如何使用属性信息生成关系[Node4j] [英] How to generate relationships using property information [Node4j]

问题描述

我已经导入了一个CSV，其中每个节点包含3列. id，parent_id和标题.这是我在mysql中使用的简单树结构.现在，我需要考虑parent_id数据来创建这些节点之间的关系.因此，每个节点到节点将具有2个父级和子级关系.我对node4j真的很陌生，并提出建议吗?

I have imported a CSV where each Node contains 3 columns. id, parent_id, and title. This is a simple tree structure i had in mysql. Now i need to create the relationships between those nodes considering the parent_id data. So each node to node will have 2 relationships as parent and child. Im really new to node4j and suggestions ?

我尝试跟随，但没有运气

i tried following, but no luck

MATCH (b:Branch {id}), (bb:Branch {parent_id})
CREATE (b)-[:PARENT]->(bb)

推荐答案

似乎您的密码非常接近.您要做的第一件事是在标签Branch的id和parent_id属性上创建一个索引.

It seems as though your cypher is very close. The first thing you are going to want to do is create an index on the id and parent_id properties for the label Branch.

CREATE INDEX ON :Branch(id)

CREATE INDEX ON :Branch(parent_id)

一旦创建了索引，您想要将所有带有标签Branch的节点进行匹配(我将使用一个特定的值来限制它，以开始确保您确实创建了想要的对象)，并为每个节点找到对应的父节点通过匹配索引属性.

Once you have indexes created you want to match all of the nodes with the label Branch (I would limit this with a specific value to start to make sure you create exactly what you want) and for each find the corresponding parent by matching on your indexed attributes.

MATCH (b:Branch), (bb:Branch)
WHERE b.id = ???
  AND b.parent_id = bb.id
CREATE (b)-[:PARENT]->(bb)

一旦您已经在一个分支上证明了这一点，并且您得到了预期的结果，那么我希望我一次将它用于更多分支.您仍然可以选择分批执行此操作，具体取决于图形中的分支数量.

Once you have proved this out on one branch and you get the results you expect I would run it for more branches at once. You could still choose to do it in batches depending on the number of branches in your graph.

创建所有:PARENT关系后，可以选择删除所有parent_id属性.

After you have created all of the :PARENT relationships you could optionally remove all of the parent_id properties.

MATCH (b:Branch)-[:PARENT]->(:Branch)
WHERE exists(b.parent_id)
REMOVE b.parent_id

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