Question

见下文实体。

人员实体

@NodeEntity
public class Person {  

    @GraphId Long personId; 

    private String name; 

    private String surname;

    @Relationship(type = "ATTENDS", direction = Relationship.OUTGOING) 
    private Set<Event> events;

参加实体

@RelationshipEntity(type = "ATTENDS")
public class Attends {

    @GraphId
    private Long id;

    @StartNode
    private Person person;

    @EndNode
    private Event event;

    private Date attendingDate;

活动实体

@NodeEntity
public class Event {

    @GraphId
    private Long eventId;

    private String eventName;

    @Relationship(type = "ATTENDS", direction = Relationship.INCOMING)
    private Set<Person> persons;

这是我的API

/人/ {PERSONID} /出席

我希望返回一个所有对象的列表，其关系是关注所提供的id的人，在下面的示例中，它将是一个事件列表。

[{
"attends":{
"attendsId":"1234",
"startDate":"98098098",
"endDate":"098098098",
event:{ "eventId":"1234", "name":"ComicCon" }
},
"attends":{
"attendsId":"1235",
"startDate":"984548098",
"endDate":"45454545",
event:{ "eventId":"1235", "name":"AWS Summit" }

}]

我尝试以下查询，但未获得结果，

List<Attends> findByPersonPersonId(Long personId);

那么如何通过查询来实现这个结果呢？

请指南，感谢。

Answer 1

@NodeEntity
public class Person {

    @GraphId Long id; 

    private String personId;

    private String name; 

    private String surname;

    @Relationship(type = "ATTENDS", direction = Relationship.OUTGOING) 
    private Set<Attends> attendedEvents =new HashSet<Attends>();
}

@NodeEntity
public class Event {

    @GraphId
    private Long id;

    private String eventId;

    private String eventName;

    @Relationship(type = "ATTENDS", direction = Relationship.INCOMING)
    private List<Attends> attendedEvents = new ArrayList<Attends>();
}


@RelationshipEntity(type = "ATTENDS")
public class Attends {

    @GraphId
    private Long id;

    private String attendId;

    @StartNode
    private Person person;

    @EndNode
    private Event event;

    private Date attendingDate;
}

@Repository
public interface PersonRepository  extends GraphRepository<Person>{

     Person findByPersonId(String personId);

}


public class PersonServiceTest extends AbstractTest{

    @Autowired
    PersonRepository personRepository;

    @Test
    public void save(){

        Person person = new Person();
        person.setName("Person1");
        person.setPersonId(UUID.randomUUID().toString());
        person.setSurname("Surname1");

        Event event1 = new Event();
        event1.setEventId(UUID.randomUUID().toString());
        event1.setEventName("Event1");

        Event event2 = new Event();
        event2.setEventId(UUID.randomUUID().toString());
        event2.setEventName("Event2");

        Attends attends1 = new Attends();
        attends1.setAttendId(UUID.randomUUID().toString());
        attends1.setAttendingDate(new Date());
        attends1.setPerson(person);
        attends1.setEvent(event1);

        Attends attends2 = new Attends();
        attends2.setAttendId(UUID.randomUUID().toString());
        attends2.setAttendingDate(new Date());
        attends2.setPerson(person);
        attends2.setEvent(event2);

        person.getAttendedEvents().add(attends1);
        person.getAttendedEvents().add(attends2);

        personRepository.save(person);


    }

    @Test
    public void get(){
        Person person = personRepository.findByPersonId("ebc17b4b-2270-4e78-ac68-87bce5444ef4");

        person.getAttendedEvents().forEach(attendedEvent -> {
            System.out.println(attendedEvent.getEvent());
        });

    }


}

Answer 2

不太确定你想要什么？如果您想获得所有人的所有出席或特定人的所有出席。

尽管如此，我已经为两者提供了答案。

创建一个AttendsRepository并在其中添加一个findByPersonId（String personId）。

public interface AttendsRepository extends GraphRepository<Attends>{

    @Query("Match(p:Person {personId:{0}})-[a:ATTENDS]->(e:Event) return a")
    Iterable<Attends> findByPersonId(String personId);

}

然后您可以通过

获得所需的结果

@Test
public void getAllAttends(){

    Iterable<Attends> allAttends = attendsRepository.findAll();
    allAttends.forEach(attend -> {
        System.out.println(attend);
    });

}

@Test
public void getAttendsByPerson(){

    String personId = "283f51e9-9ade-4f46-a005-7353b5211c8b";

    Iterable<Attends> allAttends = attendsRepository.findByPersonId(personId);
    allAttends.forEach(attend -> {
        System.out.println(attend);
    });

}

使用neo4j查询的Spring数据

2 个答案: