这是最简单的方法之一

 String pathFile = "/path/to/file";
 byte[] bytes = Files.readAllBytes(Paths.get(pathFile ));

我相信这是最简单的方法:

org.apache.commons.io.FileUtils.readFileToByteArray(file);

正如有人所说，Apache Commons File Utils可能有您正在寻找的东西

public static byte[] readFileToByteArray(File file) throws IOException

示例使用(Program.java):

import org.apache.commons.io.FileUtils;
public class Program {
    public static void main(String[] args) throws IOException {
        File file = new File(args[0]);  // assume args[0] is the path to file
        byte[] data = FileUtils.readFileToByteArray(file);
        ...
    }
}

public static byte[] readBytes(InputStream inputStream) throws IOException {
    byte[] buffer = new byte[32 * 1024];
    int bufferSize = 0;
    for (;;) {
        int read = inputStream.read(buffer, bufferSize, buffer.length - bufferSize);
        if (read == -1) {
            return Arrays.copyOf(buffer, bufferSize);
        }
        bufferSize += read;
        if (bufferSize == buffer.length) {
            buffer = Arrays.copyOf(buffer, bufferSize * 2);
        }
    }
}

以下方法不仅将Java .io. file转换为byte[]，我还发现它是读取文件的最快方法，当测试许多不同的Java文件读取方法时:

java.nio.file.Files.readAllBytes ()

import java.io.File;
import java.io.IOException;
import java.nio.file.Files;

public class ReadFile_Files_ReadAllBytes {
  public static void main(String [] pArgs) throws IOException {
    String fileName = "c:\\temp\\sample-10KB.txt";
    File file = new File(fileName);

    byte [] fileBytes = Files.readAllBytes(file.toPath());
    char singleChar;
    for(byte b : fileBytes) {
      singleChar = (char) b;
      System.out.print(singleChar);
    }
  }
}

你也可以使用NIO api来实现。只要总文件大小(以字节为单位)适合int类型，我就可以用这段代码做到这一点。

File f = new File("c:\\wscp.script");
FileInputStream fin = null;
FileChannel ch = null;
try {
    fin = new FileInputStream(f);
    ch = fin.getChannel();
    int size = (int) ch.size();
    MappedByteBuffer buf = ch.map(MapMode.READ_ONLY, 0, size);
    byte[] bytes = new byte[size];
    buf.get(bytes);

} catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
} finally {
    try {
        if (fin != null) {
            fin.close();
        }
        if (ch != null) {
            ch.close();
        }
    } catch (IOException e) {
        e.printStackTrace();
    }
}

我认为它非常快，因为它使用MappedByteBuffer。