python neo4j 获取node节点本身的id

 

 

def select(con_id):
    cql = "match (c:Consult) where c.con_id='%s' return c " % con_id
    for r in graph.run(cql).data():
        node_ = r['c']  # 返回的是node
        node_id = node_.identity  # 获取node本身的id
        print(type(node_))  # 
        print(node_id)  # 117600924
        print(r)  # {'c': (_117600924:Consult {con_id: '1', con_name: '\u80fd\u5426\u79bb\u5a5a'})}