Can I use FileReader to read a file containing images and video (say an epub file) and text and is it suggested to do so with respect to perfomance.

Question

I need to parse contents of a epub file and I am trying to see what would be the most efficient way to do it. The epub file may contain images, lot of text and occasionally videos too. Should I go for a FileInputStream or a FileReader?

Answer 1

As epub uses a ZIP archive structure I would propose to handle it as such. Find a small snippet below which list the content of an epub file.

Map<String, String> env = new HashMap<>();
env.put("create", "true");

Path path = Paths.get("foobar.epub");
URI uri = URI.create("jar:" + path.toUri());
FileSystem zipFs = FileSystems.newFileSystem(uri, env);
Path root = zipFs.getPath("/");
Files.walkFileTree(root, new SimpleFileVisitor<Path>() {
    @Override
    public FileVisitResult visitFile(Path file,
            BasicFileAttributes attrs) throws IOException {
        print(file);
        return FileVisitResult.CONTINUE;
    }

    @Override
    public FileVisitResult preVisitDirectory(Path dir,
            BasicFileAttributes attrs) throws IOException {
        print(dir);
        return FileVisitResult.CONTINUE;
    }

    private void print(Path file) throws IOException {
        Date lastModifiedTime = new Date(Files.getLastModifiedTime(file).toMillis());
        System.out.printf("%td.%<tm.%<tY %<tH:%<tM:%<tS %9d %s\n", 
                lastModifiedTime, Files.size(file), file);
    }
});

sample output

01.01.1970 00:59:59         0 /META-INF/
11.02.2015 16:33:44       244 /META-INF/container.xml
11.02.2015 16:33:44      3437 /logo.jpg
...

edit If you only want to extract files based on their names you could do it like shown in this snippet for the visitFile(...) method.

public FileVisitResult visitFile(Path file,
    BasicFileAttributes attrs) throws IOException {
    // if the filename inside the epub end with "*logo.jpg"
    if (file.endsWith("logo.jpg")) {
        // extract the file in directory /tmp/
        Files.copy(file, Paths.get("/tmp/",
            file.getFileName().toString()));
    }
    return FileVisitResult.CONTINUE;
}

Depending on how you want to process the files inside the epub you might also have a look on the ZipInputStream.

try (ZipInputStream in = new ZipInputStream(new FileInputStream("foobar.epub"))) {
    for (ZipEntry entry = in.getNextEntry(); entry != null; 
        entry = in.getNextEntry()) {
        System.out.printf("%td.%<tm.%<tY %<tH:%<tM:%<tS %9d %s\n",
                new Date(entry.getTime()), entry.getSize(), entry.getName());
        if (entry.getName().endsWith("logo.jpg")) {
            try (FileOutputStream out = new FileOutputStream(entry.getName())) {
                // process the file
            }
        }
    }
}

sample output

11.02.2013 16:33:44       244 META-INF/container.xml
11.02.2013 16:33:44      3437 logo.jpg

Answer 2

The easiest way to read a whole file as bytes (and thats what you want if it's not plain text) is to use the java.nio.file.Files class:

byte[] content = Files.readAllBytes(Paths.get("example.epub"));

Advantages of this method:

• less code = code gets more readable and has less potential for errors
• java cares about opening and closing file

---

Edit:

In order to read a file really fast you can use java.nio as well. This time java.nio.channels.FileChannel:

import java.io.FileInputStream;
import java.nio.MappedByteBuffer;
import java.nio.channels.FileChannel;

// Load the file
FileChannel c = new FileInputStream("example.epub").getChannel();
MappedByteBuffer byteBuffer = c.map(FileChannel.MapMode.READ_ONLY, 0, channel.size());

// Process the data
buffer.get(myByte, 1120, 50);

// when finished
c.close();

This will not read the whole file into memory but creates a link to the file and reads (buffers) only the parts you try to access. It will also recognize changes on the file and always return the latest content.