java安全经理

Question

grant {
     permission java.io.FilePermission "C:\\class\\*", "read, write";
     permission java.lang.RuntimePermission "createClassLoader";
};

我的policy.txt

java.security.AccessControlException: access denied (java.io.FilePermission c:\class read)
    at java.security.AccessControlContext.checkPermission(Unknown Source)
    at java.security.AccessController.checkPermission(Unknown Source)
    at java.lang.SecurityManager.checkPermission(Unknown Source)
    at java.lang.SecurityManager.checkRead(Unknown Source)
    at java.io.File.isDirectory(Unknown Source)
    at java.io.File.toURI(Unknown Source)
    at loader.Main.main(Main.java:35)

我的例外.. ???

我的电话

String path = "c:\\class\\";
String app = "x.MyHTMLPrint";


File file = new File(path);
URL url = file.toURI().toURL();
URL[] urls = new URL[]{url};

ClassLoader cl = new URLClassLoader(urls);
Class cls = cl.loadClass(app);

只要我给它完整的权限它就有效.

有任何想法吗？谢谢!

Answer 1

那里似乎有两个问题.

首先,正如Yishai所说,File.toURI似乎需要检查没有尾随分隔符的文件是否是目录.这可能是一个错误.

其次,递归子目录的通配符是' - '而不是'*'.

因此,您的策略文件需要如下所示:

grant {
    permission java.io.FilePermission "C:\\class\\-", "read";
    permission java.io.FilePermission "C:\\class", "read";
    permission java.lang.RuntimePermission "createClassLoader";
};

此外,如果您使用URLClassLoader.newInstance,则不需要createClassLoader权限,并且您将获得完整的类加载器实现.