Query recipes

Answer common relationship questions with the customer and order graph from the mapped tutorial.

Dataset

These recipes use the mapped graph tutorial: three people, four orders, and three FOLLOWS relationships. Use lowercase names to match the sample rows.

Find a person’s friends

MATCH (p:Person)-[:FOLLOWS]->(friend:Person)
WHERE p.name = 'alice'
RETURN friend.name AS name
ORDER BY name

The names are bob and carol. The equivalent Gremlin traversal is:

g.V().hasLabel('Person').has('name','alice')
  .out('FOLLOWS').values('name').order()

Find customers with large orders

MATCH (p:Person)-[:ORDERED]->(o:Order)
WHERE o.total > 100.0
RETURN p.name AS name, o.total AS total
ORDER BY total

The results are alice with 120 and carol with 500.

Sum spending by customer

MATCH (p:Person)-[r:ORDERED]->(:Order)
RETURN p.name AS name, sum(r.total) AS spent
ORDER BY spent DESC

For one customer, use a traversal reduction:

g.V().hasLabel('Person').has('name','alice')
  .out('ORDERED').values('total').sum()

Find friends within two hops

MATCH (p:Person)-[:FOLLOWS*1..2]->(friend:Person)
WHERE p.name = 'alice'
RETURN DISTINCT friend.name AS name
ORDER BY name

Alice reaches bob directly and carol both directly and through bob. DISTINCT produces one row per name.

Count the people

g.V().hasLabel('Person').count()

Or use Cypher:

MATCH (p:Person)
RETURN count(p) AS people

Both return three for the sample dataset.

Filter through an RDF vocabulary

With the SPARQL ontology:

PREFIX ex: <https://example.com/>
SELECT ?name ?age WHERE {
  ?p a ex:Person ; ex:name ?name ; ex:age ?age .
  FILTER(?age > 29)
}
ORDER BY ?name

The names are alice and carol, with ages 30 and 41.

Apply a property update

Run through MappedGraphEngine::cypher_update:

MATCH (p:Person)
WHERE p.name = 'alice'
SET p.age = 31

Query p.age afterward to read the updated value. Use typed parameters when the name or new age comes from user input.