使用Neo4j的密码：在collect中订购项目，并且不包含空对象

我正在尝试编写一个密码查询，其中记录了一个人的多维数据集的所有时间。查询工作正常，但我看到的是，如果结果Time-[：RECORDED_ON]-Cube中没有任何关系，那么我将在collect中得到一条记录，两个值都为null。另外，我想在泰晤士报集合描述中订购日期

这是我的密码：

MATCH(a:App{name:'AllAboutCubes'})<-[:MEMBER_OF]-(p:Cuber) WHERE id(p) = 1234
MATCH(p)<-[:BELONGS_TO {is_active:1}]-(c:Cube)
OPTIONAL MATCH (c)-[:TYPE]->(ct:CubeType)
OPTIONAL MATCH (c)<-[:RECORDED_ON]-(t:Time)-[:RECORDED]-(p) 
WITH c,ct,COLLECT({time:t.time,date:t.date}) as times
RETURN c.name as name, c.manufacturer as manufacturer, ct.type as type, id(c) as cube_id, times as times
ORDER BY manufacturer ASC

我不想要[{time:null，date:null}]

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有什么想法吗？谢谢，让我知道

这应该按降序对日期进行排序，如果没有日期，则返回[]：

MATCH (a:App{name:'AllAboutCubes'})<-[:MEMBER_OF]-(p:Cuber) WHERE id(p) = 1234
MATCH (p)<-[:BELONGS_TO {is_active:1}]-(c:Cube)
OPTIONAL MATCH (c)<-[:RECORDED_ON]-(t:Time)-[:RECORDED]-(p)
WITH c, t ORDER BY t.date DESC
OPTIONAL MATCH (c)-[:TYPE]->(ct:CubeType)
WITH c, ct, CASE WHEN t IS NULL THEN [] ELSE COLLECT({time:t.time,date:t.date}) END as times
RETURN c.name as name, c.manufacturer as manufacturer, ct.type as type, id(c) as cube_id, times
ORDER BY manufacturer

MATCH (a:App{name:'AllAboutCubes'})<-[:MEMBER_OF]-(p:Cuber) WHERE id(p) = 1234
MATCH (p)<-[:BELONGS_TO {is_active:1}]-(c:Cube)
OPTIONAL MATCH (c)<-[:RECORDED_ON]-(t:Time)-[:RECORDED]-(p)
WITH c, t ORDER BY t.date DESC
OPTIONAL MATCH (c)-[:TYPE]->(ct:CubeType)
WITH c, ct, CASE WHEN t IS NULL THEN [] ELSE COLLECT({time:t.time,date:t.date}) END as times
RETURN c.name as name, c.manufacturer as manufacturer, ct.type as type, id(c) as cube_id, times
ORDER BY manufacturer