Question

我得到了这个，我打算在打印x.withdraw（）时打印410。

Kyle 12345 500
Traceback (most recent call last):
    File "bank.py", line 21, in <module>
        print x.withdraw()
TypeError: 'int' object is not callable

这是我的代码：

class Bank:
    def __init__(self, name, id, balance, withdraw):
        self.name = name
        self.id = id
        self.balance = balance
        self.withdraw = withdraw
    def print_info(self):
        return "%s %d %d" % (self.name, self.id, self.balance)
    def withdraw(self):
        if self.withdraw > self.balance:
            return "ERROR: Not enough funds for this transfer"
        elif self.withdraw < self.balance and self.withdraw >= 0:
            self.balance = self.balace - self.withdraw
            return self.balance
        else:
            return "Not a legitimate amount of funds"

x = Bank("Kyle", 12345, 500, 90)
print x.print_info()
print x.withdraw()

我是否需要在类中修复某些内容，或者我的方法调用有问题？

Answer 1

您在实例上设置了一个具有相同名称的属性：

self.withdraw = withdraw

这是您现在尝试调用的属性，而不是方法。 Python没有区分方法和属性，它们不在单独的命名空间中。

为属性使用不同的名称; withdrawn（过去时的退缩）会让人想起更好的属性名称：

class Bank:
    def __init__(self, name, id, balance, withdrawn):
        self.name = name
        self.id = id
        self.balance = balance
        self.withdrawn = withdrawn
    def print_info(self):
        return "%s %d %d" % (self.name, self.id, self.balance)
    def withdraw(self):
        if self.withdrawn > self.balance:
            return "ERROR: Not enough funds for this transfer"
        elif self.withdrawn < self.balance and self.withdrawn >= 0:
            self.balance = self.balance - self.withdrawn
            return self.balance
        else:
            return "Not a legitimate amount of funds"

（我还更正了一个拼写错误;您在一个您想要使用balace}的位置使用了balance。

'int'对象在python中不可调用

1 个答案: