For the Internet Explorer to bypass the security manager, set from the menu:
Tools\Internet Option\Security\Trusted Sites\Custom Level\Microsoft Vm\Java
Custom Settings button\Edit Permision\Run Unsigned content Enable and Run
signed content Enable.
Add the site from where you running the applet in the trusted sites list.
----- Original Message -----
From: "Anton Litvinenko" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, March 13, 2002 4:59 PM
Subject: AccessControlException in applet
> Hello,
>
> I am trying to run applet, that connects to Server and tries to execute
> methods via xmlrpc.
>
> The connecion establishes, but when applet tries to call a method a
> java.security.AccessControlException appears....
>
> please help me to solve this problem
>
> Exception TRace:
> java.security.AccessControlException: access denied
> (java.lang.RuntimePermission accessClassInPackage.sun.misc)
> 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.checkPackageAccess(Unknown Source)
> at sun.applet.AppletSecurity.checkPackageAccess(Unknown Source)
> at sun.applet.AppletClassLoader.loadClass(Unknown Source)
> at java.lang.ClassLoader.loadClass(Unknown Source)
> at java.lang.ClassLoader.loadClassInternal(Unknown Source)
> at
org.apache.xmlrpc.applet.SimpleXmlRpcClient.execute(SimpleXmlRpcClient.java)
> at org.apache.xmlrpc.applet.XmlRpcApplet.execute(XmlRpcApplet.java)
> at xmlrpc.init(xmlrpc.java:29)
> at sun.applet.AppletPanel.run(Unknown Source)
> at java.lang.Thread.run(Unknown Source)
>
> Applet Code:
>
> public void init() {
> try {
> jbInit();
> System.out.println("client init ");
> initClient(6666);
> Vector params = new Vector ();
> params.addElement ("Hello World");
> // this method returns a string
> System.out.println("executin the query");
> String result = (String) execute ("echo", params);
> System.out.println(result);
> textField1.setText("result");
>
> }
> catch(Exception e) {
> e.printStackTrace();
> }
> }
>
>
>
> sincerely Yours,
> ANTON Litvinenko.
