精华 Spring boot项目集成Neo4j
发布于 6 年前 作者 pangguoming 8653 次浏览 来自 分享

第一步,创建Springboot工程 使用Eclipse 创建Maven项目,并修改pom.xml文件为:

<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
   xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
   <modelVersion>4.0.0</modelVersion>

   <groupId>com.example</groupId>
   <artifactId>demo</artifactId>
   <version>0.0.1-SNAPSHOT</version>
   <packaging>jar</packaging>

   <name>demo</name>
   <description>Demo project for Spring Boot</description>

   <parent>
      <groupId>org.springframework.boot</groupId>
      <artifactId>spring-boot-starter-parent</artifactId>
      <version>1.5.9.RELEASE</version>
      <relativePath/> <!-- lookup parent from repository -->
   </parent>

   <properties>
      <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
      <project.reporting.outputEncoding>UTF-8</project.reporting.outputEncoding>
      <java.version>1.8</java.version>
   </properties>

   <dependencies>
      <dependency>
         <groupId>org.springframework.boot</groupId>
         <artifactId>spring-boot-starter-web</artifactId>
      </dependency>

      <dependency>
         <groupId>org.springframework.boot</groupId>
         <artifactId>spring-boot-starter-test</artifactId>
         <scope>test</scope>
      </dependency>

      <dependency>
         <groupId>org.springframework.boot</groupId>
         <artifactId>spring-boot-starter-data-neo4j</artifactId>
      </dependency>
   </dependencies>
   <build>
      <plugins>
         <plugin>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-maven-plugin</artifactId>
         </plugin>
      </plugins>
   </build>
</project>

第二步,修改配置文件application.properties


server.port=8080
spring.data.neo4j.uri=http://{neo4jweb端访问地址}:7474 
spring.data.neo4j.username=neo4j账号
spring.data.neo4j.password=neo4j密码

第三步,model文件—节点User和关系UserRelation,这两个文件与neo4j里面的节点和关系的属性一一对应,此处是我的示例,具体与各自的neo4j内容对于,label必须与neo4j库里名字一致

@NodeEntity(label = "User")
public class UserNode {
    @GraphId
    private Long nodeId;

    @Property(name = "userId")
    private String userId;

    @Property(name = "name")
    private String name;

    @Property(name = "age")
    private int age;


    public Long getNodeId() {
        return nodeId;
    }

    public void setNodeId(Long nodeId) {
        this.nodeId = nodeId;
    }

    public String getUserId() {
        return userId;
    }

    public void setUserId(String userId) {
        this.userId = userId;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }
}
@RelationshipEntity(type = "UserRelation")
public class UserRelation {
    @GraphId
    private Long id;

    @StartNode
    private UserNode startNode;

    @EndNode
    private UserNode endNode;

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public UserNode getStartNode() {
        return startNode;
    }

    public void setStartNode(UserNode startNode) {
        this.startNode = startNode;
    }

    public UserNode getEndNode() {
        return endNode;
    }

    public void setEndNode(UserNode endNode) {
        this.endNode = endNode;
    }
}

第四步,repository文件----UserRepository和UserRelationRepository,@Query里面编写Cypher查询


@Component
public interface UserRepository extends GraphRepository<UserNode> {

    [@Query](/user/Query)("MATCH (n:User) RETURN n ")
    List<UserNode> getUserNodeList();

    [@Query](/user/Query)("create (n:User{age:{age},name:{name}}) RETURN n ")
    List<UserNode> addUserNodeList(@Param("name") String name, @Param("age")int age);
}

@Component
public interface UserRelationRepository extends GraphRepository<UserRelation> {

    [@Query](/user/Query)("match p=(n:User)<-[r:UserRelation]->(n1:User) where n.userId={firstUserId} and n1.userId={secondUserId} return p")
    List<UserRelation> findUserRelationByEachId(@Param("firstUserId") String firstUserId, @Param("secondUserId") String secondUserId);

    [@Query](/user/Query)("match (fu:User),(su:User) where fu.userId={firstUserId} and su.userId={secondUserId} create p=(fu)-[r:UserRelation]->(su) return p")
    List<UserRelation> addUserRelation(@Param("firstUserId") String firstUserId, @Param("secondUserId") String secondUserId);

}

第五步,修改service文件

@Service
public class Neo4jService {

    @Autowired
    private UserRepository userRepository;
    @Autowired
    private UserRelationRepository userRelationRepository;

    /**
     * Always use constructor based dependency injection in your beans.
     * Always use assertions for mandatory dependencies.
     */

    public int addUser(UserNode userNode){
        userRepository.addUserNodeList(userNode.getName(),userNode.getAge());
        return 1;
    }
}

第六步,neo4j文件注入config----Neo4jConfig,注意basePackges的路径。

@Configuration
@EnableNeo4jRepositories(basePackages = "com.example.demo.repository")
public class Neo4jConfig {
}
3 回复

老师,如果查询的节点是多标签,那么后端应该如何接收这个对象

@wangsheng0324 再创建一个 实体类 ,在头部也用 @NodeEntity(label = “另一个label名”) 就可以

@pangguoming 老师,我尝试了您说的方法,但是还是会报之前同样的错误信息:org.neo4j.ogm.exception.core.AmbiguousBaseClassException: Multiple classes found in type hierarchy that map to: [ApplyUser, Phone, a_User, b_Phone, c_OtherPhone];查询a_User标签对应的节点,会过滤出几点的其他标签,无法完成映射。您有遇到过这种情况吗?

回到顶部