用Python判断手机号码的运营商

现在的APP一般在注册应用的时候，都会让用户输入手机号码，在短信验证之前首先会验证号码的真实性，如果是不存在的号码，就不会发送验证码。检验规则如下：

1. 长度不小于11位
2. 是移动，联通，电信号段中的任意一个
3. 不考虑输入非数字的情况（简化程序）
4. 移动号段，联通号段，电信号段如下：

CN_mobile = \
[134,135,136,137,138,139,150,151,152,157,158,159,182,183,184,
187,188,147,178,1705]
CN_union = [130,131,132,155,156,185,186,145,176,1709]
CN_telecom = [133,153,180,181,189,177,1700]

程序效果如下：

Enter your number :123
Invalid length, your number should be in 11 digits
Enter your number :12312345678
No such an operator
Enter your number :18812345678
Operator : China Mobile
We're sending verification code via text to your phone: 18812345678

---

分析：

输入的手机号码默认是string字符串，而题目中给出的是list列表，所以需要有一步转换，显然是转换成字符串操作比较容易。

发现号码段中同时出现了3位和4位的数字，所以要用两个字符串去匹配，匹配的代码为

key in statement

代码：

# 校验手机号码的真实性
def judge_operator(phone_num):
    id3 = phone_num[0:3]
    id4 = phone_num[0:4]
    CN_mobile = \
    [134,135,136,137,138,139,150,151,152,157,158,159,182,183,184,
    187,188,147,178,1705]
    CN_union = [130,131,132,155,156,185,186,145,176,1709]
    CN_telecom = [133,153,180,181,189,177,1700]
    if ((id3 in str(CN_mobile)) or (id4 in str(CN_mobile))):
        return "China Mobile"
    elif ((id3 in str(CN_union)) or (id4 in str(CN_union))):
        return "China Union"
    elif ((id3 in str(CN_telecom)) or (id4 in str(CN_telecom))):
        return "China Telecom"
    else:
        return 0

def verify_phone_number():
    phone_number = input("Enter your number :")
    invalid = len(phone_number)<11
    # print(invalid)
    if (invalid):
        # print(invalid)
        print("Invalid length, your number should be in 11 digits")
        verify_phone_number()    # 函数递归调用
    else:
        operator = judge_operator(phone_number)    # 返回的是字符串
        if (operator):
            print("Operator : {}".format(operator))
            print("We're sending verification code via text to your phone: {}".format(phone_number))
        else:
            print("No such an operator")
            verify_phone_number()    # 函数递归调用

verify_phone_number()