Brute force a regular expression and match it against a flag format

brute_regex.py

#!/usr/bin/env python3

import argparse
from rstr import xeger as generate_pattern

def brute_regex(
	regex: str,
	count: int = 1000,
	match: str = "",
	show: bool = False
):
	''' Test-Case '''
	try:
		_ = generate_pattern(regex)
	except Exception as E:
		print(f"Error: {E}")
		print("[!] Please verify if your regex is a valid regex.")
		exit(1)

	uniqs = []
	print("[*] Matches::")
	for i in range(count):
		pp = False
		gen = generate_pattern(regex)
		if match.lower() not in gen.lower():
			continue

		if show:
			print(gen)
			continue

		if gen not in uniqs:
			uniqs.append(gen)
			print(gen)

if __name__ == "__main__":
	parser = argparse.ArgumentParser(description='Brute Force Regular Expression based on a certain flag.')
	parser.add_argument('-r', '--regex', type=str, help='Regex to brute force', required=True)
	parser.add_argument('-m', '--match', type=str, help="Match a specific string in flag", default="")
	parser.add_argument('-c', '--count', type=int, help="Maximum number of patterns to be generated (if the regex allows)", default=1000)
	parser.add_argument('--show-all', action='store_true', dest='show', help="Shows all the found matches (Default: Show only unique)")

	args = parser.parse_args()
	brute_regex(regex=args.regex, count=args.count, match=args.match, show=args.show)