让我们说我们有一个对象抽奖： class Draw {
private int num1, num2, num3;
public Draw (int num1, int num2, int num3) {
this.num1 = num1;
this.num2 = num2;
this.num3 = num3;
}
public int getNum1(
考虑，看起来像这样的图表： (node) -> (related nodes)
a1 -> [b1, b2, b3]
a2 -> [b1]
a3 -> []
我想回到扩展与邻国的列表中的所有节点。我现在查询是以下 MATCH (x0:A)
MATCH (x1:B)
WHERE ((x0)-[:B]->(x1))
WITH x0,collect(x1) as x2
WITH x0,