诡异的python类的实例方法，返回值为什么是None？？

[popwbob]

今天遇到的python类实例方法返回值为None的问题，同样写了个简单类来验证，可是问题不能找到原因。

#参照类，工作正常，返回值如预期
class test:
    def __init__(self, a, b):
        self.a = a
        self.b = b
        self.s = None
    def ser(self, c):
        self.s = [self.a ,self.b , c]
        try:
            123 + '123'
        except:
            return self.s

#下面类的实例方法为什么返回值为None？？？？？？？？？
#测试用数据
#(城市ID，上级城市ID，城市序号，城市名称
departments = [(2, 1, 1, '北京'),(3, 1, 2, '上海'),(4, 1, 3, '广州'),(5, 1, 4, '湖南'),(6, 5, 1, '长沙'),(7, 1, 5, '河北')]
#(用户ID,用户所在城市ID，用户序号，用户姓名)
users = [(1, 2, 1, 'qq'),(2, 3, 2, 'aa'),(3, 4, 3, 'zz'),(4, 5, 4, 'ww'),(5, 6, 5, 'ss'),(6, 7, 6, 'xx')]
#通信录的根节点
root=(1,None,1,'中国')

class StructureTree:
    def __init__(self, deps, users):
        self.deps = deps
        self.users = users
        self.node_order = []
        self.stack = []
    def select_in_tuple(self, list, value,index):
        ck = []
        for node in list:
            if node[index] == value:
                ck.append(node)
        ck.sort(key=lambda node: node[2])
        return ck
    def pile_tree(self, node):
        users=self.select_in_tuple(self.users, node[0],1)
        self.node_order.extend(users)
        children = self.select_in_tuple(self.deps, node[0],1)
        if len(children) == 0:
            try:
                next_node=self.stack.pop()
                self.node_order.append(next_node)
                self.pile_tree(next_node)
            except:
                print('return self.node_order->',self.node_order)   #返回前打印值为正常，返回后却成了None
                return self.node_order
        elif len(children)==1:
            self.node_order.append(children[0])
            self.pile_tree(children[0])
        else:
            self.stack.extend(children[:0:-1])
            self.node_order.append(children[0])
            self.pile_tree(children[0])

a = test(1, 2)
print(a.ser(3))

st=StructureTree(departments,users)
print(st.pile_tree(root))

输出：：

>>>[1, 2, 3]
>>>return self.node_order-> [(2, 1, 1, '北京'), (1, 2, 1, 'qq'), (3, 1, 2, '上海'), (2, 3, 2, 'aa'), (4, 1, 3, '广州'), (3, 4, 3, 'zz'), (5, 1, 4, '湖南'), (4, 5, 4, 'ww'), (6, 5, 1, '长沙'), (5, 6, 5, 'ss'), (7, 1, 5, '河北'), (6, 7, 6, 'xx')]
>>>None

[混沌鳄鱼]

这个多出来的None是你的外层print(st.pile_tree(root)) 打印出来的。 这个是你定义方法 def pile_tree(self, node): users=self.select_in_tuple(self.users, node[0],1) self.node_order.extend(users) children = self.select_in_tuple(self.deps, node[0],1) if len(children) == 0: try: next_node=self.stack.pop() self.node_order.append(next_node) self.pile_tree(next_node) except: print('return self.node_order->',self.node_order) #返回前打印值为正常，返回后却成了None return self.node_order elif len(children)==1: self.node_order.append(children[0]) self.pile_tree(children[0]) else: self.stack.extend(children[:0:-1]) self.node_order.append(children[0]) self.pile_tree(children[0]) 你这个方法（函数）没有写显示的return语句， 在python中会隐式地返回一个None return_value = st.pile_tree(root) print(return_value ) 这样肯定会多打印出一个None的 你调用的实例方法的时候既然已经在方法里面打印输出了，就直接调用吧。 或者在在方法中return要打印的值，在外层打印。