Android VOIP application without access to interne

2019-03-13 22:20发布

I need to develop the VOIP application between 2 android devices.
As I know there is a SIP protocol used for this purpose but it requires registation to SIP server and access to internet for SIP signaling.
Is any way to create VOIP application in android without internet access?

标签: android sip voip
5条回答
\"骚年 ilove
2楼-- · 2019-03-13 22:38

I think you can use JITSI for p2p voip service on multiple platforms including Andriod.

These are my findings about this project:-

  1. Do not need any server or internet connectivity.
  2. Users must be under a same network.
  3. Open Source.
  4. