Reputation: 537
I have a neo4j database set up with the following nodes;
(:game)-[:teamscore]-(:team)
The 'game' node would contain an id
, a date
and a name
The 'team' node link would contain an id
, a teamname
and an score
.
I am looking to match nodes where teamname
= "TeamA" but only where there isn't a linked node that exists with a different team for that 'game'.
I want to return all nodes relating to a game where only 'TeamA' has scored (and not conceded).
MATCH(g:game)-[:teamscore]-(t:team)
WHERE g.team = "TeamA"
RETURN g, t
I thought about adding some kind of NOT (g)-[:teamscore]-(:team)
to the WHERE clause but I'm not sure how I could do this in a way which doesn't eliminate the node which I do want to match.
Thanks in advance for any help on this!
Upvotes: 0
Views: 451
Reputation: 20185
So, if the assumption is correct, there can be more than one (teamA)-[:teamscore]->(game) for the same game.
If yes, then you can make sure that all connections from game to a team node are to teamA.
Given the following graph :
CREATE (g:Game {id: 1})-[:teamscore]->(:Team {id: 1, name: "TeamA"})
CREATE (g)-[:teamscore]->(:Team {id:2, name: "TeamA"})
CREATE (g)-[:teamscore]->(:Team {id:3, name: "TeamA"})
CREATE (g2:Game {id: 2})-[:teamscore]->(:Team {name: "TeamA"})
CREATE (g2)-[:teamscore]->(:Team {name: "TeamB"})
You can use pattern comprehensions and the ALL
predicate :
MATCH (n:Game)
WHERE ALL(x IN [(n)-[:teamscore]->(t) | t.name] WHERE x = "TeamA")
RETURN n
It will return Game with id 1
Upvotes: 0
Reputation: 4052
You can a match team by name and then make sure only one team is linked to that game by specifying the number of teamscore
relationships to that game is 1.
MATCH(g:game)-[:teamscore]-(t:team)
WHERE t.teamname = "TeamA" AND size((g)-[:teamscore]-(:team))=1
RETURN g, t
Upvotes: 1