如何查询neo4j的父节点实体中的子节点？

Question

For example, I have an entity like this:

@Data
@NodeEntity(label = "Subject")
public class Subject {
     @GraphId
     private Long id;
     private String name;

     @Relationship(type = "HAVE_S", direction = Relationship.OUTGOING)
     private Set<Subject> children = new HashSet<>();
}

Then I need to query a 'Subject' by graphId;

@Query("MATCH (s:Subject)-[r:HAVE_S*1]->(c:Subject) WHERE ID(s) = {graphId} RETURN s, r, c;")
Subject findById(@Param("graphId") Long graphId);

The result I want just like the following json:

{
    "id": 62
    "name": "Java"
    "children": [
        {
            "name": "Collection",
            "id": 105
        },
        {
            "name": "MultipleThreads",
            "id": 0
        }
    ]
}

But when I executed the cypher above through Spring Data, an error comes out and says "Result not of expected size. Expected 1 row but found 3".

I hope someone can help me with this problem, thanks.

Answer 1

As your model suggests A subject have multiple children so when you are fetching it by id it's returning a cartesian product of your subject* children. In this case you need to collect your children(subjects and then return)

@Query("MATCH (s:Subject)-[r:HAVE_S*1]->(c:Subject) WHERE ID(s) = {graphId} RETURN s,  collect(c) as childrens;")

or simply you can use findById() of your repository the SDN will create query by itself

Answer 2

You can try like this.

MATCH (k:LabelsTree {name:'465afe3c118589de357745a709c0441f'}) CALL apoc.path.spanningTree(k,{labelFilter:'+LabelsTree', maxLevel:3, optional:true, filterStartNode:true}) yield path return path

Refer to the connection