match (v:person)-[e:officer]->(v2)<-[e1:officer]-(v3:person)
where v.person.NAME contains ‘任’
return id(v) as id,properties(v).NAME as name,e.REL as rel,properties(v2).CORP_NAME as corpName, id(v2) as corpKey ,properties(v3).NAME as partners,id(v3) as partnersId
order by id asc, corpKey asc ,partnersId asc
// 如果 Person Officer ... 有对应的 PersonDao OfficerDao... 继承了 NebulaDaoBasic, 此时已构建起 tag 与 java 实体的关系,可直接使用 Map
@Data
public BizPojo {
private Person v;
private Officer e;
private Map v2;
private Officer e2;
private Person v3;
}
ExampleDao.java
List<BizPojo> selectExample();
ExampleDao.xml
<mapper namespace="com.example.ExampleDao">
<select id="selectExample" resultType="com.example.model.BizPojo">
match (v:person)-[e:officer]->(v2)<-[e1:officer]-(v3:person)
where v.person.NAME contains ‘任’
return
v, e, v2, v3
order by v asc, v2 asc ,v3 asc
</select>
</mapper>