在同一查询中计数neo4j中节点的总数和小计

Question

我想在cypher中创建一个all in one语句来填充数据表。两个字段计算Sample的数量:一个显示总数，另一个显示没有附加:Ghost标签的数字：

 MATCH (a:Person)-[:OWNER]->(b:Project)-[:PROJECT]->(c:Import)-[:IMPORT|:INPUT|:OUTPUT*]->(d:Sample)
WITH a,b,c,d,d AS e WHERE NOT d:Ghost
RETURN DISTINCT b.Name,(a.`First Name` + " " + a.`Last Name`),b.Description,b.Date,count(DISTINCT c),count(DISTINCT d),count(DISTINCT e)

问题是d和e得出了相同的值，而实际上e应该小于d。我怀疑d和e以某种方式指向同一个地址。我可以通过MATCH再次查询图形来解决这个问题，但是很明显，这是很昂贵的，我宁愿只做一次。有可能吗？

Answer 1

WHERE过滤WITH返回的内容，并且只得到非Ghost节点。将CASE用于条件结果：

WITH a, b, c, d,
     CASE
         WHEN d:Ghost THEN null
         ELSE d
     END AS e

您还可以为结果创建不同的示例节点，然后获取原始列表和筛选列表的大小(因此不会重复两次)：

WITH DISTINCT b.Name AS projectName,
     (a.`First Name` + " " + a.`Last Name`) AS fullName,
     b.Description AS projectDescription,
     b.Date AS projectDate,
     count(DISTINCT c) AS importCount,
     collect(DISTINCT d) AS samples
RETURN projectName, fullName, projectDescription, projectDate, importCount,
       size(samples) AS sampleCount,
       size([s IN samples WHERE NOT s:Ghost]) as nonGhostCount