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On May 1, 2006, at 9:27 AM, Prasad Rajagopal wrote: Sorry on this one. There is a java.security.AccessControlException class and I didn't find it earlier for some reason... showc java.security.AccessControlException showc: invoked with: java.security.AccessControlException public class java.security.AccessControlException extends java.lang.SecurityException { // Constructors public java.security.AccessControlException(java.lang.String); public java.security.AccessControlException(java.lang.String, java.security.Permission); // Fields private static final long serialVersionUID; private java.security.Permission perm; // Methods public java.security.Permission getPermission(); } If you could catch the exception here and indicate what permission is required?
This looks to be the only permission you actually have. Assuming you can't do that and this is 1.5 I did just discover you can do this (With Safari, untested FireFox). Start java preferences /Applications/Utilities/Java/J2SE 5.0/Java Preferences First make sure you get the console when an applet runs. From the Advanced settings "Java console" should have "Show console" selected. Then what I discovered is that from the "General" settings you can set the security debug property in the "Java Applet Runtime Parameters" text area. For example -Djava.security.debug=all or -Djava.security.debug=access,failure Then when your applet runs the security debug information will appear in the java console. Again if you can determine exactly what Permission you don't have you would then understand the source of your problem. How you fix it I don't know, but knowing that would probably have to be first? Mike Hall mikehall at spacestar dot net |
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| >Re: Java Applet Security Problem (From: "Prasad Rajagopal" <email@hidden>) |
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