问题
Say I have 3 things in my graph database and each of these 3 things have 0 or more subthings attached to them. How do I make a query to retrieve the top 2 subthings for each of the things in the graph (according to some arbitrary ordering).
Here's an example setup:
CREATE (t1:Thing)-[:X]->(a1:SubThing),
(t1)-[:X]->(a2:SubThing),
(t1)-[:X]->(a3:SubThing),
(t1)-[:X]->(a4:SubThing),
(t2:Thing)-[:X]->(b1:SubThing),
(t2)-[:X]->(b2:SubThing),
(t2)-[:X]->(b3:SubThing),
(t3:Thing);
What match command could I run to receive a table like this:
t s
(0:Thing) (3:SubThing)
(0:Thing) (4:SubThing)
(1:Thing) (7:SubThing)
(1:Thing) (8:SubThing)
(2:Thing) null
Here's a console.neo4j.org I was using to work it out.
http://console.neo4j.org/r/k32uyy
回答1:
This is probably the best way to do it. I can't unwind the collection or else I'll lose the empty :Thing
.
MATCH (t:Thing)
OPTIONAL MATCH (t)-->(s:SubThing)
WITH t, s
ORDER BY id(s)
RETURN t, collect(s)[0..2]
ORDER BY id(t)
And it returns this:
t collect(s)[0..2]
(0:Thing) [(3:SubThing), (4:SubThing)]
(1:Thing) [(7:SubThing), (8:SubThing)]
(2:Thing) []
回答2:
You need to use an OPTIONAL MATCH
here, aggregate by same start node and sort the result by the count of subthings:
MATCH (t:Thing)
OPTIONAL MATCH (t)-[:X]->(s)
RETURN t, count(s) AS degree
ORDER BY degree DESC
LIMIT 2
来源:https://stackoverflow.com/questions/29023703/how-to-limit-a-subquery-in-cypher