How to return single node with Multiple match for different labels in neo4j?

Question

I am using neo4j 3.2.x. What I am try to do is write a query that will update relation between two nodes (User and Gender) and return single User node with InterestedInGender property pulled from relation as array. There is a problem with the below query as right now it returns multiple records with single value for interestedInGender. Relations are created properly but when returning data it is returning multiple records. I just want to return User node. Is there any way we can fix this query to just return single user node.

MATCH (u:User {_id:"1234"}) 
MATCH (ig:Gender) WHERE ig.value IN ["male", "female"]
WITH u, ig
OPTIONAL MATCH (u)-[igr:INTERESTED_IN_GENDER]->()
DELETE igr
with u, ig
MERGE (u)-[:INTERESTED_IN_GENDER]->(ig)
RETURN u{
  ._id,
  interestedInGender: [(u)-[:INTERESTED_IN_GENDER]->(ig:Gender) | ig.value]
}

InverseFalcon InverseFalcon · Accepted Answer · 2017-12-17T12:21:49

The reason you're getting multiple records (rows) is because your ig match to gender matches to two :Gender nodes...two rows where both rows have the same u node, but different ig nodes. That cardinality remains throughout the rest of your query, and so you get two rows back.

You need to shrink the cardinality of u back down to 1 after you MERGE the relationship, so add this after your MERGE but before your RETURN:

WITH distinct u

How to return single node with Multiple match for different labels in neo4j?

1 Answers