基于地图参数的 Neo4j Cypher 过滤

Question

根据上一个问题：

Neo4j Cypher query structure and performance optimization Neo4j Cypher node filtering by pattern comprehension

最后我将查询重构为以下内容：

MATCH (parentD)-[:CONTAINS]->(childD:Decision) 
    WHERE id(parentD) = {decisionId} 
        MATCH (childD)<-[:SET_FOR]-(equalFilterValue)-[:SET_ON]->(equalFilterCharacteristic) 
            WHERE ALL(key IN keys({equalFilters}) WHERE id(equalFilterCharacteristic) = toInt(key) AND equalFilterValue.value = ({equalFilters}[key]))  
    WITH DISTINCT childD  
        MATCH (childD)<-[:SET_FOR]-(rangeFilterValue)-[:SET_ON]->(rangeFilterCharacteristic) 
            WHERE ALL(key IN keys({rangeFilters}) WHERE id(rangeFilterCharacteristic) = toInt(key) AND ({rangeFilters}[key])[0] <= rangeFilterValue.value <= ({rangeFilters}[key])[1]) 
WITH * MATCH (childD)-[ru:CREATED_BY]->(u:User)  
RETURN ru, u, childD AS decision 
SKIP 0 LIMIT 100

如果每个*filter type(map) 只有一个键，则此查询可以正常工作，例如：

queries.add(new InQuery(integerCharacteristic.getId(), 30));

或

queries.add(new InQuery(stringCharacteristic.getId(), "Two"));

但当我添加 2 个或更多条件时失败，例如：

queries.add(new InQuery(integerCharacteristic.getId(), 30));
queries.add(new InQuery(stringCharacteristic.getId(), "Two"));

以下查询未按预期工作，我的测试断言失败：

MATCH (parentD)-[:CONTAINS]->(childD:Decision) 
WHERE id(parentD) = {decisionId} 
    MATCH (childD)<-[:SET_FOR]-(inFilterValue)-[:SET_ON]->(inFilterCharacteristic) 
        WHERE ALL(key IN keys({inFilters}) WHERE id(inFilterCharacteristic) = toInt(key) AND ({inFilters}[key]) IN inFilterValue.value) 
WITH * MATCH (childD)-[ru:CREATED_BY]->(u:User)  
RETURN ru, u, childD AS decision 
SKIP 0 LIMIT 100

参数：

inFilters = {3153=30, 3151=Two}

为什么当inFilters map 包含 2 个或更多键时它不起作用以及如何使它起作用？

Answer 1

因此，由于这仅适用于 1 个键，让我们回顾一下当有 2 个键时，WHERE ALL 在做什么。

WITH inFilters = {3153=30, 3151=Two}
....
WHERE ALL(key IN keys({equalFilters}) 
    WHERE id(equalFilterCharacteristic) = toInt(key) 
    AND equalFilterValue.value = ({equalFilters}[key]))

相当于

WHERE id(equalFilterCharacteristic) = toInt(3153) 
AND equalFilterValue.value = ({equalFilters}[3153]) 
AND id(equalFilterCharacteristic) = toInt(3151) 
AND equalFilterValue.value = ({equalFilters}[3151])

问题在于，现在我们正在检查每个 equalFilterCharacteristic 的节点 id 是否同时等于 3153 和 3151。由于 Neo4j 仅使用实数作为节点 id，因此当有超过 1 个键时，上述语句基本上最终会减少到昂贵的 WHERE FALSE。因此，在上述情况下，WHERE ALL 永远不会为真。但是，如果至少 1 个检查组为真，则 WHERE ANY 将评估为真，并且等效于

WHERE ( 
    id(equalFilterCharacteristic) = toInt(3153) 
    AND equalFilterValue.value = ({equalFilters}[3153]) 
)
OR 
(
    id(equalFilterCharacteristic) = toInt(3151) 
    AND equalFilterValue.value = ({equalFilters}[3151]) 
)

当然，既然您知道要匹配的密钥，您可以跳过 ALL 并直接执行（不要认为您需要对 id() 进行 toInt()，但 id() 很长）

WHERE equalFilterValue.value = ({equalFilters}[id(equalFilterCharacteristic)])