Neo4J 和 Cypher 查询

Question

我是 Neo4j 和 Cypher 查询的新手。我的创建查询就像每个商店有 2 个冷水机，每个冷水机有 2 个 PLC，每个冷水机有 2 个传感器。

创建如下

Create(:SHOP{name:"Shop1"})-[:hasChiller]->(:CHILLER{name:"Chiller1"})
Create(:SHOP{name:"Shop1"})-[:hasChiller]->(:CHILLER{name:"Chiller2"})

Create(:SHOP{name:"Shop2"})-[:hasChiller]->(:CHILLER{name:"Chiller3"})
Create(:SHOP{name:"Shop2"})-[:hasChiller]->(:CHILLER{name:"Chiller4"})

Create(:CHILLER{name:"Chiller1"})-[:hasPLC]->(:PLC{name:"Plc1"})
Create(:CHILLER{name:"Chiller1"})-[:hasPLC]->(:PLC{name:"Plc2"})

Create(:CHILLER{name:"Chiller2"})-[:hasPLC]->(:PLC{name:"Plc3"})
Create(:CHILLER{name:"Chiller2"})-[:hasPLC]->(:PLC{name:"Plc4"})

Create(:CHILLER{name:"Chiller3"})-[:hasPLC]->(:PLC{name:"Plc5"})
Create(:CHILLER{name:"Chiller3"})-[:hasPLC]->(:PLC{name:"Plc6"})

Create(:CHILLER{name:"Chiller4"})-[:hasPLC]->(:PLC{name:"Plc7"})
Create(:CHILLER{name:"Chiller4"})-[:hasPLC]->(:PLC{name:"Plc8"})

Create(:PLC{name:"Plc1"})-[:hasSensor]->(:SENSOR{name:"Sensor1"})
Create(:PLC{name:"Plc1"})-[:hasSensor]->(:SENSOR{name:"Sensor2"})
Create(:PLC{name:"Plc2"})-[:hasSensor]->(:SENSOR{name:"Sensor3"})
Create(:PLC{name:"Plc2"})-[:hasSensor]->(:SENSOR{name:"Sensor4"})
Create(:PLC{name:"Plc3"})-[:hasSensor]->(:SENSOR{name:"Sensor5"})
Create(:PLC{name:"Plc3"})-[:hasSensor]->(:SENSOR{name:"Sensor6"})
Create(:PLC{name:"Plc4"})-[:hasSensor]->(:SENSOR{name:"Sensor7"})
Create(:PLC{name:"Plc4"})-[:hasSensor]->(:SENSOR{name:"Sensor8"})
Create(:PLC{name:"Plc5"})-[:hasSensor]->(:SENSOR{name:"Sensor9"})
Create(:PLC{name:"Plc5"})-[:hasSensor]->(:SENSOR{name:"Sensor10"})
Create(:PLC{name:"Plc6"})-[:hasSensor]->(:SENSOR{name:"Sensor11"})
Create(:PLC{name:"Plc6"})-[:hasSensor]->(:SENSOR{name:"Sensor12"})
Create(:PLC{name:"Plc7"})-[:hasSensor]->(:SENSOR{name:"Sensor13"})
Create(:PLC{name:"Plc7"})-[:hasSensor]->(:SENSOR{name:"Sensor14"})
Create(:PLC{name:"Plc8"})-[:hasSensor]->(:SENSOR{name:"Sensor15"})
Create(:PLC{name:"Plc8"})-[:hasSensor]->(:SENSOR{name:"Sensor16"})

然而，在 SHOP1 下获得传感器的匹配

MATCH(s:SHOP{name:"Shop1"})-[:hasChiller]->(cc:CHILLER)-[:hasPLC]->(pp:PLC)-[:hasSensor]->(ss:SENSOR) return ss.name

什么都不返回。说没有变化，也没有数据。

我正在 Neo4J 沙盒环境中尝试这个。我这样做是基于我在 SQL SERVER GRAPH 2019 中使用 match 子句的理解。

谁能指出我哪里出错了？

Answer 1

您不正确地创建了“相同”节点的多个实例。 您应该创建每个节点一次，然后在需要创建涉及该节点的关系时使用其绑定的变量名称。

删除所有数据并改为遵循此模式（您必须填写“...”部分）：

CREATE
  (sh1:SHOP{name:"Shop1"}), (sh2:SHOP{name:"Shop1"}),
  (c1:CHILLER{name:"Chiller1"}), (c2:CHILLER{name:"Chiller2"}),(c3:CHILLER{name:"Chiller3"}), (c4:CHILLER{name:"Chiller4"}),
  (p1:PLC{name:"Plc1"}), ..., (p8:PLC{name:"Plc8"}),
  (se1:SENSOR{name:"Sensor1"}), ..., (se16:SENSOR{name:"Sensor16"}),

  (sh1)-[:hasChiller]->(c1), (sh1)-[:hasChiller]->(c2),
  ... // create remaining relationships using bound variable names for nodes