Question

大家好！我有一个Neo4j数据库，其中包含有关在特定时刻访问多个检查点的访问者的信息。访问者，访问和检查点表示为节点。

每个访问者都由他/她的设备MAC地址标识，每个检查点都有唯一的ID。

每个访问节点都包含一个EnterTime属性，该属性是在相应检查点附近首次发现具有某些特定MAC地址的访问者时的时间（自1970年以来的毫秒数）。

注意：访问的持续时间应计算为下次访问的EnterTime与此次访问的EnterTime之间的差异。每个检查点可能会被访问多次。

使用此密码查询创建数据库：

//Create visitors:
CREATE (Visitor1: Visitor {MAC:'00:0a:95:9d:68:16'})
CREATE (Visitor2: Visitor {MAC:'00:0a:95:9d:68:17'})
CREATE (Visitor3: Visitor {MAC:'00:0a:95:9d:68:18'})

//Create CheckPoints:
CREATE (CheckPoint1: CheckPoint {CheckPointId: 1})
CREATE (CheckPoint2: CheckPoint {CheckPointId: 2})
CREATE (CheckPoint3: CheckPoint {CheckPointId: 3})
CREATE (CheckPoint4: CheckPoint {CheckPointId: 4})
CREATE (CheckPoint5: CheckPoint {CheckPointId: 5})


//Create visits:
//by visitor 1:
CREATE  (Visitor1)-[:MAKES]->(Visit1: Visit {EnterTime: 1488358800000})-[:TO]->(CheckPoint1)
CREATE  (Visitor1)-[:MAKES]->(Visit2: Visit {EnterTime: 1488359400000})-[:TO]->(CheckPoint2)
CREATE  (Visitor1)-[:MAKES]->(Visit3: Visit {EnterTime: 1488361200000})-[:TO]->(CheckPoint3)
CREATE  (Visitor1)-[:MAKES]->(Visit4: Visit {EnterTime: 1488363600000})-[:TO]->(CheckPoint4)
CREATE  (Visitor1)-[:MAKES]->(Visit5: Visit {EnterTime: 1488364800000})-[:TO]->(CheckPoint5)
CREATE  (Visitor1)-[:MAKES]->(Visit6: Visit {EnterTime: 1488365400000})-[:TO]->(CheckPoint1)

//by visitor 2:
CREATE  (Visitor2)-[:MAKES]->(Visit7: Visit {EnterTime: 1488358800000})-[:TO]->(CheckPoint1)
CREATE  (Visitor2)-[:MAKES]->(Visit8: Visit {EnterTime: 1488360300000})-[:TO]->(CheckPoint4)
CREATE  (Visitor2)-[:MAKES]->(Visit9: Visit {EnterTime: 1488362400000})-[:TO]->(CheckPoint2)
CREATE  (Visitor2)-[:MAKES]->(Visit10: Visit {EnterTime: 1488363000000})-[:TO]->(CheckPoint1)

//by visitor 3:
CREATE  (Visitor3)-[:MAKES]->(Visit11: Visit {EnterTime: 1488353820000})-[:TO]->(CheckPoint1)
CREATE  (Visitor3)-[:MAKES]->(Visit12: Visit {EnterTime: 1488354600000})-[:TO]->(CheckPoint4)
CREATE  (Visitor3)-[:MAKES]->(Visit13: Visit {EnterTime: 1488358200000})-[:TO]->(CheckPoint3)
CREATE  (Visitor3)-[:MAKES]->(Visit14: Visit {EnterTime: 1488359700000})-[:TO]->(CheckPoint1)

我正在编写一个密码查询来查找每个检查点的总访问时间。我正在尝试这个：

match (vr: Visitor)-->(v1: Visit)-->(cp1: CheckPoint),
 (vr)-->(v2: Visit)-->()
where v2.EnterTime > v1.EnterTime
with cp1.CheckPointId as CheckPointId,
     v1.EnterTime as EnterTime,
     min((v2.EnterTime - v1.EnterTime) / 60000) as visit_duration
return CheckPointId, sum(visit_duration) as total_visit_duration
order by CheckPointId;

结果应为：

cp1  - 48 min
cp2 – 40 min
cp3 – 65 min
cp4 – 115 min
cp5 – 10 min

但是我的查询返回了错误的结果：

cp1  - 23 min
cp2 – 40 min
cp3 – 65 min
cp4 – 115 min
cp5 – 10 min

显然，错误是访问者（mac == 00：0a：95：9d：68：17）访问id == 1的检查点的长度没有被考虑在内。但这是我的猜测。

我的查询有什么问题？如何改进以返回正确的结果？

非常感谢您的帮助！谢谢！

Answer 1

我认为你在寻找的是：

match ()<--(v2:Visit)<--(vr: Visitor)-->(v1: Visit)-->(cp1: CheckPoint)
where v2.EnterTime > v1.EnterTime
with vr,
 cp1.CheckPointId as CheckPointId,
 v1.EnterTime as EnterTime,
 min((v2.EnterTime - v1.EnterTime) / 60000) as visit_duration
return CheckPointId, sum(visit_duration) as total_visit_duration
order by CheckPointId;

我在vr声明中添加了with。 Cypher根据您在with语句中输入的所有值进行分组。您的代码无法工作的原因是因为访问者1和2的检查点1的v1.EnterTime是相同的。因为EnterTime语句中只有CheckpointID和with，所以cypher按EnterTime分组，所以因为你有min(v2.EnterTime - v1.EnterTime)它只是拿了较小的一个并且做了Visitor不要同时使用。现在我们在with语句中引入soundPlayer = new Audio(soundName).play();，如果两个人同时启动相同的检查点，它也将按访问者分组，这解决了我们的问题。

希望这有帮助

带聚合函数的Neo4j cypher查询返回不完整的结果

1 个答案: