如何在neo4j-ogm或spring-data-neo4j中动态更改实体类型?

问题描述 投票:0回答:2

有一个关于“如何动态地向Neo4j中的节点添加标签”的问题。有没有办法动态更改实体类型?

举个例子:

    @NodeEntity
public class User {

   @Properties(prefix = "custom")
   private Map userProperties;

}

我从https://neo4j.com/blog/spring-data-neo4j-5-0-release/看到我可以创建动态属性。我也可以在运行时使用动态类型吗?我想在需要时动态地将“用户”类型更改为“消费者”/“管理员”/“生产者”。实体类型并非详尽无遗。

提前致谢! :)

neo4j spring-data-neo4j spring-data-neo4j-4 neo4j-ogm
2个回答
1
投票

@Labels上有一个Set<String>注释,除了类和接口的主类型之外,它还存储/管理。

见:https://docs.spring.io/spring-data/neo4j/docs/current/reference/html/#reference:annotating-entities:node-entity:runtime-managed-labels


0
投票

@Labels机制很棒,对于许多用例来说,这是我所说的最佳解决方案。

如果你想从你的存储库中取出另一个类,那么确实需要做更多的工作。

我在音乐相关项目中这样做。我有Artist(不抽象,完全可用于任何我不知道是否是乐队的乐队)和BandSoloArtist从Artist延伸,附加标签:

@NodeEntity
public class Artist {}

@NodeEntity
public class Band extends Artist{}

我在自定义存储库扩展中知道的是:

interface ArtistRepository<T extends Artist> extends Repository<T, Long>, ArtistRepositoryExt {

    Optional<T> findOneByName(String name);

    // Specifying the relationships is necessary here because the generic queries won't recognize that
    // Band has a relationship to country that _should_ be loaded with default depth of 1.

    @Query("MATCH (n:Artist) WITH n MATCH p=(n)-[*0..1]-(m) RETURN p ORDER BY n.name")
    List<T> findAllOrderedByName();

    @Query("MATCH (n:Artist) WHERE id(n) = $id WITH n MATCH p=(n)-[*0..1]-(m) RETURN p")
    Optional<T> findById(@Param("id") Long id);

    <S extends T> S save(S artist);
}

interface ArtistRepositoryExt {
    Band markAsBand(Artist artist);

    SoloArtist markAsSoloArtist(Artist artist);
}

class ArtistRepositoryExtImpl implements ArtistRepositoryExt {

    private static final String CYPHER_MARK_AS_BAND = String.format(
        "MATCH (n) WHERE id(n) = $id\n" +
            "OPTIONAL MATCH (n) - [f:BORN_IN] -> (:Country)\n" +
            "REMOVE n:%s SET n:%s\n" +
            "DELETE f",
        SoloArtist.class.getSimpleName(),
        Band.class.getSimpleName());
    private static final String CYPHER_MARK_AS_SOLO_ARTIST = String.format(
        "MATCH (n) WHERE id(n) = $id\n" +
            "OPTIONAL MATCH (n) - [f:FOUNDED_IN] -> (:Country)\n" +
            "REMOVE n:%s SET n:%s\n" +
            "DELETE f",
        Band.class.getSimpleName(),
        SoloArtist.class.getSimpleName());

    private final Session session;

    public ArtistRepositoryExtImpl(Session session) {
        this.session = session;
    }

    @Override
    public Band markAsBand(Artist artist) {
        session.query(CYPHER_MARK_AS_BAND, Map.of("id", artist.getId()));
        // Needs to clear the mapping context at this point because this shared session
        // will know the node only as class Artist in this transaction otherwise.
        session.clear();
        return session.load(Band.class, artist.getId());
    }

    @Override
    public SoloArtist markAsSoloArtist(Artist artist) {
        session.query(CYPHER_MARK_AS_SOLO_ARTIST, Map.of("id", artist.getId()));
        // See above
        session.clear();
        return session.load(SoloArtist.class, artist.getId());
    }
}

虽然这很好用,但我会在更深层次的嵌套类场景中获得努力的想法。此外,如果您想以多态方式使用存储库,那么您可以按照我的要求执行redeclare派生的finder方法。

我也保留了专用的存储库。

如果这个问题仍然与您有关,请告诉我是否也适合您。你会在这里找到整个项目:

https://github.com/michael-simons/bootiful-music

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