How to Copy a Part of a File with Java

Learn how to efficiently copy the content of a ZIP file that follows a header in Java. This guide explains the steps and tips for effective file manipulation in Java.
---

Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: Copy a part of the file with Java

If anything seems off to you, please feel free to write me at vlogize [AT] gmail [DOT] com.
---
How to Copy a Part of a File with Java: A Simple Guide

When working with binary files in Java, you might find yourself in a situation where you need to skip a certain number of bytes at the beginning of a file in order to access its content. This is commonly the case with files that have headers, like a ZIP file that includes a significant amount of metadata or other non-essential information at the start. In this guide, we’ll tackle the question: How can you skip a header of 0x270 bytes and copy the rest of the ZIP content into a new file?

The Challenge

Imagine you have a data file consisting of two parts:

A header that is 0x270 bytes long

A ZIP file that follows immediately after the header

If you're looking to copy only the ZIP content to a new file, you need to efficiently bypass that header and work specifically with the bytes that belong to the ZIP file. Let’s explore how to do this using Java.

The Solution

Here is a simple Java code snippet that achieves this goal:

[[See Video to Reveal this Text or Code Snippet]]

Code Breakdown

Setting Up File Input Streams:

Initialize a FileInputStream to read from your source file.

Use a DataInputStream which facilitates reading primitive data types from the underlying input stream.

Skip the Header:

Use the skipBytes method to bypass the header (0x270 bytes). This is crucial in ensuring you only get the ZIP content.

Prepare the Destination File:

Create a new File object for the destination file where you will store the ZIP data, and ensure the file is created.

Buffer for Reading Data:

Use a byte array buffer to read data. While the original code used a 16-byte buffer, it is recommended to use a larger size (e.g., 64K bytes) to improve efficiency and reduce system call overheads.

Writing the Data:

Use a loop to read the data into the buffer until the end of the file is reached. Write the read bytes directly into the destination file.

Closing Streams:

Always ensure you close both input and output streams to free up system resources.

Important Considerations

Buffer Size Matters:

Instead of using a buffer of size 16 bytes, opt for 64K. This will drastically improve the performance of file copying due to reduced system calls, especially with larger files.

Avoid Unnecessary Classes:

While using DataInputStream is possible, Java's FileInputStream has a skip(long) method that could simplify your code by eliminating the need for DataInputStream.

Testing Your Code

If you're unsure about how the code will perform without a file to test against, you can simulate the file structure you need. For example:

On Linux, you can create a stream containing 0x270 zero bytes and append a regular ZIP file using commands like dd and cat. This allows you to experiment with your code without needing the actual file.

Conclusion

Copying a part of a file in Java, especially when it involves skipping headers, is a straightforward process if done correctly. With the code and considerations discussed in this post, you should be well-equipped to handle similar file manipulation tasks efficiently. Remember to always test your implementation and optimize for performance to enhance the user experience.