Question

我尝试了以下查询@ http://goo.gl/Ou2GZG

START s=node(1), t=node(4)
MATCH p=s-[*]-pt--t
WHERE SINGLE (n1 IN nodes(p) 
              WHERE id(n1)=id(t))
WITH DISTINCT pt AS pts, t
MATCH p=t-[*]-pfn
WHERE NONE (n IN nodes(p) 
            WHERE id(n)=3 OR id(n)=7)
RETURN DISTINCT pfn AS pf

但我不想在倒数第二行硬编码3和7，其中3和7是（pts）中包含的节点。我尝试了以下但是我得到了“Unclosed parenthesis”错误

START s=node(1), t=node(4) 
MATCH p=s-[*]-pt--t 
WHERE SINGLE (n1 IN nodes(p) 
              WHERE id(n1)=id(t)) 
WITH DISTINCT pt AS pts, t 
MATCH p=t-[*]-pfn FOREACH(pt in pts : 
                          WHERE NONE (n IN nodes(p) 
                                      WHERE id(n)=id(pt))) 
RETURN DISTINCT pfn AS pf

Answer 1

我认为你可以使用ALL谓词来确保路径p中的每个节点n都不存在pt中与节点n具有相同id的节点，

START s=node(1), t=node(4) 
MATCH p=s-[*]-pt--t 
WHERE SINGLE (n1 IN nodes(p) 
              WHERE id(n1)=id(t)) 
WITH DISTINCT collect(id(pt)) AS pts, t 
MATCH p=t-[*]-pfn 
WHERE ALL (n IN nodes(p) 
           WHERE NONE (pt IN pts 
                       WHERE id(n)= pt)) 
RETURN DISTINCT pfn AS pf

如何在Cypher查询中实现以下功能？

1 个答案: