春季数据|neo4j |以正确的顺序查询路径

Question

版本：

<dependency>
            <groupId>org.neo4j</groupId>
            <artifactId>neo4j-ogm-core</artifactId>
            <version>2.1.1</version>
        </dependency>
        <dependency> <!-- If you're using the HTTP driver -->
            <groupId>org.neo4j</groupId>
            <artifactId>neo4j-ogm-http-driver</artifactId>
            <version>2.1.1</version>
        </dependency>
        <!-- https://mvnrepository.com/artifact/org.springframework.data/spring-data-neo4j -->
        <dependency>
            <groupId>org.springframework.data</groupId>
            <artifactId>spring-data-neo4j</artifactId>
            <version>4.2.0.RELEASE</version>
        </dependency>

这是我的实体：

@Data
@NodeEntity
@EqualsAndHashCode(exclude = {"operatedByBuses"})
@ToString(of = {"name"})
public class BusStop {
    @GraphId
    private Long graphId;
    @Index(unique = true, primary = true)
    private String name;
    private String pincode;
    private String landmark;
    private String[] latlong;
    @Relationship(type = "OPERATED_BY")
    private Set<OperatedByBus> operatedByBuses = new HashSet<>();
}
@Data
@RelationshipEntity(type = "OPERATED_BY")
@ToString(of = "displayName")
public class OperatedByBus {
    @GraphId
    private Long id;
    @StartNode
    private BusStop origin;
    @EndNode
    private BusStop destination;
}

我正在尝试以正确的顺序获得A，B，C和D的A和D之间的路线，然后我将公共汽车在每个对象中获取。

这是我的Cypher：

String findBuses = "MATCH p=shortestPath((o:BusStop)-[buses*1..40]->(d:BusStop))n" +
                "WHERE o.name =~ '(?i).*ABC Bus Stand.*'n" +
                "  AND d.name =~ '(?i).*XYZ Bus Stand.*'n" +
                "RETURN p";
        Iterable<BusStop> busstops = session.query(BusStop.class, findBuses, Collections.emptyMap());
        System.out.println(busstops);
        for (BusStop busStop : busstops) {
            System.out.println(busStop.getName());
            System.out.println("t " + busStop.getOperatedByBuses());
        }

但结果不正确。我认为结果为d，c，a，b（或某些随机顺序），而不是a，b，c，d。

我能想到的一种方法是向操作的bbybus say int legid添加一个属性，然后在我的查询中按LEGID订购。不确定这是否是最好的方法。

我缺少什么？

Answer 1

修改查询如下：

MATCH p=shortestPath((o:BusStop)-[buses*1..40]->(d:BusStop))
WHERE 
  o.name =~ '(?i).*ABC Bus Stand.*' AND
  d.name =~ '(?i).*XYZ Bus Stand.*'
RETURN nodes(p) as busStops,relationships(p)

请注意，当您的查询可能返回多个行，当时子句匹配多个节点。

然后使用session.query(findBuses, Collections.emptyMap());。结果类型是org.neo4j.ogm.model.Result，使用以下来获得单个结果：

Iterable<Map<String, Object>> result.queryResults();
.. iterate over results
   // to get a `busStops` for single path as collection, which should be in order
   .. (Collection<BusStop>) item.get("busStops");