在 Cypher Neo4j 中找到正确的路径

Question

I'm working with Flight Analyzer database ( https://neo4j.com/graphgist/flight-analyzer ). We have there few nodes and relationships types. Nodes: Airport

(SEA:Airport { name:'SEA' })

Flight

(f0:Flight { date:'11/30/2015 04:24:12', duration:218, distance:1721, airline:'19977' })

Ticket

(t1f0:Ticket { class:'economy', price:1344.75 })

Relationships Destination

(f0)-[:DESTINATION]->(ORD)

Origin

(f0)-[:ORIGIN]->(SEA)

Assign

(t1f0)-[:ASSIGN]->(f0)

Now I need to find some path and I have problem with that connection ORIGIN - FLIGHT - DESTINATION. I need to find all airports that are connected to LAX airport with sum of ticket prices < 3000.

I tried

MATCH path = (origin:Airport { name:"LAX" })<-[r:ORIGIN|DESTINATION*..5]->(destination:Airport)
WHERE REDUCE(s = 0, n IN [x IN NODES(path) WHERE 'Flight' IN LABELS(x)] |
s + [(n)<-[:ASSIGN]-(ticket) | ticket.price][0]
) < 3000
RETURN path

but in this solution LAX can be ORIGIN and DESTINATION too. I only want to chose paths that always have the same order aiport1 <- origin - flight1 - destination -> airport2 <- origin - flight2 - destination -> aiport etc..

I need to include departure and arrive time so flight1 date + duration < flight2 date then flight2 date + duration < flight3 date etc...

Answer 1

[UPDATED]

This query should check that:

• matched paths have alternating ORIGIN/DESTINATION relationships, and
• every departing flight lands at least 30 minutes before the next departing flight (if any), and
• the sum of the ticket prices of the Flight nodes (which are every other node starting at the second one) < 3000

MATCH p = (origin:Airport {name: 'LAX'})-[:ORIGIN|DESTINATION*..5]-(destination:Airport)
WHERE
  ALL(i IN RANGE(0, LENGTH(p)-1) WHERE
    TYPE(RELATIONSHIPS(p)[i]) = ['ORIGIN', 'DESTINATION'][i] AND
    (i%4 <> 1 OR (i + 2) > LENGTH(p) OR
      (apoc.date.parse(NODES(p)[i].date,'m','MM/dd/yyyy hh:mm:ss') + NODES(p)[i].duration + 30) < apoc.date.parse(NODES(p)[i+2].date,'m','MM/dd/yyyy hh:mm:ss'))
  ) AND
  REDUCE(s = 0, n IN [k IN RANGE(1, LENGTH(p), 2) | NODES(p)[k]] |
    s + [(n)<-[:ASSIGN]-(ticket) | ticket.price][0]
  ) < 3000
RETURN p

The query uses the apoc.date.parse function to convert each date into the number of epoch minutes, so that a duration (assumed to also be in minutes) can be added to it.

Answer 2

I believe, you should create new relationships like flyto from an airport to an airport with ticket price and ticket class. it can be useful. then you can find flights easier.

 match 
(a:Airport )<-[:ORIGIN]-(f:Flight)-[:DESTINATION ]->(b:Airport ),
(f)-[:ASSIGN]-(t:Ticket)
CREATE (a)-[r:FLY_TO {price:t.price,Class:t.class} ]->(b)