How do you link two nodes together as parent child in neo4j - neo4j

I have a set of nodes which are part of a hierarchy. One node can be related to other node by virtue of child having a parentKey which links to another node. In relational land this would be represented as a 'pigs ear' in an ER diagram.
How can I can generate this relationship between the nodes in neo4j?
I'm quite new to graphs so apologies if I haven't explain it very well.
Thanks

If I understand you correctly, you want to link a "child" node to a "parent" node. That is very easy to do. For instance:
CREATE (child:Person)-[:HAS_PARENT]->(parent:Person)
In this sample data model, we have a Person node label, and a HAS_PARENT relationship type. HAS_PARENT relationships are used to link Person nodes to represent the hierarchy.

If you're talking about already existing nodes, you can match the existing nodes and then use merge to create a relationship.
MATCH (child:SomeLabel) MATCH (parent:SomeOtherLabel)
MERGE (parent)-[:HAS_CHILD]->(child)
You can also use merge when creating new nodes.
See http://neo4j.com/docs/stable/query-merge.html

Related

Create Compositional vs Aggregational relationship in Cypher/Neo4J

Just learning about Graph Databases and NEO4J. I assume that the same distinction between compositional and aggregational relationships applies in Graph databases as in other databases.
Creating an aggregational relationship in NEO4J/Cypher, assuming we already have a Country node called 'Italy' in the database, the below query will create a Currency node called 'Euro' if one doesn't already exist and then create a relationship from Italy to Euro if that relationship doesn't already exist...
MATCH (co:Country {name:'Italy'})
MERGE (cu:Currency {name:'Euro'})
MERGE (co)-[cc:COUNTRYCURRENCY]->(cu)
RETURN cc
If there is not already a relationship Italy->Euro but there is already a Currency node 'Euro' (for example because we had already created the Euro for it to be used by another country, eg 'France'), then the above query would not create a second duplicate node called 'Euro'. it would reuse the existing 'Euro' node and create a new relationship to it from 'Italy'. This is correct behaviour for an aggregational relationship.
So say I want to merge a compositional relationship instead. IE If a parent already has a child with specified name/properties then I don't want to create a duplicate so I need to use MERGE. But if the specified parent doesn't have this child yet and there is already a child node for a different parent with the same name/properties as this child node then I want to create a new one in the context of this parent rather than reuse the existing one. So I need to make the whole operation conditional on whether a relationship already exists between the parent and a child having given properties. Is this syntactically possible in a one-liner?
In Neo4j any node is allowed to exist independently, there is no enforcement of a constraint that says the child can only exist while the parent does.
However you can:
Create a whole path as an atomic operation.
Create a child, only if the parent exists.
Delete a parent and any/all of its child relationships.
Meanwhile a single relationship/edge can of course only exist between two nodes/vertices.
Question:
If I understood your question correctly:
The parent definitely exists.
The child maybe exists, for that parent.
Another child that looks the same, but does not belong to this parent, isn't the parent's child.
Solution:
MERGE (p:Person {name: 'Jasper'})
MERGE (p)-[r:HAS_CHILD]-(c:Child {name: 'Kris'})
What happened:
Regardless of whether there was an existing child named 'Kris', with or without a parent, a new child will be created.
Explanation:
Merge matches a whole pattern, so if that exact pattern does not exist, it will be treated as a create.
We have a knowledge base article on Understanding how MERGE works that covers several cases, including the one for this question.
The section in the article starting with "MERGE using combinations of bound and unbound variables for different use cases" covers what you're after. You want to MATCH or MERGE on the parent node, and then MERGE the relationship to the child node:
MERGE (p:Person {name: 'Jasper'})
MERGE (p)-[r:HAS_CHILD]->(c:Person {name: 'Kris'})
...
MERGE is like a MATCH, and if the MATCH fails, then a CREATE. When we already have bound variables (p is bound to a node because of the MERGE on the first line) then that existing bound node will be used, the node for p won't be recreated.

