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AccessControlException

Ronnie Phelps
Ranch Hand

Joined: Mar 12, 2001
Posts: 329
I'm having trouble starting my server. Here are the contents of my policy file.
java.policy:

grant {
java.security.AllPermission;
};

This is how I execute the command:

java -Djava.security.policy=java.policy learning.rmi.Request

Here's the exception I get

java.security.Policy: error parsing file:C:/learning/rmi/java.policy
java.security.Policy: line 2: expected permission entry
Exception in thread "main" java.security.AccessControlException: access denied (
java.net.SocketPermission 127.0.0.1:1099 connect,resolve)
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.checkConnect(Unknown Source)
at java.net.Socket.<init>(Unknown Source)
at java.net.Socket.<init>(Unknown Source)
at sun.rmi.transport.proxy.RMIDirectSocketFactory.createSocket(Unknown S
ource)
at sun.rmi.transport.proxy.RMIMasterSocketFactory.createSocket(Unknown S
ource)
at sun.rmi.transport.tcp.TCPEndpoint.newSocket(Unknown Source)
at sun.rmi.transport.tcp.TCPChannel.createConnection(Unknown Source)
at sun.rmi.transport.tcp.TCPChannel.newConnection(Unknown Source)
at sun.rmi.server.UnicastRef.newCall(Unknown Source)
at sun.rmi.registry.RegistryImpl_Stub.rebind(Unknown Source)
at java.rmi.Naming.rebind(Unknown Source)
at learning.rmi.Request.main(Request.java:29)
john d
Greenhorn

Joined: Jul 11, 2002
Posts: 9
you need to put "permission" in front of the call like this!
grant {
// Allow everything for now
permission java.security.AllPermission;
};
hope it helps
Ronnie Phelps
Ranch Hand

Joined: Mar 12, 2001
Posts: 329
Thanks John. I had already figured it out but I guess i chosed the wrong forum to post it in, because "Distributed java" doesn't seem to be very popular. Thanks anyway dude.
 
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subject: AccessControlException