I need a MATCH where either relationship is true. I understand the (person1)-[:r1|:r2]-(person2). The problem I am having is that one of the MATCH traverse through another node. IE:
(p1:person)-[:FRIEND]-(p2:person)-[:FRIEND]-(p3:person)
So I want this kind of logic. The enemy of my enemy is my friend. And my friend is my friend. Output list of all the names who are my friend. I also limit the relationship to a particular value.
Something like:
MATCH (p1:Person)-[:ENEMY{type:'human'}]-(myEnemy:Person)-[enemy2:ENEMY{type:'human'}]-(myFriend:Person)
OR (p1:Person)-[friend:FRIEND{type:'human'}]-(myFriend:Person)
RETURN p1.name, myFriend.name
I need one list that I can then do aggregation on.
This is my first posting....so if my question is a mess...hit me with your feedback and I will clarify :)
You can use the UNION clause to combine 2 queries and also remove duplicate results:
MATCH (p:Person)-[:ENEMY{type:'human'}]-(:Person)-[:ENEMY{type:'human'}]-(f:Person)
WHERE ID(p) < ID(f)
RETURN p.name AS pName, f.name AS fName
UNION
MATCH (p:Person)-[:FRIEND{type:'human'}]-(f:Person)
WHERE ID(p) < ID(f)
RETURN p.name AS pName, f.name AS fName
The ID(p) < ID(f) filtering is done to avoid having the same pair of Person names being returned twice (in reverse order).
[UPDATE]
To get a count of how many friends each Person has, you can take advantage of the new CALL subquery syntax (in neo4j 4.0) to do post-union processing:
CALL {
MATCH (p:Person)-[:ENEMY{type:'human'}]-(:Person)-[:ENEMY{type:'human'}]-(f:Person)
WHERE ID(p) < ID(f)
RETURN p.name AS pName, f
UNION
MATCH (p:Person)-[:FRIEND{type:'human'}]-(f:Person)
WHERE ID(p) < ID(f)
RETURN p.name AS pName, f
}
RETURN pName, COUNT(f) AS friendCount
Related
i'm using neo4j. what i'd like to do is to create a root node for search result and to create relationships from root node to search result nodes. and I'd like to set incremental number to each relationship's property.
if possible, with one query.
Sorry for not explaining enough.
This is what I'd like to do.
Any more concise way?
// create test data
WITH RANGE(0, 99) AS indexes,
['Paul', 'Bley', 'Bill', 'Evans', 'Robert', 'Glasper', 'Chihiro', 'Yamanaka', 'Fred', 'Hersch'] AS names
UNWIND indexes AS index
CREATE (p:Person { index: index, name: (names[index%10] + toString(index)) });
// create 'Results' node with relationships to search result 'Person' nodes.
// 'SEARCH_RESULT' relationships have 'order' and 'orderBy' properties.
CREATE(x:Results{ts: TIMESTAMP()})
WITH x
MATCH(p:Person)
WHERE p.name contains '1'
MERGE(x)-[r:SEARCH_RESULT]->(p)
WITH x, r, p
MATCH (x)-[r]->(p)
WITH x, r, p
ORDER BY p.name desc
WITH RANGE(0, COUNT(r)-1) AS indexes, COLLECT(r) AS rels
UNWIND indexes AS i
SET (rels[i]).order = i
SET (rels[i]).orderBy = 'name'
RETURN rels;
// validate
MATCH(x:Results)-[r:SEARCH_RESULT]->(p:Person)
RETURN r, p.name ORDER BY r.order;
I have the following Cypher query:
MATCH (genre:Genre)<-[:BELONGS_TO]-(t:Title)
WHERE genre.name IN ["Comedy", "Drama"]
RETURN t
Which returns titles that belong to Comedy OR Drama genres.
How to change this query in order to return all titles that belong to Comedy AND Drama genres?
SIZE is your friend.
