Geek asks and answers

How to play sound on many devices at the same time

I want to play sound on three external sound cards at the same time, I mean, when I press a button, I hear sound from three speakers that are connected to my three sound cards. I have a function, but it plays sound only on one device, the first thing it finds, I mean that in this code the first device is number 0, so it plays sound in it, but if you write the device number 1 first , it will reproduce the sound in it, as the output that it reproduces sound in only one device, it does not work for all devices at the same time. This is his code:

public void playAllAvailableDevices()
{
//create a new class for each wav file & output etc.
WaveOut waveOut1 = new WaveOut();
WaveFileReader waveReader1 = new WaveFileReader(fileName);
WaveOut waveOut2 = new WaveOut();
WaveFileReader waveReader2 = new WaveFileReader(fileName);
WaveOut waveOut3 = new WaveOut();
WaveFileReader waveReader3 = new WaveFileReader(fileName);

switch (waveOutDevices)
{
case 1: 
waveOut1.Init(waveReader1);
waveOut1.DeviceNumber = 0;
waveOut1.Play();
break;
case 2: 
waveOut1.Init(waveReader1);
waveOut1.DeviceNumber = 0;
waveOut1.Play();

waveOut2.Init(waveReader2);
waveOut2.DeviceNumber = 1;
waveOut2.Play();
break;
case 3:
waveOut1.Init(waveReader1);
waveOut1.DeviceNumber = 0;
waveOut1.Play();

waveOut2.Init(waveReader2);
waveOut2.DeviceNumber = 1;
waveOut2.Play();

waveOut3.Init(waveReader3);
waveOut3.DeviceNumber = 2;
waveOut3.Play();
break;
}}

fileName is the name of the sound file I want to play, in my code I get this name from darabase:

private void btnAlarm1_Click(object sender, EventArgs e)
    {

        OdbcConnection cn = new OdbcConnection("DSN=cp1");
        cn.Open();
        OdbcCommand cmd1 = new OdbcCommand("select chemin from alarme where code_alarme=41", cn);
        cmd1.Connection = cn;
        fileName = cmd1.ExecuteScalar().ToString();
        wave = new WaveOut();
        playAllAvailableDevices();
    }

, ???? . .

+3
2

Init DeviceNumber WaveOut. , :

    private void PlaySoundInDevice(int deviceNumber, string fileName)
    {
        if (outputDevices.ContainsKey(deviceNumber))
        {
            outputDevices[deviceNumber].WaveOut.Dispose();
            outputDevices[deviceNumber].WaveStream.Dispose();
        }
        var waveOut = new WaveOut();
        waveOut.DeviceNumber = deviceNumber;
        WaveStream waveReader = new WaveFileReader(fileName);
        // hold onto the WaveOut and  WaveStream so we can dispose them later
        outputDevices[deviceNumber] = new PlaybackSession { WaveOut=waveOut, WaveStream=waveReader };

        waveOut.Init(waveReader);
        waveOut.Play();
    }

    private Dictionary<int, PlaybackSession> outputDevices = new Dictionary<int, PlaybackSession>();

    class PlaybackSession
    {
        public IWavePlayer WaveOut { get; set; }
        public WaveStream WaveStream { get; set; }
    }

WaveOut, , , . , , , :

    private void DisposeAll()
    {
        foreach (var playbackSession in outputDevices.Values)
        {
            playbackSession.WaveOut.Dispose();
            playbackSession.WaveStream.Dispose();
        }
    }

, for switch, :

    public void PlayInAllAvailableDevices(string fileName)
    {
        int waveOutDevices = WaveOut.DeviceCount;
        for (int n = 0; n < waveOutDevices; n++)
        {
            PlaySoundInDevice(n, fileName);
        }
    }
+7

More articles:


All Articles