如何用Cypher查询部门树中两节点间的节点及关联用户

Got it, let's work through this problem step by step. First, I'll make some reasonable assumptions about your graph model (since you didn't specify it explicitly):

• Department nodes have a label Department with a name property (e.g., 'D1', 'D1.A1', 'D1.A1.B1')
• Parent departments connect to child departments via a PARENT_OF relationship (so (parent:Department)-[:PARENT_OF]->(child:Department))
• Departments link to their users via a HAS_USER relationship (so (dept:Department)-[:HAS_USER]->(user:User)), with users having a name property like 'U1', 'U2', 'U4'

Here's the Cypher query that will deliver exactly the results you need:

// Match the full hierarchy path from root department D1 to target department D1.A1.B1
MATCH path=(root:Department {name: 'D1'})-[:PARENT_OF*]->(target:Department {name: 'D1.A1.B1'})
// Extract all department nodes along the matched path
WITH nodes(path) AS pathDepartments
// Unwind the list to process each department individually
UNWIND pathDepartments AS dept
// Match all users associated with each department in the path
MATCH (dept)-[:HAS_USER]->(user:User)
// Collect unique departments and users to avoid duplicate entries
WITH COLLECT(DISTINCT dept.name) AS departments, COLLECT(DISTINCT user.name) AS users
// Return the final result sets
RETURN departments, users

Breakdown of how this works: #

1. Path Matching: The [:PARENT_OF*] matches any number of consecutive PARENT_OF relationships, capturing the full hierarchy from D1 down to D1.A1.B1. The nodes(path) function pulls out all departments on this exact path.
2. Unwind & User Matching: We break down the list of departments to find every user linked to each department in the path.
3. Collect Results: Using COLLECT(DISTINCT) ensures we don't get duplicate entries (a safe practice even if your data doesn't have overlaps right now).

Quick adjustments for different graph models: #

• If your parent-child relationship is reversed (child departments point to parents with a PARENT relationship), update the path match to:

  MATCH path=(target:Department {name: 'D1.A1.B1'})-[:PARENT*]->(root:Department {name: 'D1'})
  

• If your department uses an identifier other than name (like id), swap out the property in the MATCH clauses.
• If users point to departments via a BELONGS_TO relationship instead, adjust the user match to (user:User)-[:BELONGS_TO]->(dept).

内容的提问来源于stack exchange，提问作者Yogi