Find node using another node and relationship

Question

How to check whether a node exists using relationship and other node of the relationship?

 A ->(IN) B

I want to check if B is present with node A having relationship IN using py2neo

I tried this cypher query:

MATCH (a { name:'xyz' })<-[:IN]-(b)
Return b

But I was looking for something in py2neo like find function?

Are you using cypher? what are your labels? what's the context? what have you tried (as Stefan already said). You should take a look at this: http://stackoverflow.com/help/how-to-ask — Supamiu, Dec 10 '15 at 09:11

Supamiu · Answer 1 · 2015-12-10T19:04:56.540

0

I'll assume you are using cypher, so you should try this:

OPTIONAL Match (A:Foo)-[:IN]->(B:Bar)
return RETURN B IS NOT NULL AS exists

Using case allows you to return a boolean based on the existence of B using an optional match.

edited Dec 10 '15 at 19:04

answered Dec 10 '15 at 09:10

Supamiu

Your `RETURN` clause can be simplified to: `RETURN B IS NOT NULL AS exists`. – cybersam Dec 10 '15 at 17:56
I was not sure if it was correct, so I did a CASE, thank you, will edit asap :) – Supamiu Dec 10 '15 at 18:37

score 0 · Answer 2 · answered Dec 10 '15 at 17:53

0

You can use the py2neo function match-one() (documented on this page). It will return one such relationship, should any exist.

If you want to see all such relationships, you can use the match() function instead.

answered Dec 10 '15 at 17:53

cybersam

2 Answers2