Neo4j Cypher-如何使用Cypher有效地计算多个属性值并正确分页

我正在努力获得正确的密码，既有效又允许通过跳过和限制分页

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下面是一个简单的场景：我有相关的节点（公司），您可以将

与

和聚合一起使用，还可以选择映射结果

match (c:Company {id: 'MY.co'})<-[:type_of]-(s:Set)<-[:job_for]-(j:Job) 
with s, j.Status as Status,count(*) as StatusCount
return s.Description, collect([Status,StatusCount]);

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match（c:Company{id:'MY.co'}）这非常有效！谢谢我曾试图找出如何以这种方式使用collect，但没有成功。这仍然是有效的，并允许对集合进行分页。
s.Description        j.Status       StatusCount
-------------------+--------------+----------------
Set 1              | Unassigned   | 10
Set 1              | Completed    | 2
Set 2              | Unassigned   | 3
Set 1              | Reviewed     | 10
Set 3              | Completed    | 4
Set 2              | Reviewed     | 7

s.Description        Unassigned      Completed    Reviewed
-------------------+--------------+-------------+----------
Set 1              | 10           | 2           | 10
Set 2              | 3            | 0           | 7
Set 3              | 0            | 4           | 0

match (c:Company {id: 'MY.co'})<-[:type_of]-(s:Set)<-[:job_for]-(j:Job) 
with s, j.Status as Status,count(*) as StatusCount
return s.Description, collect({Status:Status,StatusCount:StatusCount]) as StatusCounts;


   List<Object> statusMaps =  (List<Object>) row.get("StatusCounts");
   for(Object statusEntry : statusMaps ) {
     Map<String,Object> statusMap = (Map<String,Object>) statusEntry;
     String status = (String) statusMap.get("Status");
     Number count = statusMap.get("StatusCount");
    }

match (c:Company {id: 'MY.co'})<-[:type_of]-(s:Set)<-[:job_for]-(j:Job) 
with s, j.Status as Status,count(*) as StatusCount
return s.Description, collect([Status,StatusCount]);

match (c:Company {id: 'MY.co'})<-[:type_of]-(s:Set)<-[:job_for]-(j:Job) 
with s, j.Status as Status,count(*) as StatusCount
return s.Description, collect({Status:Status,StatusCount:StatusCount]);