How to specify more than two relationships to the same node in a `MERGE`

Question

I have this situation

CREATE (a:X { id: 1 })
CREATE (b:Y { id: 2 })
CREATE (c:Z { id: 3 })
CREATE (:Z { id: 4 })
CREATE (a)-[:HAS]->(d:W { id: 5 })<-[:HAS]-(b)
CREATE (c)-[:HAS]->(d)

And I want to run this query

MATCH (a:X { id: 1 })
MATCH (b:Y { id: 2 })
MATCH (c:Z { id: 4 }) // The one with no relationships yet
MERGE (a)-[:HAS]->(d:W)<-[:HAS]-(b), (d)<-[:HAS]-(c)
ON CREATE SET d.id = 6

The problem is that I'm not allowed to put more than one path in a single MERGE, how can I do this?

This is the result I want to achieve

Notice that I can't just do this

MATCH (a:X { id: 1 })
MATCH (b:Y { id: 2 })
MATCH (c:Z { id: 4 })
MERGE (a)-[:HAS]->(d:W)<-[:HAS]-(b) // This path already exists, but it uses the other `Z`
ON CREATE SET d.id = 6
MERGE (d)<-[:HAS]-(c) // Separate merge

Because the result would be

Finbar Good · Accepted Answer · 2025-11-14 10:49:35Z

2

In order to create three relationships connecting three existing nodes to a new node with label W — but only if those nodes are not already so connected — you can do it using CREATE:

MATCH (a:X { id: 1 }), (b:Y { id: 2 }), (c:Z { id: 4 })
  WHERE NOT EXISTS { (a)-[:HAS]->(d:W)<-[:HAS]-(b), (d)<-[:HAS]-(c) }
CREATE (a)-[:HAS]->(d:W)<-[:HAS]-(b), (d)<-[:HAS]-(c)

If the pattern in NOT EXISTS already exists, then MATCH produces no rows for CREATE to execute over i.e. nothing is created.

edited Nov 14 at 10:49

answered Nov 13 at 22:30

Finbar Good

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5 Comments

AFatNiBBa Nov 13 at 23:09

Sorry if I wasn't clear: The id was just to recognize the node, the real "id" of my node are the 3 relationships

Finbar Good Nov 14 at 10:50

OK I think I get it now. Answer updated.

AFatNiBBa Nov 14 at 19:24

Is there a way to make it so that I can return the new/existing "W" node from the query?

Finbar Good Nov 17 at 16:11

The node d (and a and b and c) is available for projecting after CREATE, so you can just add the line RETURN d to return that node in the query results.

AFatNiBBa Nov 20 at 16:41

Wouldn't your query skip the ones that already exist?

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