres(1, "Romance"));

        // 1. 将分类数据按id分组，提取分类名称
        Map<Integer, List<String>> genreMap = genres.stream()
                .collect(Collectors.groupingBy(
                        Genres::getId,
                        Collectors.mapping(Genres::getName, Collectors.toList())
                ));

        // 2. 关联电影与分类，生成结果列表
        List<MovieWithGenres> result = movies.stream()
                .map(movie -> new MovieWithGenres(
                        movie.getId(),
                        movie.getName(),
                        genreMap.getOrDefault(movie.getId(), Collections.emptyList())
                ))
                .collect(Collectors.toList());

        // 输出结果
        System.out.println("result: [");
        result.forEach(item -> System.out.println("                " + item + ","));
        System.out.println("        ]");
    }
}

代码说明 #

• Collectors.groupingBy配合Collectors.mapping直接完成分类数据的分组与名称提取，简化后续处理。
• Map.getOrDefault确保无匹配分类时返回空列表，避免空指针异常。
• 重写toString方法让输出格式与期望的JSON结构一致，方便查看结果。

内容的提问来源于stack exchange，提问作者footlessbird