Neo4j apoc.convert.toTree 不显示具有唯一关系的重复子项

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英文:

Neo4j apoc.convert.toTree is not showing duplicate children with unique relationship

问题

I have two nodes A and B (A->B) connected via two relationships. Those 2 relationships are unique. For example, you can think of the two rels as two paths from A to B at two different times. Converting them to one relationship is not viable at this time.

For displaying in UI, query result should be a tree structure. Used following query to get the tree

the resulting tree is

Preferred result is

How can I get a tree with duplicate children is more than one relationship exists between same two nodes?

Thanks.

英文:

I have two nodes A and B (A->B) connected via two relationships. Those 2 relationships are unique. For example, you can think of the two rels as two paths from A to B at two different times. Converting them to one relationship is not viable at this time.
Neo4j apoc.convert.toTree 不显示具有唯一关系的重复子项

For displaying in UI, query result should be a tree structure. Used following query to get the tree

  1. MATCH p=(n:Label1 {name:'main'})-[:calls*..2]->(m)
  2. WITH COLLECT(p) AS ps 
  3. CALL apoc.convert.toTree(ps) yield value
  4. RETURN value;

the resulting tree is

Neo4j apoc.convert.toTree 不显示具有唯一关系的重复子项

Preferred result is

Neo4j apoc.convert.toTree 不显示具有唯一关系的重复子项

How can I get a tree with duplicate children is more than one relationship exists between same two nodes?

Thanks.

答案1

得分: 0

我在这个SO问题上提供了一个解决方法。虽然它不完全等同于apoc函数生成的嵌套文档，但我认为它可以满足您的需求。

  1. MATCH p=(n:Label1 {name:'main'})-[:calls*..2]->(m)
  2. // 每个路径都将被返回
  3. WITH COLLECT(p) AS ps, p, n
  4. CALL apoc.convert.toTree(ps) YIELD value
  5. // 作为嵌套文档返回节点n及其关系
  6. RETURN {_id: ID(n), _type: labels(n), node:n, relationships: COLLECT(value['calls'][0])} AS result

示例结果：

  1. [
  2.   {
  3.     "result": {
  4.       "_id": 31,
  5.       "_type": [
  6.         "Label1"
  7.       ],
  8.       "node": {
  9.         "name": "main"
  10.       },
  11.       "relationships": [
  12.         {
  13.           "calls.y": "prop y",
  14.           "_type": "Main",
  15.           "_id": 32
  16.         },
  17.         {
  18.           "_type": "Main",
  19.           "_id": 32,
  20.           "calls.x": "prop x"
  21.         }
  22.       ]
  23.     }
  24.   }
  25. ]

希望这对您有帮助。

英文:

I offer a hack on this SO question. Although it is not exactly the resulting nested document as the apoc function but I think it can server your purpose.

  1. MATCH p=(n:Label1 {name:'main'})-[:calls*..2]->(m)
  2. // each path will be returned
  3. WITH COLLECT(p) AS ps, p, n
  4. CALL apoc.convert.toTree(ps) yield value
  5. // return node n and it's relationships as a nested document
  6. RETURN {_id: ID(n), _type: labels(n), node:n, relationships: collect(value['calls'][0])} as result

Sample result:

  1. [
  2.   {
  3.     "result": {
  4.       "_id": 31,
  5.       "_type": [
  6.         "Label1"
  7.       ], 
  8.       "node": {
  9.         "name": "main"
  10.       },
  11.       "relationships": [
  12.         {
  13.           "calls.y": "prop y",
  14.           "_type": "Main",
  15.           "_id": 32
  16.         },
  17.         {
  18.           "_type": "Main",
  19.           "_id": 32,
  20.           "calls.x": "prop x"
  21.         }
  22.       ]
  23.     }
  24.   }
  25. ]

huangapple
  • 本文由 发表于 2023年2月6日 13:04:06
  • 转载请务必保留本文链接：https://go.coder-hub.com/75357491.html
  • neo4j
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