python3基于给定的正则表达式生成x值

Question

我试图使用rstr基于regex定义生成值。

以下是起作用的原因：

[root@localhost ~]# python3
Python 3.6.8 (default, Nov 16 2020, 16:55:22)
[GCC 4.8.5 20150623 (Red Hat 4.8.5-44)] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>>import rstr
>>>print(rstr.xeger(r'[2-9]\d{11}'))
>>>867050869842

现在，我想编写一个脚本，它将以任何正则表达式作为参数，并生成10次输出：

#!/usr/bin/python3

import sys
import rstr

if len(sys.argv) != 2:
        print("[+] Usage ./regexgen.py regex")
        exit()

regex = sys.argv[1]


for string in range(10):
        print(rstr.xeger(r'{1}')).format(regex)

但是，当执行脚本时，它会失败，出现以下错误：

[root@localhost ~]# python3 script2.py [2-9]\d{11}
Traceback (most recent call last):
  File "script2.py", line 18, in <module>
    print(rstr.xeger(r'{1}')).format(regex)
  File "/usr/local/lib/python3.6/site-packages/rstr/xeger.py", line 63, in xeger
    parsed = re.sre_parse.parse(pattern)
  File "/usr/lib64/python3.6/sre_parse.py", line 855, in parse
    p = _parse_sub(source, pattern, flags & SRE_FLAG_VERBOSE, 0)
  File "/usr/lib64/python3.6/sre_parse.py", line 416, in _parse_sub
    not nested and not items))
  File "/usr/lib64/python3.6/sre_parse.py", line 616, in _parse
    source.tell() - here + len(this))
sre_constants.error: nothing to repeat at position 0

我的猜测是，xeger无法用格式函数解析插入的regex。

Answer 1

试试这个：

import sys

import rstr

if len(sys.argv) != 2:
    print("[+] Usage ./regexgen.py regex")
    exit()

regex = sys.argv[1]  # regex passed as arg [2-9]\d{11}
for string in range(10):
    print(rstr.xeger(regex))

输出：

302799318288
303010436356
523231185691
537677558398
824580634154
398638175299
546948835775
845745020055
456616189703
579459609359