使用javax.sound.midi接收midi输入

时间:2018-06-24 20:00:34

标签: midi receiver javax.sound.midi

我希望有一个程序可以在获得midi输入时执行某些操作。

例如,当我单击控制器上的按钮1时,应打印“您单击了btn 1”,当我单击按钮2时应显示“您单击了btn 2”。

我尝试使用javax.sound.midi库,但论坛或youtube上的所有示例均无法使用。

这对我来说最有效。它打印了我PC的所有Midi设备,但没有收到任何信息。有人可以帮忙吗?

package de.snke.dev;

import javax.sound.midi.*;;



public class Main  extends Object implements Receiver{

static MidiClass myMidi;

public static void main(String[] args) throws Exception {

    MidiDevice.Info[] info =
             MidiSystem.getMidiDeviceInfo();

            for (int i=0; i < info.length; i++) {
             System.out.println(i + ") " + info[i]);
             System.out.println("Name: " + info[i].getName());
             System.out.println("Description: " +
             info[i].getDescription());

             MidiDevice device =
            MidiSystem.getMidiDevice(info[i]);
             System.out.println("Device: " + device);
            }

}

public void send(MidiMessage msg,
        long time) {
        System.out.println("Received message " + msg);
        }

        public void close() {
        System.out.println("Closing");
        }
}

编辑:现在我有

Sequencer           seq;
Transmitter         seqTrans;
Synthesizer         synth;
Receiver         synthRcvr;
try {
      seq     = MidiSystem.getSequencer();
      seqTrans = seq.getTransmitter();
      synth   = MidiSystem.getSynthesizer();
      synthRcvr = synth.getReceiver(); 
      seqTrans.setReceiver(synthRcvr);      
} catch (MidiUnavailableException e) {
      // handle or throw exception
}

我现在已经连接到我的APC Mini了吗?对不起,我是初学者。 如果是,我现在如何读取Midi输入?如果没有,我必须更改什么?

1 个答案:

答案 0 :(得分:0)

解决方案是:

package de.snke.dev;

import javax.sound.midi.*;
import java.util.ArrayList;
import java.util.List;
import java.io.*;

public class Main
{

public void Main()
{
    MidiDevice device;
    MidiDevice.Info[] infos = MidiSystem.getMidiDeviceInfo();
    for (int i = 0; i < infos.length; i++) {
        try {
        device = MidiSystem.getMidiDevice(infos[i]);
        //does the device have any transmitters?
        //if it does, add it to the device list
        System.out.println(infos[i]);

        //get all transmitters
        List<Transmitter> transmitters = device.getTransmitters();
        //and for each transmitter

        for(int j = 0; j<transmitters.size();j++) {
            //create a new receiver
            transmitters.get(j).setReceiver(
                    //using my own MidiInputReceiver
                    new MidiInputReceiver(device.getDeviceInfo().toString())
            );
        }

        Transmitter trans = device.getTransmitter();
        trans.setReceiver(new MidiInputReceiver(device.getDeviceInfo().toString()));

        //open each device
        device.open();
        //if code gets this far without throwing an exception
        //print a success message
        System.out.println(device.getDeviceInfo()+" Was Opened");


    } catch (MidiUnavailableException e) {}
}


}
//tried to write my own class. I thought the send method handles an MidiEvents sent to it
public class MidiInputReceiver implements Receiver {
public String name;
public MidiInputReceiver(String name) {
    this.name = name;
}
public void send(MidiMessage msg, long timeStamp) {


    byte[] aMsg = msg.getMessage();
    // take the MidiMessage msg and store it in a byte array

    // msg.getLength() returns the length of the message in bytes
    for(int i=0;i<msg.getLength();i++){
        System.out.println(aMsg[i]);
        // aMsg[0] is something, velocity maybe? Not 100% sure.
        // aMsg[1] is the note value as an int. This is the important one.
        // aMsg[2] is pressed or not (0/100), it sends 100 when they key goes down,  
        // and 0 when the key is back up again. With a better keyboard it could maybe
        // send continuous values between then for how quickly it's pressed? 
        // I'm only using VMPK for testing on the go, so it's either 
        // clicked or not.
    }
    System.out.println();
}
public void close() {}
}
}

这2个线程解决方案相结合,可以打开所有midi设备并打印其速度,音符值和状态(是否单击)

别忘了!

您必须致电

Main main = new Main();
main.Main();

在单独的类中启动Main类中的Main方法。