Java 弹簧数据Neo4j 4.0.0:Can';t在具有相同标签的节点之间创建关系
我将Spring数据Neo4j 4.0.0与Neo4j 2.2.1一起使用,并尝试在具有完全相同标签的两个节点之间创建关系 所以,我有一个NodeEntity类,里面有一个与类本身类型相同的变量,并将其注释为Relationship。 但是,当我使用repository对象的save()方法将对象保存到数据库时,无法创建关系 提前感谢您,我们将非常感谢您的建议 编辑 下面是节点实体类Java 弹簧数据Neo4j 4.0.0:Can';t在具有相同标签的节点之间创建关系,java,neo4j,nodes,relationship,spring-data-neo4j-4,Java,Neo4j,Nodes,Relationship,Spring Data Neo4j 4,我将Spring数据Neo4j 4.0.0与Neo4j 2.2.1一起使用,并尝试在具有完全相同标签的两个节点之间创建关系 所以,我有一个NodeEntity类,里面有一个与类本身类型相同的变量,并将其注释为Relationship。 但是,当我使用repository对象的save()方法将对象保存到数据库时,无法创建关系 提前感谢您,我们将非常感谢您的建议 编辑 下面是节点实体类 public class ArchitectureUnitState extends UnitState {
public class ArchitectureUnitState extends UnitState {
public ArchitectureUnitState()
{
super();
}
public ArchitectureUnitState(String name, String description, String parentArchitectureUnitName)
{
super(name, description);
this.parentArchitectureUnitName = parentArchitectureUnitName;
}
@Relationship(type="PART_OF", direction = Relationship.OUTGOING)
private ArchitectureUnitState architectureUnitState;
@Relationship(type="STATE_OF", direction = Relationship.OUTGOING)
private ArchitectureUnit architectureUnit;
@Transient
private String parentArchitectureUnitName;
public void partOf(ArchitectureUnitState architectureUnitState) {
this.architectureUnitState = architectureUnitState;
}
public void stateOf(ArchitectureUnit architectureUnit) {
this.architectureUnit = architectureUnit;
}
public void childOf(String parentArchitectureUnitName) {
this.parentArchitectureUnitName = parentArchitectureUnitName;
}
public String getParentName() {
return parentArchitectureUnitName;
}
}
@NodeEntity
public class UnitState {
@GraphId
protected Long id;
private String name;
private String description;
public UnitState() {
}
public UnitState(String name, String description) {
this.name = name;
this.description = description;
}
public void setName(String name) {
this.name = name;
}
public void setDescription(String description) {
this.description = description;
}
public String getName() {
return name;
}
public String getDescription() {
return description;
}
}
因此,顺序是:我创建了ArchitectureUnitState对象,相互映射,然后使用ArchitectureUnitStateRepository的save()方法保存
如果我这样做,则不会创建关系的部分,尽管我在调试中看到值在那里
我现在的解决方法是首先保存所有ArchitectureUnitState节点,然后从数据库中再次检索它们,将它们映射到另一个节点,然后再次保存。通过这种方式,可以创建关系,但我需要保存两次。这是我使用上述类的测试用例
@Test
public void testArchitectureState() {
ArchitectureUnitState state1 = new ArchitectureUnitState("one","desc one","root");
ArchitectureUnitState state2 = new ArchitectureUnitState("two","desc two","root");
ArchitectureUnit unit1 = new ArchitectureUnit("unit1");
ArchitectureUnit unit2 = new ArchitectureUnit("unit2");
state1.partOf(state2);
state1.stateOf(unit1);
state2.stateOf(unit2);
architectureUnitStateRepository.save(state1);
state1 = architectureUnitStateRepository.findByName("one");
assertEquals("two", state1.getArchitectureUnitState().getName());
assertEquals("unit1", state1.getArchitectureUnit().getName());
state2 = architectureUnitStateRepository.findByName("two");
assertNull(state2.getArchitectureUnitState());
assertEquals("unit2", state2.getArchitectureUnit().getName());
}
它确实按预期通过,并且在图中创建的节点似乎指示相同的情况
请注意,assertNull(state2.getArchitectureUnitState())
保持为true,因为关系的方向被指定为传出。state2中没有关系的传出部分,因此不会加载任何部分
如果我把测试改为
@Test
public void testArchitectureBothWays() {
ArchitectureUnitState state1 = new ArchitectureUnitState("one","desc one","root");
ArchitectureUnitState state2 = new ArchitectureUnitState("two","desc two","root");
ArchitectureUnit unit1 = new ArchitectureUnit("unit1");
ArchitectureUnit unit2 = new ArchitectureUnit("unit2");
state1.partOf(state2);
state2.partOf(state1);
state1.stateOf(unit1);
state2.stateOf(unit2);
architectureUnitStateRepository.save(state1);
state1 = architectureUnitStateRepository.findByName("one");
assertEquals("two", state1.getArchitectureUnitState().getName());
assertEquals("unit1", state1.getArchitectureUnit().getName());
state2 = architectureUnitStateRepository.findByName("two");
assertEquals("one",state2.getArchitectureUnitState().getName());
assertEquals("unit2", state2.getArchitectureUnit().getName());
}
然后我们在两个方向上都有一个关系,现在state2与state1有一个关系。
我们对这样的类进行了测试:@NodeEntity(label=“a”)公共静态类a扩展了E{@Relationship(type=“EDGE”,direction=Relationship.INCOMING)a;}请发布您的实体类和代码,以便我们提供帮助you@Luanne我已经编辑了我的帖子。谢谢