Question

尝试使用py2neo V3以事务方式添加节点和关系。我想将此人和他们所有的电影添加为一个交易。

我无法让外部和内部循环中的节点在同一个事务中工作。我很自信这种关系不是以交易方式添加的，因为我正在调用tx1.graph。

Neo4j V3.0.7 Py2Neo v3.1.2

from py2neo import Graph,Node,Relationship,authenticate, watch
from py2neo.ogm import GraphObject, Property, RelatedTo, RelatedFrom

class Movie(GraphObject):
    __primarykey__ = "title"

    title = Property()

class Person(GraphObject):
    __primarykey__ = "name"

    name = Property()

    acted_in = RelatedTo(Movie)

People = ["John","Jane","Tarzan"]
Movies = [["John","Movie1"],["John","Move2"],["Jane","Movie3"],["Jane","Movie4"],["Tarzan","Movie4"]]
graph = Graph("http://localhost:7474")

for p in People:
    print(p)
    tx = graph.begin()
    p1 = Person()
    p1.name = p
    tx.merge(p1)
    tx.commit()
    for m in Movies:
        if m[0] != p:
            continue
        print(m[1])
        tx1 = graph.begin() #did not work using original tx transaction
        m1 = Movie()
        m1.title = m[1]
        tx1.merge(m1)       
        p1.acted_in.add(m1)
        #tx1.merge(p1) #did not create relationship
        #tx1.create(p1) #did not create relationship
        tx1.graph.push(p1) # worked in adding relationship, but can't guarantee was part of the transaction
        tx1.commit()

Answer 1

尝试以下循环，它为每个Person及其关系使用单个事务：

for p in People:
    print(p)
    tx = graph.begin()
    p1 = Person()
    p1.name = p
    tx.merge(p1)
    for m in Movies:
        if m[0] != p:
            continue
        print(m[1])
        m1 = Movie()
        m1.title = m[1]
        tx.merge(m1)       
        p1.acted_in.add(m1)
        tx.graph.push(p1)
    tx.commit()

注意：使用单个事务处理多个人实际上会更有效。但是你不希望同时处理太多（取决于你的数据模型），因为这会使服务器内存不足。

Py2neo V3在循环中向事务添加关系

1 个答案: