RGPerson – 随机身份信息生成脚本

发表于 2019年08月14日
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RGPerson 随机身份信息

做测试的时候经常遇到一些模拟注册的业务场景，要填写的东西很多，一般都是临时去百度信息，这样很繁琐所以决定造轮子撸了个随机身份生成的，如果是非测试场景，可参考社会工程学相关。

该脚本生成信息：姓名\年龄\性别\身份证\手机号

生成的信息都是模拟的假数据，仅限用于业务测试使用。

本编写原理

脚本的三大函数： genMobile()、genIdCard()、genName()

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

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"}

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}

def genMobile():

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

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

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

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()

1 2	def last_name(): return GBK2312()

5.拼接

def last_name():
    return GBK2312()

1 2	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)

6位区域码爬取http://www.360doc.com/content/12/1010/21/156610_240728293.shtml，存到了districtcode.py

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。

8位出生日期码，具体Python代码如下：

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

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）。

2位顺序码

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。

1位性别码

性别码 奇数表示男性，偶数表示女性。

最后一位是校验码，这里采用的是ISO 7064:1983,MOD 11-2校验码系统。校验码为一位数，但如果最后采用校验码系统计算的校验码是“10”，碍于身份证号码为18位的规定，则以“X”代替校验码“10”。

最难的还是校验码的算法，参考师傅的解说：

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
将这17位数字和系数相乘的结果相加。
用加出来和除以11，得余数
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。
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])

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