Question

我有两个问题：

我在一个方法中运行此查询

doc.getElementByClassName()

另一种方法

String query = "MATCH (g:Grid {name:'"+gridLocation+"'})<-[r:WILL_GO]-(t:Taxi)"
                    + "WHERE r.reachedTime <= '"+userPickUp+"' RETURN t.name AS Taxi";

Result taxiWillGo = graphDb.execute(query);

两个查询都返回相同的元素（出租车），是否可以将两个执行的结果合并为一个，所以最后我有一个“表”显示所有结果。

String query2 = "MATCH p=((g:Grid {name:'"+gridLocation+"'})-[r:TO*1..2]-(g2:Grid)), (g2)-[r2:LOCATION]-(t:Taxi) "
                    + "WITH  t, p, REDUCE(totalTime = 0, x IN RELATIONSHIPS(p) | totalTime + x.time) AS totalTime "
                    + "WHERE totalTime <= 6 RETURN t.name as Taxi LIMIT 3";

Result taxiNeighbor = graphDb.execute(query2);

提前谢谢！

Answer 1

您可以使用UNION Cypher子句合并两个查询的结果：http://neo4j.com/docs/stable/query-union.html

MATCH ...
RETURN t

UNION

MATCH ...
RETURN t

cypher merge执行结果java

1 个答案: