Need

Ensuring complete and secure code implementation

Context

• Usage of Java 8 for developing applications with enhanced features and performance
• Usage of javax.servlet-api for Java Servlet development

Description

Non compliant code

        package com.example.servlet;

import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class BinaryServlet extends HttpServlet {
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        String binaryFilePath = "/path/to/binary/file";
        response.setContentType("application/octet-stream");
        response.setHeader("Content-Disposition", "attachment;filename=" + binaryFilePath);
        ServletOutputStream out = response.getOutputStream();
        FileInputStream in = new FileInputStream(binaryFilePath);
        byte[] buffer = new byte[4096];
        int length;
        while ((length = in.read(buffer)) > 0){
           out.write(buffer, 0, length);
        }
        in.close();
        out.flush();
    }
}
        
        

The above code is a simple servlet that serves a binary file from a specified path when the doGet method is called.

The problem with the above code is that it serves binary files without any checks or validations. This can lead to several security vulnerabilities:

1. Unencrypted sensitive information: The binary file could contain sensitive information that is not encrypted. In this case, anyone who can access this servlet could potentially download and decompile the binary file to extract this sensitive information.

2. Potential for serving malicious files: Since there is no validation on what type of file is being served, this servlet could potentially be used to serve malicious binary files. This can lead to serious security breaches if these files are executed on the client's machine.

3. No source code for binary files: As per the best practices, all the binary files should have their corresponding source code in the repository. This is important for transparency, maintainability, and security. Without the source code, it's impossible to know what the binary file does, which makes it harder to maintain and more susceptible to hidden vulnerabilities or malicious code.

In conclusion, this code is vulnerable due to the lack of validations and checks when serving binary files and the potential for serving files with unencrypted sensitive information or malicious content.

Steps

• Avoid storing sensitive information in binary files.
• Encrypt sensitive information before storing it in binary files.
• Ensure that the binary files in the repository have their corresponding source code.
• Regularly review and update the binary files to address any vulnerabilities.
• Implement access controls to restrict unauthorized access to the binary files.
• Consider using secure file storage solutions to protect the binary files.
• Implement secure file download mechanisms to prevent unauthorized access to the binary files.
• Perform regular security audits and penetration testing to identify any vulnerabilities in the binary files.

Compliant code

        package com.example.servlet;

import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;
import javax.crypto.Cipher;
import javax.crypto.spec.SecretKeySpec;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class BinaryServlet extends HttpServlet {
    private static final String SECRET_KEY = "my-secret-key"; // Replace with your secret key

    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        String binaryFilePath = "/path/to/binary/file";
        response.setContentType("application/octet-stream");
        response.setHeader("Content-Disposition", "attachment;filename=" + binaryFilePath);
        ServletOutputStream out = response.getOutputStream();

        // Encrypt the file before sending
        byte[] fileContent = Files.readAllBytes(Paths.get(binaryFilePath));
        byte[] encryptedContent = encrypt(fileContent, SECRET_KEY);
        out.write(encryptedContent);

        out.flush();
    }

    private byte[] encrypt(byte[] data, String key) {
        try {
            SecretKeySpec skey = new SecretKeySpec(key.getBytes(), "AES");
            Cipher cipher = Cipher.getInstance("AES");
            cipher.init(Cipher.ENCRYPT_MODE, skey);
            return cipher.doFinal(data);
        } catch (Exception e) {
            throw new RuntimeException(e);
        }
    }
}
        
        

The updated code now includes encryption for the binary file before it is sent to the client. This is done using the AES encryption algorithm. The encrypt method takes the binary file data and a secret key as input, and returns the encrypted data.

The secret key is used to initialize a SecretKeySpec object, which is then used to initialize a Cipher object in encryption mode. The doFinal method of the Cipher class is then used to encrypt the data.

This ensures that even if the binary file contains sensitive information, it is not exposed in its raw form to the client. Instead, the client receives the encrypted data, which can only be decrypted using the correct secret key.

This solution also implies that the secret key must be securely managed. It should not be hard-coded as shown in this example, but instead should be securely stored and retrieved, for example, from a secure vault.

Please note that this is a basic example of file encryption. Depending on the specific requirements and the sensitivity of the data, a more complex encryption solution may be needed. This could include using a stronger encryption algorithm, implementing key rotation, or using a hardware security module (HSM) for key management.

References

• 233. Incomplete funcional code