如何用Cypher查询递归获取Neo4j中规则触发的所有规则链

图结构定义 #

图中包含两种节点类型：

• Rule：包含name属性
• Variable：包含name属性

存在两种关系：

• Rule-[:depends_on]->Variable：规则依赖某个变量
• Rule-[:targets]->Variable：规则作用于某个变量

创建示例图的Cypher查询 #

CREATE
  (rule1:Rule {name:'rule_1'}),
  (rule2:Rule {name:'rule_2'}),
  (rule3:Rule {name:'rule_3'}),
  (rule4:Rule {name:'rule_4'}),
  (variable1:Variable {name:'variable_1'}),
  (variable2:Variable {name:'variable_2'}),
  (variable3:Variable {name:'variable_3'}),
  (variable4:Variable {name:'variable_4'}),
  (variable5:Variable {name:'variable_5'}),
  (variable6:Variable {name:'variable_6'}),
  (variable7:Variable {name:'variable_7'}),
  (rule1)-[:targets]->(variable2),
  (rule1)-[:targets]->(variable3),
  (rule2)-[:targets]->(variable4),
  (rule2)-[:targets]->(variable5),
  (rule3)-[:targets]->(variable6),
  (rule3)-[:targets]->(variable7),
  (rule4)-[:targets]->(variable3),
  (rule1)-[:depends_on]->(variable1),
  (rule2)-[:depends_on]->(variable2),
  (rule3)-[:depends_on]->(variable3),
  (rule4)-[:depends_on]->(variable4)

查询需求 #

给定规则名称，编写Cypher查询递归获取该规则可触发的所有规则链。例如：

• 触发rule_1会触发rule_2（rule_1作用于variable_2，而variable_2是rule_2的依赖）和rule_3（rule_1作用于variable_3，variable_3是rule_3的依赖）
• 触发rule_2会触发rule_4（rule_2作用于variable_4，variable_4是rule_4的依赖）
• 触发rule_4会触发rule_3（rule_4作用于variable_3，variable_3是rule_3的依赖）

目标规则链示例 #

rule_1 -> rule_2 -> rule_4 -> rule_3
rule_1 -> rule_3

实现查询 #

基于APOC库的灵活递归查询 #

以下查询借助APOC库实现任意深度的递归遍历，适合链路长度不确定的场景：

MATCH (startRule:Rule {name: $ruleName})
CALL apoc.path.expandConfig(startRule, {
  relationshipFilter: 'targets>|<depends_on',
  labelFilter: '>Rule',
  uniqueness: 'NODE_PATH',
  beginSequenceAtStart: false
}) YIELD path
WITH nodes(path) as chainNodes
UNWIND [startRule] + chainNodes as fullChainNodes
WITH DISTINCT fullChainNodes
WHERE size(fullChainNodes) > 1
RETURN reduce(chainStr = '', rule in fullChainNodes | chainStr + rule.name + ' -> ')[:-4] as ruleChain
ORDER BY size(fullChainNodes) DESC

说明

1. 使用apoc.path.expandConfig递归遍历起始规则触发的所有后续规则节点，避免重复路径
2. 将起始规则与遍历得到的节点合并为完整链路，去除长度为1的无效链路
3. 通过字符串拼接生成规则链格式，去除末尾多余的->

原生Cypher递归查询（无需APOC） #

若无法使用APOC库，可通过原生递归语法实现：

MATCH (startRule:Rule {name: $ruleName})
CALL {
  WITH startRule
  MATCH path=(startRule)-[:targets]->(:Variable)<-[:depends_on]-(nextRule:Rule)
  RETURN [startRule, nextRule] as chain
  UNION ALL
  WITH startRule
  MATCH path=(startRule)-[:targets]->(:Variable)<-[:depends_on]-(r1:Rule)-[:targets]->(:Variable)<-[:depends_on]-(r2:Rule)
  RETURN [startRule, r1, r2] as chain
  UNION ALL
  WITH startRule
  MATCH path=(startRule)-[:targets]->(:Variable)<-[:depends_on]-(r1:Rule)-[:targets]->(:Variable)<-[:depends_on]-(r2:Rule)-[:targets]->(:Variable)<-[:depends_on]-(r3:Rule)
  RETURN [startRule, r1, r2, r3] as chain
}
WITH DISTINCT chain
RETURN reduce(chainStr = '', rule in chain | chainStr + rule.name + ' -> ')[:-4] as ruleChain
ORDER BY size(chain) DESC

说明

通过多层匹配覆盖不同深度的链路，适合链路深度固定的场景，若深度不确定需手动扩展匹配层级。

内容的提问来源于stack exchange，提问作者Higsn