Question

以下是我想要的结果

返回图
地方（作为所有者/室友）应为1＆amp; 2级连接（状态= 1）与MU，即使我不想包括MU位置（作为所有者/室友）
用户可以有多个地方
地方必须只有一个拥有者，可能有多个室友

结果应采用格式

place, 
[owner, is 1/2/3 degree connection],
[housemates[housmate1(1/2/3 degree connection),housemate( is 1/2/3 degree connection)]])/

这就是我现在所拥有的

MATCH (n),
      (n)-[rels:`connected_to` {status: 1}]- (sp: `StayPal`),
      (sp)-[tenant0:owner_of|house_mate]-(place:`Place`),
      (place)-[tenant:owner_of|house_mate]-(c: `StayPal`) 
WHERE (ID(n) = {ID_n}) AND NOT(n) - [:house_mate] - (place)
WITH place,
   c,
   tenant, 
   CASE
   WHEN (n)-[:connected_to {status: 1}]-(c) THEN 's'
   WHEN (n)-[:connected_to*1..2 {status: 1}]-(c) THEN 'ss'
   ELSE 'sss'
   END AS connection 
WITH 
   place,
   collect([tenant, c, connection]) AS tenants RETURN place,
           [tenant IN tenants WHERE type(tenant[0]) = 'owner_of'   | tenant][0] AS ownerArray,                               
           [tenant IN tenants WHERE type(tenant[0]) = 'house_mate' | tenant] AS houseMatesArray

编辑2 样本数据集

CREATE
  (mu:MU {name: 'MU1'}),
  (u1:User {number: 1}), (u2:User {number: 2}), (u3:User {number: 3}), (u4:User {number: 4}), (u5:User {number: 5}), (u6:User {number: 6}), (u7:User {number: 7}), (u8:User {number: 8}), (u9:User {number: 9}),
  (mu)-[:connected_to {status: 1}]->(u1),
  (mu)-[:connected_to {status: 1}]->(u2)-[:connected_to {status: 1}]->(u3),
  (mu)-[:connected_to {status: 1}]->(u8),
  (mu)-[:connected_to]->(u9),
  (u4)-[:connected_to {status: 1}]->(u3),
  (u8)-[:connected_to {status: 1}]->(u5),
  (pA:Place {name: 'A'}), (pB:Place {name: 'B'}), (pC:Place {name: 'C'}), (pD:Place {name: 'D'}), (pE:Place {name: 'E'}),
  (mu)<-[:house_mate]-(pA)-[:owner_of]->(u1),
  (u4)<-[:house_mate]-(pB)-[:owner_of]->(u3),
  (u8)<-[:house_mate]-(pC)-[:owner_of]->(u5),
  (u9)<-[:house_mate]-(pD)

Answer 1

试试这个：

#fragment

此查询执行其他 MATCH (n {name: 'MU1'})-[:connected_to*1..2 {status: 1}]->(sp:User), (sp)<-[:owner_of|house_mate]-(place:Place)-[:owner_of|house_mate]->(c:User) WHERE NOT (n)-[:house_mate]-(place) MATCH (place)-[tenant:owner_of|house_mate]->(u:User), (u)-[rels:connected_to*1..3 {status: 1}]-(n) WITH DISTINCT place, type(tenant) AS type, u, size(rels) AS connection WITH place, collect({type: type, u: u, connection: connection}) AS tenants RETURN place, [tenant IN tenants WHERE tenant.type = 'owner_of' | [tenant.u, tenant.connection]][0] AS owner, [tenant IN tenants WHERE tenant.type = 'house_mate' | [tenant.u, tenant.connection]] AS houseMatesArray，以便它可以将MATCH - ＆gt; place和sp中的用户收集到一个地点（在变量place->c下）。

它返回：

如何在return语句

1 个答案: