security

AccessController example

In this example we shall show you how to create and check an access permission to a file, using the AccessController Class. To create a new permission to a file and check the access permission of a file, one should perform the following steps:

  • Create a new FilePermission with the specified path of the file and action mode of the permission.
  • Use the checkPermission(Permission perm) API method of the AccessController so as to check whether the access request indicated by the specified permission should be allowed or denied. If a requested access is allowed, checkPermission returns quietly. If denied, an AccessControlException is thrown.

Let’s take a look at the code snippet that follows:

package com.javacodegeeks.snippets.core;

import java.io.FilePermission;
import java.security.AccessControlException;
import java.security.AccessController;

public class Main {

    public static void main(String args[]) {

  FilePermission fp = new FilePermission("C:/Users/nikos7/Desktop/output.txt", "read");

  try {


AccessController.checkPermission(fp);

  } catch (AccessControlException ex) {


System.out.println("Access denied");

  

  }
    }
}

Output:

Access denied

 
This was an example of how to create and check access permission to a file in Java.

Byron Kiourtzoglou

Byron is a master software engineer working in the IT and Telecom domains. He is an applications developer in a wide variety of applications/services. He is currently acting as the team leader and technical architect for a proprietary service creation and integration platform for both the IT and Telecom industries in addition to a in-house big data real-time analytics solution. He is always fascinated by SOA, middleware services and mobile development. Byron is co-founder and Executive Editor at Java Code Geeks.
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