Neo4j 如何获取（DISTINCT）所有节点和至少有一个节点的关系？

Question

我最近一直在尝试使用 neo4j 和 cypher，但我不太能正确地获得以下查询。我想 return 与给定员工共享至少一个项目的所有员工以及他们从事的项目。举个例子可能更简单：

对于输入为员工 1 的上图，期望的结果是：

Employee	Project
0	A
1	A
1	B
4	A

我尝试了以下查询，但它 return 存在重复关系：

MATCH (a0:Employee {name:1})-[:WORKS]->(b0:Project) 
MATCH (b0)<-[:WORKS]-(a:Employee) 
MATCH (a)-[:WORKS]->(b:Project) 
RETURN a.name AS employee, b.name AS project 
ORDER BY employee, project

Employee	Project
0	A
1	A
1	A
1	B
1	B
4	A

预先感谢您的帮助。

注意：这些查询可用于创建上图

CREATE (:Employee {name: 0}), 
       (:Employee {name: 1}), 
       (:Employee {name: 2}),  
       (:Employee {name: 3}), 
       (:Employee {name: 4}), 
       (:Project {name: 'A'}), 
       (:Project {name: 'B'}), 
       (:Project {name: 'C'})
MATCH (e:Employee), (p:Project) WHERE e.name=0 AND p.name='A' CREATE (e)-[:WORKS]->(p)
MATCH (e:Employee), (p:Project) WHERE e.name=1 AND p.name='A' CREATE (e)-[:WORKS]->(p)
MATCH (e:Employee), (p:Project) WHERE e.name=1 AND p.name='B' CREATE (e)-[:WORKS]->(p)
MATCH (e:Employee), (p:Project) WHERE e.name=4 AND p.name='A' CREATE (e)-[:WORKS]->(p)
MATCH (e:Employee), (p:Project) WHERE e.name=3 AND p.name='C' CREATE (e)-[:WORKS]->(p)

Answer 1

经过几个小时的努力，我找到了解决方案。我不确定它是否有效。在某种程度上，答案已经在问题中 - 不同。此查询完成工作：

MATCH (a0:Employee {name:0})-[:WORKS]->(b0:Project) 
MATCH (b0)<-[:WORKS]-(a:Employee)
WITH DISTINCT(a)
MATCH (a)-[:WORKS]->(b:Project) 
RETURN a.name AS employee, b.name AS project 
ORDER BY employee, project

Answer 2

您似乎想要问题的答案：给定某个员工从事的项目，找出所有从事这些项目的员工。

你可以用一个简单的查询来回答这个问题：

MATCH (:Employee {name:1})-[:WORKS]->(b:Project) 
MATCH (a:Employee)-[:WORKS]->(b)
RETURN a.name AS employee, b.name AS project 
ORDER BY employee, project

Neo4j 如何获取（DISTINCT）所有节点和至少有一个节点的关系？

Neo4j how to get (DISTINCT) all nodes and relationships with at least one node in common?

neo4j

cypher