随机产生一个手机号，保证不重复

Posted 2022-12-16 wenm1128

import random
import os.path

def get_mobile_no(mobile_no_file_path):

    mobile_no =""
    prifix =[135,136,137,138,150,159,188]

    while 1:
       #随机选择一个开头
        mobile_prifix =random.choice(prifix)
       #随机生成后八位
        mobile_postfix =random.randint(10000000,99999999)
       #拼接
        mobile_no =str(mobile_prifix)+str(mobile_postfix)
       #判断文件是否存在
        try:
            with open(mobile_no_file_path,"a+") as fp:
                #判重逻辑
                if mobile_no in fp.read():
                    continue
                #写入文件  
                fp.write(mobile_no+"\n")
                break
        except  Exception as e:
            with open("no.txt","a+") as fp:
                if mobile_no in fp.read():
                    continue
                fp.write(mobile_no+"\n")
                break
    return mobile_no

print(get_mobile_no("d:\\no2.txt"))

 

 

生成八位随机数也可以用zfill

Python zfill() 方法返回指定长度的字符串，原字符串右对齐，前面填充0。

代码改动：

import random
import os.path

def get_mobile_no(mobile_no_file_path):

    mobile_no =""
    prifix =[135,136,137,138,150,159,188]

    while 1:
        mobile_prifix =random.choice(prifix)
        mobile_postfix =str(random.randint(0,99999999)).zfill(8)#从0开始，用zfill生成八位数
        mobile_no =str(mobile_prifix)+str(mobile_postfix)
        try:
            with open(mobile_no_file_path,"a+") as fp:
                if mobile_no in fp.read():
                    continue
                fp.write(mobile_no+"\n")
                break
        except  Exception as e:
            with open("no.txt","a+") as fp:
                if mobile_no in fp.read():
                    continue
                fp.write(mobile_no+"\n")
                break
    return mobile_no

print(get_mobile_no("d:\\no2.txt"))

 

 

遗留问题：用列表序列化做？