Merge nodes with different labels into new node with both labels

I have a Neo4J database with a bunch of employee and consultant nodes, with a relationship consults pointing from a consultant to an employee node. A consultant can consult many employees and an employee can have multiple consultants.
My issue is that some (not all!) of the consultants are employees as well. How do I go about merging nodes to have two labels to specify those consultants that are employees?
I exported my data from Postgres and imported it to Neo so I have a bunch of nodes like the examples below:
The name field on all the nodes is unique.
Is there a way to match nodes with the same name, create a new node with the new title, and delete the old nodes?
(c:Consultant {name:“Consultant1”})
(e:Employee {name:“Consultant1"})
Desired fix:
(p:Consultant:Employee {name:“Consultant1”)
The APOC procedure apoc.refactor.mergeNodes should work for your use case.
It merges multiple nodes from a list into the first node, and also merges all their relationships as well.

store files in Neo4j : neostore.nodestore.db

I'm reading about Neo4j underlying infrastructure in it's book and I think I found a contradiction .Here In the text it is mentioned that :"The next four
bytes represent the ID of the first relationship connected to the node, and the following
four bytes represent the ID of the first property for the node" :
but as you can see in the figure 6-4 : if you look at the photo it is Nextrelid! which one is correct? and if we only store first relationship in the nodestore file, what happen to the other relationship?
From the point of view of the node, the next relationship id is the same thing as "the id of the first relationship connected to the node". They're different ways of describing the same thing.
The pattern here is that relationships are stored as a chain. To iterate over all relationships, from the node, you use the id of the first relationship to jump to that relationship in memory, then jump to the area in memory on that relationship where the next rel id is stored and pointer chase across the rest of the chain.
That said, when relationships reach a particular density (I think it's 50 rels per node) then the structure is somewhat different, a new entity is present between the node and its relationships to allow for more efficient navigation of its relationships.

How does Neo4j Traverse a graph using relationship ID's in nodestore file

Something has confused me a lot I was Wondering If you could help me with this please
According to Neo4j graph database book, there are 4 bytes in node store file contains the ID of the nodes relationship . If the node has 100 relationship (and all of them are the node's first relationship in the relationship chain) how does neo4j understand which id to choose??? for example I wrote Match(a:user{Name:'a')-[r:Has-skill]->(b:skill)
Imagine The user node has lot's of relationship but we are interested in [has_skill] relationship how does neo4j understand which id in related to this relationship?
The relationship chain that you are talking about is not the same as a "path". A node does not have more than one relationship that is the first in the chain.
The chain of relationships is a doubly-linked list that contains that Node's relationships. Given that Neo4J already has found the first user in the pattern, it will perform the following steps (or something similar):
Follow the pointer from the node record to the first element of the linked list that contains all of that node's relationships (this first element is the "first relationship in the chain").
For each element of the linked list:
Check if matches the criteria for the searched-for relationship (here, it would be that it has the type HAS_SKILL).
If it does match the criteria, the relationship is kept for future following; if it does not match, it is discarded.
Follow the pointer to the next element in the relationship linked-list (in the "chain"); if at the last element already, exit the loop.
For each of the relationships retrieved by scanning the linked list, follow them to the node they point to and continue evaluating the pattern.
The actual algorithm may differ slightly; e.g. it may use depth-first traversal instead of breadth-first, or it may be optimized in a different way, but the end result is the same.
From Graph Databases, 2nd Edition by Ian Robinson, Jim Webber and Emil Eifrem, page 154:
To find a relationship for a node, we follow that node’s relationship pointer to its first relationship (the LIKES relation‐ ship in this example). From here, we then follow the doubly linked list of relation‐ ships for that particular node (that is, either the start node doubly linked list, or the end node doubly linked list) until we find the relationship we’re interested in.
Finally, #InverseFalcon points out that this will be implemented differently for densely-related nodes, by their estimate at around 50+ relationships. At this point, a slightly different structure is used which groups by types and direction so the cost to search through is reduced.

Dynamic Nodes and relationship creation in Neo4j

I am storing Hierarchy data in Neo4j. I want to store history of the Node. Consider I have a label called GROUP and the earlier name was "MARKETING" now it has been changed to "MARKET123". So i want to create a new node where the name will be MARKET123 and the create a relationship with other connected node same as for the older node named "MARKETING"...
But all this i want to do dynamically instead of passing the other Nodes name and the relationship value in the cypher query.
Please suggest me how it can be done.
You can add versioning to your graph nodes.
Here is a graph gist about time-based versioning that you may adapt to your needs.
http://www.neo4j.org/graphgist?608bf0701e3306a23e77

Resources