指定漏洞时无法利用，显示暂不支持的解决方法（猜测） #36

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on Jul 15, 2025

第1502行if name not in s2_list:应改为if name not in s2_dict:
判断时检测的应是字符串，却变成了类
导致字符串与类相比较，判断不存在该漏洞

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