AQL：如何取回所有边必须具有属性（数组）并且该数组必须具有至少一个特定值的路径

Question

Suppose I have an graph like this:

A---edge1---B property P [P1,P2]

B---edge2---C property P [P2,P3]

C---edge3---D property P [P2,P3]

B---edge4---D property P [P1,P3]

And each edge have an property P which is an array of string [P1,P2,P3]. Each edge has their own value of P Now I would like to return all vertexes that:

1. For Each vertex, check if exist a path where each edges, all of them must have at least P2 in property
2. Depth smaller or equals then 2

**

For entity in entities
For v,e,p in Collection in 0..2 ANY entity 
Filter //What should I do here? I tried p.edges[*].P[*] ANY =="P2" 
Collect v._key into groups 
return {key:v._key,group:groups} //Get the vertex that satisfy the condition

**

Answer 1

p.edges[*].P[*] is an array of arrays. To get ANY == to compare array to individual element would require a flattened array, but then it cannot be determined whether each array has "P2" .

One of the following conditions should work however:

• FILTER p.edges[* RETURN "P2" IN CURRENT.P] ALL == true

• FILTER (FOR arr IN p.edges[*].P RETURN "P2" IN arr) ALL == true

• FILTER LENGTH(p.edges[* FILTER "P2" IN CURRENT.P]) == LENGTH(p.edges)