如何在特定大小的Java中生成二进制Blob对象？

Question

I am trying to build an opaque Binary Blob 64-bit encoded with any garbage/random content, also trying to make it in such a way so that I can specify the size in Kb for the blob of data to generate into it.

The purpose of it is to bench-mark the write throughput of a DB application which support the blob datatype (eg C*). The blob does not need to be unique, hence it can be constructed once and just reuse it to avoid latency client side due to GC and object creation.

I have naively tried using a ByteBuffer.allocateDirect(bSize) thinking I could just use whatever garbage it will pickup from memory, however I am doubtful that this approach works.

Any suggestions as to how to achieve that?

Answer 1

If you are using Linux, this little command line trick creates dummy files quickly:

 dd if=/dev/zero of=FileName bs=1024 count=1000

count * bs = filesize

Then:

String sql = "INSERT INTO table (blob) VALUES (?)";
PreparedStatement stmt = conn.prepareStatement(sql);
File blob = new File("path_to_FileName");
FileInputStream   fis = new FileInputStream(blob);
stmt.setBinaryStream(1, fis, (int) blob.length());
stmt.execute();

Answer 2

If you need the data to be random you can use Random.nextBytes(byte[]); however it is rare for such systems to care about the contents so you can just use all zeros.

byte[] blob = new byte[byteCount];

// optionally
// new Random().nextBytes(blob);

stmt.setBinaryStream(1, new ByteArrayInputStream(blob), blob.length);

Answer 3

Make a custom implementation of InputStream to produce random data:

import java.io.IOException;
import java.io.InputStream;
import java.util.Random;

public class RandomInputStream extends InputStream
{
    private int available;
    private Random random = new Random();

    public RandomInputStream(final int totalSize)
    {
        this.available = totalSize;
    }

    @Override
    public int read() throws IOException
    {
        if(available == 0)
            return -1;
        --available;
        return random.nextInt(256);
    }


    @Override
    public int available() throws IOException
    {
        return available;
    }
}

And then just use it as usual:

InputStream is = new RandomInputStream(1024*1024);
stmt.setBinaryStream(1, is, is.available());

This would be the fastest solution possible.

Answer 4

byte[] blob = new byte[byteCount];

// optionally // new Random().nextBytes(blob);

stmt.setBinaryStream(1, new ByteArrayInputStream(blob), blob.length);