MATCH (t:Title)-[:BELONGS_TO]->(g:Gender)
WHERE g.name IN ["Comedy", "Drama"]
WITH t, COLLECT(g) AS g
WHERE SIZE(g) >= x
RETURN t
x - is number of elements in IN clause
I have a following Neo4j Cypher query that checks if relationship exists between User and entity and returns boolean result:
MATCH (u:User) WHERE u.id = {userId} MATCH (entity) WHERE id(entity) = {entityGraphId} RETURN EXISTS( (u)<-[:OWNED_BY]-(entity) )
Please help to rewrite this query in order to be able to accept a collection of {entityGraphIds} instead of a single {entityGraphId} and check if a relationship exists between User and any entities with these {entityGraphIds}.
For example, I have user1 and entity1, entity2. user1 has a relationship with entity2. I'll pass {user.id} like {userId} and {entity1.id, entity2.id} like {entityGraphIds} and this query should return true.
I believe you can simply use the IN operator. Considering these parameters:
:params {userId: 1, entityGraphIds : [2,3,4]}
Then, the query:
MATCH (u:User) WHERE u.id = {userId}
MATCH (entity) WHERE id(entity) IN ({entityGraphIds})
RETURN EXISTS( (u)<-[:OWNED_BY]-(entity) )
EDIT:
If you are trying to return true when :User is connected to at least 1 entity, then you can simplify your query to:
OPTIONAL MATCH (u:User)<-[:OWNED_BY]-(entity:Entity)
WHERE u.id = {userId} AND id(entity) IN ({entityGraphIds})
RETURN u IS NOT NULL
I am having 2 nodes lets say of 2 type 'Student' and 'Class'
Student have {id, name}.
Class have {id, name}.
Student can have optional relationship with Class node as 'ATTENDS'.
(s:Student)-[r:ATTENDS]->(c:Class).
[r:ATTENDS] - Optional relationship. (present or may not present)
I want student record as it's all properties. If present relationship then class_name will match with present "Class" node else class_name will be null.
{student_id,student_name,class_name}
I tried by cypher query, but not getting result. Please help.
OPTIONAL MATCH (s:Student)-[:ATTENDS]->(c:Class) WHERE s.id =1
RETURN s.id AS student_id , s.name as student_name, c.name as class_name
By this query, if relationship exists then all values, if no relationship exists then all values are null.
If you don't care about the type of relation, you could run
MATCH (student:Student {id :1})
OPTIONAL MATCH (s)-->(class:Class)
RETURN student.id, student.name, class.name
and you'll have no need to set aliases.
Got solution to this problem by trying different queries.
MATCH (s:Student {id :1})
OPTIONAL MATCH (s)-[:ATTENDS]->(c:Class)
RETURN s.id AS student_id , s.name as student_name, c.name as class_name
Need to first match required criteria and then optional match. If anyone have simpler solution then please post.
Wrote a graph-gist for this at http://gist.neo4j.org/?11110772
The short answer is:
MATCH (s:Student) OPTIONAL MATCH (s)-->(c:Course)
RETURN s.name, c.name
Read the gist for more details. http://gist.neo4j.org/?11110772
Note that you cannot ignore the first MATCH. If the entire query is optional, nothing will be retrieved. In SQL you also have a non optional query on one table and then a left join to the second, optional, table.
How can I do an match in clause in cypher
e.g. I'd like to find movies with ids 1, 2, or 3.
match (m:movie {movie_id:("1","2","3")}) return m
if you were going against an auto index the syntax was
START n=node:node_auto_index('movie_id:("123", "456", "789")')
how is this different against a match clause
The idea is that you can do:
MATCH (m:movie)
WHERE m.movie_id in ["1", "2", "3"]
However, this will not use the index as of 2.0.1. This is a missing feature in the new label indexes that I hope will be resolved soon. https://github.com/neo4j/neo4j/issues/861
I've found a (somewhat ugly) temporary workaround for this.
The following query doesn't make use of an index on Person(name):
match (p:Person)... where p.name in ['JOHN', 'BOB'] return ...;
So one option is to repeat the entire query n times:
match (p:Person)... where p.name = 'JOHN' return ...
union
match (p:Person)... where p.name = 'BOB' return ...
If this is undesirable then another option is to repeat just a small query for the id n times:
match (p:Person) where p.name ='JOHN' return id(p)
union
match (p:Person) where p.name ='BOB' return id(p);
and then perform a second query using the results of the first:
match (p:Person)... where id(p) in [8,16,75,7] return ...;
Is there a way to combine these into a single query? Can a union be nested inside another query?