neo4j 查找父节点是否存在关系 [英] neo4j find parent node is relationship exists

问题描述

我正在尝试开发像 Google Drive 这样的产品.现在是第二阶段.我被困在进行密码查询，我需要帮助.

I am trying to develop a product like a google drive. This is the second phase now. I am stuck in making cypher query and I need help.

用户 A 可以与群组或其他用户 B 共享文件/文件夹.

a user A can share a file/folder with a group or another user B.

case 1:用户A，共享了包含其他子文件夹的文件夹.用户 B 将看到共享文件夹.但是，现在用户 A 共享了一个子文件夹，而用户 B 应该只能看到根文件夹.

case 1: user A, shared the folder which consists of other subfolders. user B would see the shared folder. however, now user A shared a subfolder, and again user B should see only the root folder.

案例2:用户A，共享了唯一的子文件夹，那么用户B将只能看到一个共享的子文件夹

case 2: user A, shared the only subfolder then, user B will only see a shared subfolder

基本上我需要从底部遍历节点到它的父节点和它们的父节点，如果有的话，并在具有这种关系的第一个节点处停止

basically I need to traverse the node from bottom to its parent and their parent with CAN_ACCESS_DIR relationship if any and stop at the first node with this relationship

请帮忙

附加图片场景 ->找到JAVA2的根(JAVA)并检查它是否已与同一组或用户共享(CAN_ACCESS_DIR).如果是，那么我们将只发送 JAVA 文件夹.如果没有，则只发送 JAVA2 文件夹

attached image scenario -> find the root (JAVA) of JAVA2 and check if it has been shared (CAN_ACCESS_DIR) with the same group or a user. If yes, then we will send only the JAVA folder. If not then send only JAVA2 folder

match (g:group) where g.name = "group1" 
match p = (d:directory { name: "JAVA2"})<-[:LEAF | CONTAINS*0..]-(a:directory) where (d)<-[:CAN_ACCESS_DIR]-(g)
unwind nodes(p) as i
return collect(distinct(i)) as result

match (u:user) where u.name = "John"
match (t:tower) where t.name = "Genome"
match (d:directory) where d.name = "Root"
match (u)-[:MEMBER_OF_GROUP]->(g:group)
match (d)-[:CONTAINS*0..]->(s) where (s)<-[:CAN_ACCESS_DIR]-(g)<-[:MEMBER_OF_GROUP]-(u)

不是neo4j的专家，请指导我.谢谢

not an expert in neo4j, please guide me out. thanks

推荐答案

[UPDATED]

此查询应返回最接近根的组可访问的a 目录，从特定(子)目录开始:

This query should return the group-accessible a directory that is closest to the root, starting at a specific (sub)directory:

MATCH p=(g:group)-[:CAN_ACCESS_DIR]->(d:directory)<-[:CONTAINS*0..]-(a)
WHERE g.name = "group1" AND d.name = "JAVA2"
MATCH (a)<-[:CAN_ACCESS_DIR]-(g)
RETURN p, a
ORDER BY LENGTH(p) DESC
LIMIT 1

由于单个 MATCH 子句会过滤掉同一关系的多次使用，此查询使用 2 个 MATCH 子句来允许相同的 CAN_ACCESS_DIR匹配两次的关系，以防 d 是唯一可访问的目录(因此与 a 相同).

Since a single MATCH clause would filter out multiple uses of the same relationship, this query uses 2 MATCH clauses to allow the same CAN_ACCESS_DIR relationship to match twice, in case d is the only accessible directory (and therefore the same as a).

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