python中的迭代器，应打印10个有效id号

Question

下面是我目前为止的代码，代码应该实现一个迭代器，在check_id函数的条件下打印10个有效ID。

https://www.codepile.net/pile/b1PVdbBq

输出应该是：123456782 123456790 123456808 123456816 123456824 123456832 123456840 123456857 123456865 123456873但不知何故我的输出是错误的...帮助...

import string
letters = string.ascii_letters
digits = string.digits


class NumNotNineLong(Exception):

    def __init__(self):
        pass
    def __str__(self):
        return "The number you provided is not nine digits long."

class NotNumber(Exception):

    def __init__(self):
        pass

    def __str__(self):
        return "The input you provided is not an integer"

class IDIterator():

    increment = 1

    def __init__(self,_id):
        self._id = _id  

    def __iter__(self):
       return self 

    def __next__(self):

        while check_id_valid(str(self._id)[-9::]) == False:
            self._id *= 2
            self._id += IDIterator.increment
            IDIterator.increment += 1
        if check_id_valid(str(self._id)[-9::]):
            result = str(self._id)[-9::]
            self._id *= 2
            self._id += 1
            IDIterator.increment = 2
            return result



def check_id_valid(id_number): #A function that checks whether ID number are valid or not
    for letter in str(id_number):
        if letter not in string.digits:
            raise NotNumber
    numbers = [int(i) for i in str(id_number)]
    if len(numbers) != 9:
        raise NumNotNineLong
    set_numbers = []
    for i in range(len(numbers)):
        if i % 2 == 0:
            set_numbers.append(numbers[i])
        else:
            set_numbers.append(numbers[i] * 2)
    true_numbers = []
    for num in set_numbers:
        if num > 9:
            temp = [int(i) for i in str(num)]
            true_numbers.append(sum(temp))
        else:
            true_numbers.append(num)
    if sum(true_numbers) % 10 == 0:
        return True
    else:
        return False

def main():
    result = IDIterator(123456780)
    for _ in range(10):
        print(result.__next__())


if __name__ == "__main__":
    main()

练习如下所示：一个名为check\u id\u valid的布尔函数，定义为：def check\u id\u valid（id\u number）：该函数作为id number（integer）参数获取，并返回真布尔值（true），否则返回假值（false）。当check\u id函数是自包含的时，它工作得很好

然后应该实现一个类来表示名为IDIterator的迭代器。

此类有一个名为id_的属性，表示0到999999999范围内的ID号（此属性在创建迭代器实例时初始化）。

在上述类中必须使用返回迭代器实例的iter方法。

在上述类中，必须使用下一个方法，该方法返回ID\ uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。

要检查您编写的代码是否正常工作，请使用您创建的类中的迭代器编写一个主程序。

具有下一个ID值的迭代器必须初始化为123456780，生成10个新ID并将其打印到屏幕上。以下是您应该收到的输出：

Answer 1

我不明白你想在这里做什么：

def __next__(self):

    while check_id_valid(str(self._id)[-9::]) == False:
        self._id *= 2
        self._id += IDIterator.increment
        IDIterator.increment += 1
    if check_id_valid(str(self._id)[-9::]):
        result = str(self._id)[-9::]
        self._id *= 2
        self._id += 1
        IDIterator.increment = 2
        return result

下面是您尝试使用的代码（但我不建议使用它，因为这不是他们要求的！）：

class IDIterator:

    def __init__(self, _id):
        self._id = _id
        self.ok = ""

    def __iter__(self):
        return self

    def __next__(self):
        x = self.ok
        while not check_id_valid(str(self._id)[-9:]):
            self._id += 1
        else:
            self.ok = self._id
            self._id += 1
            return x

我建议您使用以下代码：

class IDIterator:

    def __init__(self, _id=0, range=999999999):
        self._id = _id
        self.range = range
        self.count = 0

    def __iter__(self):
        return self

    def __next__(self):
        while self.count < self.range:
            x = self._id
            if not check_id_valid(x):
                self._id += 1
            else:
                self._id += 1
                self.count += 1
                return x
        else:
            raise StopIteration