[英]How to solve java.security.AccessControlException
我需要有關java.security.AccessControlException
建議,我在執行以下代碼時會得到建議。 (我在這里曾咨詢過類似的問題,但未能成功實現)
這是我的服務器代碼:
public class GetPageInfos extends UnicastRemoteObject implements RemoteGetInfo{
private static final String url="http://www.lemonde.fr/";
public class GetPageInfos extends UnicastRemoteObject implements RemoteGetInfo{
private static final String url="http://www.lemonde.fr/";
public GetPageInfos() throws RemoteException{
}
public String getSiteInfos() throws RemoteException {
Document doc;
try {
doc = Jsoup.connect(url).get();
String title = doc.title();
return "title is "+title;
} catch (IOException e) {
System.out.println("Faild! "+e.getMessage());
return "not found";
}
}
public static void main(String[] args){
try {
GetPageInfos infos= new GetPageInfos();
//System.setProperty("java.rmi.server.hostname","5lq04x1.gemalto.com");
Naming.rebind("RemoteGetInfo", infos);
/*GetPageInfos obj=new GetPageInfos();
RemoteGetInfo stub = (RemoteGetInfo) UnicastRemoteObject.exportObject(obj, 0);
Registry registry = LocateRegistry.getRegistry();
registry.bind("RemoteGetInfo", stub);
*/
System.out.println("server ready");
} catch (RemoteException e) {
System.out.println("GetPageInfos "+e.getMessage());
}
catch (MalformedURLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
這是我的客戶代碼:
//RMI Client
public class PrintSiteInfos {
public static void main(String arg[])
{
System.setSecurityManager(new RMISecurityManager());
try
{
/*String host=null;
Registry registry = LocateRegistry.getRegistry(host);
RemoteGetInfo stub = (RemoteGetInfo) registry.lookup("RemoteGetInfo");
String response = stub.getSiteInfos();
System.out.println(response); */
RemoteGetInfo obj = (RemoteGetInfo) Naming.lookup( "RemoteGetInfo");
System.out.println(obj.getSiteInfos());
}
catch (Exception e)
{
System.out.println("PrintSiteInfos exception: " + e.getMessage());
e.printStackTrace();
}
}
}
所以我得到了
exception: access denied ("java.net.SocketPermission" "127.0.0.1:1099" "connect,resolve")
我發現我必須傳遞一個我喜歡的策略文件:
grant {
permission java.security.AllPermission;};
但是如何? 還有其他建議嗎?
您可以只授予套接字權限,而不能授予所有權限(這可能會帶來安全風險)。 因此類似:
grant {
permission java.net.SocketPermission "127.0.0.1:1099", "connect, resolve";
};
有兩種方法:
1)作為命令行的參數
java -Djava.security.policy=mypolicyfile PrintSiteInfo
2)在JRE環境中:
在JRE_HOME / lib / security / java.policy文件中添加權限
擺脫安全管理器。 除非您正在使用RMI代碼庫功能,否則不需要它。
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