neo4j - Cypher 选择其任何邻居不包含属性的顶点

Question

{
    "identity": 7,
    "labels": [
        "Parent"
    ],
    "properties": {
        "name": "foo1"
    }
}, 
{
    "identity": 8,
    "labels": [
        "Child"
    ],
    "properties": {
        "name": "bar2"
    }
}, 
{
    "identity": 9,
    "labels": [
        "Child"
    ],
    "properties": {
        "name": "bar1"
    }
}, 
{
    "identity": 10,
    "labels": [
        "Parent"
    ],
    "properties": {
        "name": "foo2"
    }
}

我想选择 Parents 没有任何 child 与 name='abc'

预期行为:我应该得到两个 parent (foo1 和 foo2)作为结果

查询 1:

Match (x1:Parent) with x1 
optional Match (x1)-[:CHILD]-(x2:Child) with x1 , collect(x2) as x3 
UNWIND x3 as x2 
WITH x1 , x2 , x3 
where none (x IN  x3 where x.name IN ['abc']) 
return DISTINCT x1 

此查询仅返回 1 个父级(foo1)，但它应该返回两个父级，而第二个父级(foo2)未连接到任何子级。

PS :使用 UNWIND 的原因是将变量用于变量 x2 上的进一步 WHERE 子句。

Answer 1

这将得到所有没有 child 名字的 parent :'abc' 或没有任何 child

MATCH (p:Parent)
WHERE NOT EXISTS((p)-[:CHILD]->(:Child {name: 'abc'}))
RETURN p 

Check if the given parent, p has a child not named: 'abc'

==========编辑:

我个人不喜欢这个查询，因为它先收集 x2，然后再展开。但是为了回答 SO 问题，

Match (x1:Parent) with x1 
optional Match (x1)-[:CHILD]->(x2:Child) with x1 , collect({x1:x1, x2:x2}) as x3 
UNWIND x3 as x2 
WITH x2.x1 as x1, x2.x2 as x2, x3 
where NOT x2.name IN ['abc'] OR x2 is NULL 
RETURN distinct x1

I collected BOTH x1 and x2 then UNWIND the collection by getting x1and x2 during unwind. This will include other nodes that has no x2(without a CHILD). I also simplified the where condition because x2 is also accessible rather than looping again on x3