RGPerson - 随机身份生成脚本

RGPerson

项目地址:https://github.com/gh0stkey/RGPerson

RGPerson - 随机身份生成

环境:python3

使用方法:python3 RGPerson.py

rgperson

为什么需要Ta

相信很多师傅们在做测试的时候经常遇到一些注册的业务功能,要填写的东西很多,我一般都是临时去百度用的信息,这样很繁琐所以决定造轮子撸了个随机身份生成的。

介绍

该脚本生成信息:姓名\年龄\性别\身份证\手机号\组织机构代码\统一社会信用代码

脚本编写原理

脚本的函数: genMobile()、genIdCard()、genName()、genOrgCode()、genCreditCode()

genMobile() 为随机生成手机号的函数

genName() 为随机生成姓名的函数

genIdCard() 为随机生成身份证的函数

genOrgCode() 为随机生成组织机构代码的函数

genCreditCode() 为随机生成统一社会信用代码的函数

genMobile()

随机生成手机号:需要知道国内手机号的构成

1.长度为十一位

2.前三位表示运营商

现在我们只需要做到收集手机号号段的前三位以及对应的运营商:

prelist = {"133":"电信","149":"电信","153":"电信","173":"电信","177":"电信","180":"电信","181":"电信","189":"电信","199":"电信","130":"联通","131":"联通","132":"联通","145":"联通","155":"联通","156":"联通","166":"联通","171":"联通","175":"联通","176":"联通","185":"联通","186":"联通","166":"联通","134":"移动","135":"移动","136":"移动","137":"移动","138":"移动","139":"移动","147":"移动","150":"移动","151":"移动","152":"移动","157":"移动","158":"移动","159":"移动","172":"移动","178":"移动","182":"移动","183":"移动","184":"移动","187":"移动","188":"移动","198":"移动"}

获取该数组的长度:len(prelist) -> 42

随机生成下标获取三位数:prelist.keys()[random.randint(0,41)]

然后再随机填补后8位即可:

def genMobile():
    prelist = {"133":"电信","149":"电信","153":"电信","173":"电信","177":"电信","180":"电信","181":"电信","189":"电信","199":"电信","130":"联通","131":"联通","132":"联通","145":"联通","155":"联通","156":"联通","166":"联通","171":"联通","175":"联通","176":"联通","185":"联通","186":"联通","166":"联通","134":"移动","135":"移动","136":"移动","137":"移动","138":"移动","139":"移动","147":"移动","150":"移动","151":"移动","152":"移动","157":"移动","158":"移动","159":"移动","172":"移动","178":"移动","182":"移动","183":"移动","184":"移动","187":"移动","188":"移动","198":"移动"}
    three = list(prelist.keys())[random.randint(0,len(prelist)-1)]
    mobile = three + "".join(random.choice("0123456789") for i in range(8))
    op = prelist[three]
    return {mobile:op}

genName()

随机生成姓名:中文名字通常为2、3位汉字组成

1.收集常用的姓氏随机取其一个:

def first_name():
    first_name_list = ['赵', '钱', '孙', '李', '周', '吴', '郑', '王', '冯', '陈', '褚', '卫', '蒋', '沈', '韩', '杨', '朱', '秦', '尤', '许',
                '何', '吕', '施', '张', '孔', '曹', '严', '华', '金', '魏', '陶', '姜', '戚', '谢', '邹', '喻', '柏', '水', '窦', '章',
                '云', '苏', '潘', '葛', '奚', '范', '彭', '郎', '鲁', '韦', '昌', '马', '苗', '凤', '花', '方', '俞', '任', '袁', '柳',
                '酆', '鲍', '史', '唐', '费', '廉', '岑', '薛', '雷', '贺', '倪', '汤', '滕', '殷', '罗', '毕', '郝', '邬', '安', '常',
                '乐', '于', '时', '傅', '皮', '卞', '齐', '康', '伍', '余', '元', '卜', '顾', '孟', '平', '黄', '和', '穆', '萧', '尹',
                '姚', '邵', '堪', '汪', '祁', '毛', '禹', '狄', '米', '贝', '明', '臧', '计', '伏', '成', '戴', '谈', '宋', '茅', '庞',
                '熊', '纪', '舒', '屈', '项', '祝', '董', '梁']
    n = random.randint(0, len(first_name_list) - 1)
    f_name = first_name_list[n]
    return f_name

2.这里一开始想搜罗常用的名字,但参考了其他师傅的代码发现随机生成中文字符更好一点:

def GBK2312():
    head = random.randint(0xb0, 0xf7)
    body = random.randint(0xa1, 0xf9)
    val = f'{head:x}{body:x}'
    st = bytes.fromhex(val).decode('gb2312')
    return st

3.随机生成名字的第二个字:(这里用一个list做一个空值,随机取生成的汉字或空值,用于成为随机生成2位名字或3位名字)

def second_name():
    second_name_list = [GBK2312(), '']
    n = random.randint(0, 1)
    s_name = second_name_list[n]
    return s_name

4.随机生成名字的最后一个字:(用于满足三个汉字的名字)

def last_name():
    return GBK2312()

5.拼接

def last_name():
    return GBK2312()

genIdCard()

随机生成身份证:公民身份号码是由17位数字码和1位校验码组成

18位数字组合的方式是:

1 1 0 1 0 2 Y Y Y Y M M D D 8 8 8 X
区域码(6位) 出生日期码(8位) 顺序码(2位) 性别码(1位) 校验码(1位)

区域码 指的是公民常住户口所在县(市、镇、区)的行政区划代码,如110102是北京市-西城区。但港澳台地区居民的身份号码只精确到省级。

age = random.randint(16,60) #可调整生成的年龄范围(身份证),这边是16-60岁
y = date.today().year - age #生成的年份
m = date(y, 1, 1) #生成的月份,初始值为1月1日
d = timedelta(days=random.randint(0, 364)) #随机生成的天数
datestring = str(m + d) #加天数得到最终值

出生日期码 表示公民出生的公历年(4位)、月(2位)、日(2位)。

顺序码 表示在同一区域码所标识的区域范围内,对同年、同月、同日出生的人编定的顺序号。

性别码 奇数表示男性,偶数表示女性。

最难的还是校验码的算法,参考师傅的解说:

1.将前面的身份证号码17位数分别乘以不同的系数。从第一位到第十七位的系数分别为:7 9 10 5 8 4 2 1 6 3 7 9 10 5 8 4 2

2.将这17位数字和系数相乘的结果相加。

3.用加出来和除以11,得余数

4.余数只可能是0 1 2 3 4 5 6 7 8 9 10这11个数字,其分别对应的最后一位身份证的号码为1 0 X 9 8 7 6 5 4 3 2。

5.通过上面得知如果余数是2,就会在身份证的第18位数字上出现罗马数字的Ⅹ,如果余数是10,身份证的最后一位号码就是2。

测试代码如下,取了几个真实的身份证号码发现可用:

def test(id_num):
	id_code_list = [7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2]
	check_code_list = [1, 0, 'X', 9, 8, 7, 6, 5, 4, 3, 2]
	a = 0
	print(len(id_num))
	for i in range(17):
		a = a + (int(id_num[i]) * id_code_list[int(i)])
	print(check_code_list[a % 11])

整合一下(Copy)就变成了如下完整的代码:

def genIdCard(age,gender):
    area_code = ('%s' % random.choice(list(area_dict.keys())))
    id_code_list = [7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2]
    check_code_list = [1, 0, 'X', 9, 8, 7, 6, 5, 4, 3, 2]
    if str(area_code) not in area_dict.keys():
        return None
    datestring = str(date(date.today().year - age, 1, 1) + timedelta(days=random.randint(0, 364))).replace("-", "")
    rd = random.randint(0, 999)
    if gender == 0:
        gender_num = rd if rd % 2 == 0 else rd + 1
    else:
        gender_num = rd if rd % 2 == 1 else rd - 1
    result = str(area_code) + datestring + str(gender_num).zfill(3)
    b = result + str(check_code_list[sum([a * b for a, b in zip(id_code_list, [int(a) for a in result])]) % 11])
    return b

参考

https://www.cnblogs.com/evening/archive/2012/04/19/2457440.html

https://www.cnblogs.com/thunderLL/p/7682148.html

https://blog.csdn.net/ak739105231/article/details/83932151

https://github.com/jayknoxqu/id-number-util

https://blog.csdn.net/tobacco5648/article/details/50613025

https://github.com/xbeginagain/generator