Question

我正在寻找在我的密码查询中进行某种多重JOIN的最佳方法。我找到了解决方案，但对我来说似乎有点太复杂了。此查询的目标是从第一个节点开始，检索ID和名称＆＃39;来自与第一个节点相关的许多其他节点的属性，通过多种类型的关系（参见下文）

START c=node(72)
OPTIONAL MATCH (c)<-[:MANAGEMENT_TEAM]-(from_board)
OPTIONAL MATCH (c)<-[:SHAREHOLDER]-(shareholder)
OPTIONAL MATCH (c)-[:PRODUCT_OFFERED]->(product)
OPTIONAL MATCH (c)<-[:CUSTOMER]-(customer)-[:CORE_BUSINESS]->(industry_sector)
OPTIONAL MATCH (c)-[:ASSIGNEE]->(patent)
RETURN [x IN collect(DISTINCT(from_board))| {id:id(x), name:x.name}] AS Management,
       [x IN collect(DISTINCT(shareholder))| {id:id(x), name:x.name}] AS Shareholders,
       [x IN collect(DISTINCT(product))| {id:id(x), name:x.name}] AS Products,
       [x IN collect(DISTINCT(customer))| {id:id(x), name:x.name}] AS Customers,
       [x IN collect(DISTINCT(industry_sector))| {id:id(x), name:x.name}] AS Industry_sectors,
       [x IN collect(DISTINCT(patent))| {id:id(x), name:x.name}] AS Patent

此查询返回6列，每列包含元组列表（id＆amp; name）。你看到它有多改进吗？（尤其是丑陋的部分＆＃34;提取/收集/分离＆＃34;）或者，你知道另一种方式吗？感谢

Answer 1

一个选项：

MATCH (c)
WHERE id(c) = 72
OPTIONAL MATCH (c)<-[:MANAGEMENT_TEAM]-(from_board)
WITH c, collect(distinct {id:id(from_board), name:from_board.name}) as Management
OPTIONAL MATCH (c)<-[:SHAREHOLDER]-(shareholder)
WITH c, Management, collect(distinct {id:id(shareholder), name:shareholder.name}) as Shareholders
....
RETURN c, Management, Shareholders, ...

另一个选项

MATCH (c)
WHERE id(c) = 72
OPTIONAL MATCH (c)-[r]-(x)
RETURN c,type(r) as type, collect(distinct {id:id(x), name:x.name}) as connected

Answer 2

尝试使用schema indexes进行搜索查询。

它被称为INDEXING

索引使数据库搜索效率提高。用于创建INDEX

CREATE INDEX属性（属性应该是唯一的）

Answer 3

感谢@Micheal，你的第二个选项让cypher查询变得更加清晰，现在它看起来像是：

START c=node({companyId})
      OPTIONAL MATCH (c)<-[management_rel:MANAGEMENT_TEAM]-(management)
      OPTIONAL MATCH (c)<-[shareholder_rel:SHAREHOLDER]-(shareholder)
      OPTIONAL MATCH (c)-[:PRODUCT_OFFERED]->(product)
      OPTIONAL MATCH (c)<-[:CUSTOMER]-(customer)-[:CORE_BUSINESS]->(industry_sector)
      OPTIONAL MATCH (c)-[:ASSIGNEE]->(patent)
    RETURN
      c AS company,
      collect(distinct({id:id(management), name:management.name, role:management_rel.role})) AS managements,
      collect(distinct({id:id(shareholder), name:shareholder.name, percentage:shareholder_rel.percentage  })) AS shareholders,
      collect(distinct({id:id(product), name:product.name, description:product.description})) AS products,
      collect(distinct({id:id(customer), name:customer.name, sector_id:id(industry_sector), sector:industry_sector.name}))  AS customers,
      collect(distinct({id:id(patent), name:patent.name, title:patent.title, website:patent.website, description:patent.description})) AS patents

如您所见，请求有更多加入，例如'managaments'元组中添加的管理关系。此请求需要183 ms才能由EC2小实例上的REST API执行。地图创建，独特和收集的组合非常强大:)。

如何在Cypher中改进多个JOIN查询

3 个答案: