英文:
current/previous node relations in spring data neo4j (LinkedList)
问题
我正试图创建一个具有类似于下图的 CURRENT/PREVIOUS 关系的节点链表。
我不确定我的解决方案是否是处理这种情况的正确方式，但为了实现这一目标，我使用了一个方法来填充新消息，创建了两个带有关系的节点，如下所示：
@Builder@Data@NoArgsConstructorpublic class Person {@Id@GeneratedValueprivate Long id;private String name;@Relationship(type = "LATEST")private Message message;void newMessage(Message newMessage) {newMessage.setPrevious(message);message = newMessage;}}@Builder@Data@NoArgsConstructorpublic class Message {@Id@GeneratedValueprivate Long id;private String text;@Relationship(type = "PREVIOUS")private Message previous;}
我还创建了一个用于测试这个解决方案的示例代码：
@SpringBootApplicationpublic class NewsFeedApplication {public static void main(String[] args) {SpringApplication.run(NewsFeedApplication.class, args);}@BeanCommandLineRunner init(PersonRepository personRepository) {return args -> {Person personToAdd1 = Person.builder().name("John").build();personToAdd1.newMessage(Message.builder().text("first message").build());personToAdd1.newMessage(Message.builder().text("second message").build());personToAdd1.newMessage(Message.builder().text("third message").build());personRepository.save(personToAdd1);personToAdd1.newMessage(Message.builder().text("New message.").build());personRepository.save(personToAdd1);};}}
我觉得我已经接近成功，但我不知道如何重置先前的 CURRENT 关系，因此我的解决方案产生了如下输出：
因此问题是：
- 如果我的方法可行，如何移除先前的
CURRENT关系。 - 如果这种方法是错误的，如何正确地为节点实现具有 CURRENT/PREVIOUS 关系的链表。
英文:
I’m trying to create linked list of nodes with CURRENT/PREVIOUS relation similar to the picture below.
I'm not sure if my solution is the right way to handle this scenario, but to achieve that I created two nodes with a single method to populate new messages as below:
@Builder@Data@NoArgsConstructorpublic class Person {@Id@GeneratedValueprivate Long id;private String name;@Relationship(type = "LATEST")private Message message;void newMessage(Message newMessage) {newMessage.setPrevious(message);message = newMessage;}}@Builder@Data@NoArgsConstructorpublic class Message {@Id@GeneratedValueprivate Long id;private String text;@Relationship(type = "PREVIOUS")private Message previous;}
I also created a sample code to test this solution:
@SpringBootApplicationpublic class NewsFeedApplication {public static void main(String[] args) {SpringApplication.run(NewsFeedApplication.class, args);}@BeanCommandLineRunner init(PersonRepository personRepository) {return args -> {Person personToAdd1 = Person.builder().name("John").build();personToAdd1.newMessage(Message.builder().text("first message").build());personToAdd1.newMessage(Message.builder().text("second message").build());personToAdd1.newMessage(Message.builder().text("third message").build());personRepository.save(personToAdd1);personToAdd1.newMessage(Message.builder().text("New message.").build());personRepository.save(personToAdd1);};}}
I feel like I'm close, but I don't know how to reset the previous CURRENT relation and my solution produces output as:
So the question is:
-
If my approach is okay, how could I remove previous
CURRENTrelation. -
If this approach is wrong, how could I implement linked list with CURRENT/PREVIOUS relations for nodes correctly.
答案1
得分: 0
我找到了丢失的拼图，即分离供应商关系。我不知道为什么事先我假设这个关系应该由 Spring Data 仓库的 "save" 方法自动删除。
有效解决方案：
public interface PersonRepository extends Neo4jRepository<Supplier, Long> {@Query("MATCH (n:Person {name: $name})-[r:LATEST]->() DELETE r")void detachLatestFromPerson(String name);}
void newMessage(PersonRepository personRepository, Message newMessage) {personRepository.detachLatestFromPerson(name);newMessage.setPrevious(message);message = newMessage;}
附：我仍然有疑虑，因为我不确定这是否是处理这种情况的好方法，所以如果你知道更好的解决方案，可以随意发布，我们总是可以更换 "最佳答案" 
英文:
I found the missing puzzle, i.e. detaching the supplier relation. I don't know why I assumed in advance that this relation should be deleted automatically by the spring data repository "save" method.
Working solution:
public interface PersonRepository extends Neo4jRepository<Supplier, Long> {@Query("MATCH (n:Person {name: $name})-[r:LATEST]->() DELETE r")void detachLatestFromPerson(String name);}
void newMessage(PersonRepository personRepository, Message newMessage) {personRepository.detachLatestFromPerson(name);newMessage.setPrevious(message);message = newMessage;}
PS. I still have doubts, as I'm not sure if that's a good approach to handle this scenario, so if you know a better solution, go ahead and post it, we can always swap the 'best answer'
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