如何使用密碼返回neo4j中多節點或邊的集合

Question

我想要可視化neo4j的數據，我需要返回某個子圖的所有節點和邊。 （：User） - [：RATED] - >（：Movie） - [：HAS_GENRE] - >（：Genre），（：User） - [：SIMILAIRITY] - >（ ：用戶）。我想要顯示中心是指定電影的子圖。

該代碼的相關部分如下所示。

@Query("match (u:User)-[r:RATED]->(m:Movie)-[hg:HAS_GENRE]->(g:Genre) where m.id={0} return id(u) as id, case labels(u)[0] when 'Genre' then u.name when 'User' then u.id when 'Movie' then u.title end as caption, labels(u)[0] as type " 
     + "union " 
     + "match (u:User)-[r:RATED]->(m:Movie)-[hg:HAS_GENRE]->(g:Genre) where m.id={0} return id(m) as id, case labels(m)[0] when 'Genre' then m.name when 'User' then m.id when 'Movie' then m.title end as caption, labels(m)[0] as type " 
     + "union " 
     + "match (u:User)-[r:RATED]->(m:Movie)-[hg:HAS_GENRE]->(g:Genre) where m.id={0} return id(g) as id, case labels(g)[0] when 'Genre' then g.name when 'User' then g.id when 'Movie' then g.title end as caption, labels(g)[0] as type") 
    List<NodeInfo> findMovieRelevantNodes(int movieId); 

    @Query("match (u:User)-[r:RATED]->(m:Movie)-[hg:HAS_GENRE]->(g:Genre) where m.id={0} return id(u) as source,type(r) as type, id(m) as target, case type(r) when 'HAS_GENRE' then r.probability when 'SIMILARITY' then r.similarity when 'RATED' then r.rate end as caption " 
     + "union " 
     + "match (u:User)-[r:RATED]->(m:Movie)-[hg:HAS_GENRE]->(g:Genre) where m.id={0} return id(m) as source,type(hg) as type, id(g) as target, case type(hg) when 'HAS_GENRE' then r.probability when 'SIMILARITY' then r.similarity when 'RATED' then r.rate end as caption") 
    List<EdgeInfo> findMovieRelevantEdges(int movieId); 

@QueryResult 
public interface NodeInfo { 
    @ResultColumn("id") 
    int getId(); 

    @ResultColumn("caption") 
    String getCaption(); 

    @ResultColumn("type") 
    String getType(); 
} 

@QueryResult 
public interface EdgeInfo { 
    @ResultColumn("source") 
    int getSource(); 

    @ResultColumn("type") 
    String getType(); 

    @ResultColumn("target") 
    int getTarget(); 

    @ResultColumn("caption") 
    String getCaption(); 
} 

它看起來像冗餘和性能差。那麼是否有更好的方法來重寫這些密碼查詢。

Answer 1

試試這個：

match (u:User)-[r:RATED]->(m:Movie)-[hg:HAS_GENRE]->(g:Genre) 
where m.id={0} 
return {id:id(u), caption:u.id} as user, 
     {id:id(m), caption:m.title} as movie, 
     {id:id(g), caption:g.name} as genre;