I have two nodes.
How do I express the pattern "nodes without relationships"? Cannot find it in official doc!
A picture to illustrate.
MATCH (n)
WHERE NOT (n)--()
RETURN n;
Related
You are given two arrays one for some labels and another for some relationships and you are asked to return the nodes and their relationships which are found only in the arrays you where given. I tried different approach to it but I couldn't get a better cipher to return the graph with respect to both arrays
MATCH (n)-[r]-(m) where n in ["username"] and r in ["knows"] return n,r
The code above, I know its completely wrong but it kinda shows the idea, share your thoughts š
This should work:
MATCH (n)-[r]-(m)
WHERE ANY(l IN labels(n) WHERE l IN ['username','label2'])
AND type(r) IN ['knows','relType2']
RETURN n,r,m
I have a one direction tree as described as:
Node_C [is_a_method_of] Node_A
Node_D [is_related_to] Node_B
Node_C [is_related_to] Node_D
Node_A and Node_B are root nodes because they are not related or are a method of other nodes. How can I return them?
I saw in another post:
MATCH (n)
WHERE NOT (n)--()
RETURN n;
But that returns orphan nodes.
You would probably want to add some relationship type and direction to your query:
MATCH (n)
WHERE NOT (n)-[:is_a_method_of]->() AND NOT (n)-[:is_related_to]->()
RETURN n;
You haven't really specified the direction of relationships in your graph, so this is only my assumption. You could adapt this query to work on your graph schema if my assumptions are wrong.
I am trying to query using Neo4j.
I would like to print result of obtaining information while AUTO-COMPLETE is ON in Neo4j.
For example, suppose query that creating 3 nodes as shown below.
create (david:Person {name: 'david'}), (mike:Person {name: 'mike'}), (book:Book {title:'book'}), (david)-[:KNOWS]->(mike), (david)-[:WRITE]->(book), (mike)-[:WRITE]->(book)
Here are 2 images:
Auto-complete on
Auto-complete off
Figure is shown after query, and I would like to obtain all relating nodeās relationships based on starting node ('book' node).
I used this query as shown below.
match (book:Book)-[r]-(person) return book, r, person
Whether AUTO-COMPLETE is ON or OFF, I expect to obtain all nodeās relationships including āDavid knows Mikeā, but system says otherwise.
I studied a lot of Syntax structure at neo4j website, and somehow it is very difficult for me. So, I upload this post to acquire assistance for you.
You have to return all the data that you need yourself explicitly. It would be bad for Neo4j to automatically return all the relationships for a super node with thousands of relationships for example, as it would mean lots of I/O, possibly for nothing.
MATCH (book:Book)-[r]-(person)-[r2]-()
RETURN book, r, person, collect(r2) AS r2
Thanks to InverseFalcon, this is my query that works.
MATCH p = (book:Book)-[r]-(person:Person)
UNWIND nodes(p) as allnodes WITH COLLECT(ID(allnodes)) AS ALLID
MATCH (a)-[r2]-(b)
WHERE ID(a) IN ALLID AND ID(b) IN ALLID
WITH DISTINCT r2
RETURN startNode(r2), r2, endNode(r2)
My problem is very close to this topic: Cypher query: Finding all paths between two nodes filtered by relationship properties but what I'm trying to do is to find path which has increasing value of relationship property along the path. So in that previos topics example solution paths (from A to D) would be:
A->D and A->B->D
I used solution from previous topic
START a=node(1), d=node(4)
MATCH p=a-[r:ACTIVATES*..]->d
WITH head(relationships(p)) as r1,p
WHERE all(r2 in relationships(p)
where r2.temperature > r1.temperature)
return p;
and it works for this example. Problem is when there is path with more then 2 relationships, for example:
activates:50 activates:70 activates:60
(A)-------------->(B)-------------->(C)-------------->(D)
this path unfortunately matches too.
Is there way how to write this query in cypher or I'll have to use gremlin instead?
Thanks for any suggestion.
Update: What I need is some construction like (in pseudo programing language):
WITH head(relationships(p)) as r1,p
FOREACH(r2 in tail(relationships(p)):
r1.temperature < r2.temperature, r1 = r2)
but in cypher if it's posible.
This one worked in my example
START a=node(1), d=node(4)
MATCH p=a-[r:ACTIVATES*..]-d
WITH head(relationships(p))as r1,last(relationships(p))as r2,p
WHERE all(r3 in relationships(p)
where r2.temperature > r1.temperature AND NOT r3.temperature < r1.temperature)
return p;
Update: Is there any way to do that with node properties?
How can I show all nodes and relationships in Data Browser tab?
What are sample index queries that I can type in in search field?
You may also want to try a cypher query such as:
START n=node(*) RETURN n;
It's very obvious, and it will return all the existing nodes in the database.
EDITĀ : the following displays the nodes and the relationships :
STARTĀ n=node(*) MATCH (n)-[r]->(m) RETURN n,r,m;
More simple way is
MATCH (n) RETURN (n)
MATCH (n) OPTIONAL MATCH (n)-[r]-() RETURN n, r;
You can show everything with simple MATCH (n) RETURN n, as offical documentation suggests.
START n=node(*) RETURN n from Neo4j 2.0 is deprecated:
The START clause should only be used when accessing legacy indexes
(see Chapter 34, Legacy Indexing). In all other cases, use MATCH
instead (see Section 10.1, āMatchā).
There is a little help icon beside the search field, if you hoover over it it shows the syntax.
If a property of your nodes and relationships is indexed you can search for all of them like this.
node:index:indexname:fieldname:*
rels:index:indexname:fieldname:*
I found that this worked, retrieving all nodes including orphans, and all relationships:
MATCH (n) MATCH ()-[r]->() RETURN n, r
Other good way for get ALL nodes (and nodes without relationship) :
MATCH (n) RETURN n UNION START n = rel(*) return n;