Python 3.x py2neo：从关系和所有传入的不带密码的关系中获取结束节点

我很难找到与节点有关系的节点。 我希望能够从选定节点中找到与选定节点具有关系的节点

下面是示例图的代码：

from py2neo import Node, Relationship, Graph, NodeSelector, Path
from py2neo.ogm import *
graph = Graph(user = 'neo4j', password = 'neo4j')
graph.delete_all()

class Man(GraphObject):
    __primarykey__ = "name"
    name = Property("name")
    likes = RelatedTo("Woman", "LIKES")

class Woman(GraphObject):
    __primarykey__ = "name"
    name = Property("name")
    likes = RelatedTo("Man", "LIKES")

new_man = Man()
new_man.name = "John"
graph.push(new_man)

new_woman = Woman()
new_woman.name = "Sarah"
new_woman.likes.add(Man.select(graph, primary_value="John").first())
graph.push(new_woman)

new_man = Man()
new_man.name = "Joe"
new_man.likes.add(Woman.select(graph, primary_value="Sarah").first())
graph.push(new_man)

我试图说出莎拉喜欢的人的名字：

sarah = Woman.select(graph, primary_value="Sarah").first()
sarah.likes._related_objects[0][0].name
# returns "John"
# or
list(sarah.__ogm__.related.values())[0]._related_objects[0][0].name
# returns "John"

在没有查看其他节点的情况下，我无法找到任何方法来获得喜欢Sarah的人的名字。这是可能的还是我在浪费时间？有没有更好的方法来做这些？ 我是否被困在：

def get_who_likes_sarah():
    names = []
    for m in Man.select(graph):
        try:
            name = m.likes._related_objects[0][0].name
            if name == "Sarah":
                names.append(m.name)
        except:
            pass
    return names

---

你应该这样做：

for rel in graph.match(start_node=sarah, rel_type="LIKES"):
    names.append(rel.end_node()["name"])

---

你有办法解决这个问题吗？放弃了我所拥有的。我的代码最后是这样的

[m.name代表m中的Man.select（graph）代表w中的m.likes.\u相关的\u对象，如果w[0]。name==“Sarah”]