武器文本提取报错

[ulysses9910]

    def process_weapons(self):
        """add story context and 5-level weapon skill descriptions"""
        all_readable_files = list(
            glob.iglob(os.path.join(self.repo, "Readable", self.lang, "*"))
        )
        files = [f for f in all_readable_files if "Weapon" in f]
        # load story
        for weapon_file in files:
            idx = int(re.match(".+Weapon(\d+).txt", weapon_file).group(1))
            with open(weapon_file, "r", encoding="utf-8") as g:
                story = "".join(g.readlines()).strip("\n")
            self.map_weaponId_to_info[idx]["ReadableText"] = story

上面这段处理武器过程的时候， idx = int(re.match(".+Weapon(\d+).txt", weapon_file).group(1)) 这一句会报错，因为武器描述文件名称可能有Weapon11513_1.txt; Weapon11513_2.txt这样的形式（内容也不一样，应该是不同精炼等级的不同描述？） 稍微做了一点修改，以识别同一件武器的不同描述：

        for weapon_file in files:
            # 使用单个正则表达式匹配两种格式
            match = re.match(r".+Weapon(\d+)(?:_(\d))?.txt", weapon_file)
            if not match:
                continue  # 如果文件名不匹配，跳过
        
            idx1 = int(match.group(1))
            idx2 = match.group(2)  # idx2 可能是 None

            # 读取文件内容
            with open(weapon_file, "r", encoding="utf-8") as g:
                story = "".join(g.readlines()).strip("\n")
        
            # 根据 idx2 的存在与否决定字典键名
            readable_text_key = f"ReadableText{idx2}" if idx2 else "ReadableText1"

            # 更新字典中的内容
            self.map_weaponId_to_info[idx1][readable_text_key] = story