如何选择仅与给定B节点或仅与其子集相关的节点

Question

我的图表看起来像这样

A1 -> B1 & B2 & B3
A2 -> B2 & B3
A3 -> B3
A4 -> B4

如果在cypher中，我可以要求所有与 关系到给定B节点或仅其子集的A节点？

实施例

A nodes for B1 & B2 & B3 & B4 would be [A1, A2, A3, A4]
A nodes for B1 & B2 & B3 would be [A1, A2, A3]
A nodes for B2 & B3 would be [A2, A3]
A nodes for B3 would be [A3] 
A nodes for B4 would be [A4]
A nodes for B1 & B2 would be []
A nodes for B2 would be []

Answer 1

如果function f3() { return Math.random() > 0.5 ? Promise.resolve(true): 'naaah' } f3 = f3.toString().replace(/Math\.random\(\)\s[>]\s0\.5/, function(match) { return new Function(`return ${match}`)() }); // here we can visually determine whether `true` or `false` is returned; // though as vision is based on perspective, we make sure of the type // using `RegExp` at defining `type` console.log(f3); // here we determine `true` or `false` using `RegExp` let type = /return\strue/.test(f3) ? "Promise" : "String"; console.log(`f3 return type: ${type}`); f3 = new Function(`return ${f3}`); console.log(f3()());个节点的B属性值为“B1”，“B2”等，则此查询应该有效（假设包含列表的name参数传递B节点名称字符串）：

$list

Answer 2

添加我的答案，该答案建立在cybersam的答案之上，但优化了索引查找（您需要:B(name)上的索引才能利用此功能）

MATCH (a:A)-->(b:B)
WHERE b.name in $list
WITH a, COLLECT(b.name) AS bNames
WHERE ALL(n IN bNames WHERE n IN $list)
RETURN a;

这里的优点是第二行中的WHERE子句将执行索引查找：B节点并扩展到连接：A节点。换句话说，您可以立即考虑更相关的节点子集，而不是从所有连接开始：A和：B节点并过滤到WHERE ALL()中的相关：B节点，这些节点不会使用索引。

尝试对两个查询进行分析（在添加索引之后），您应该会看到db hits的减少。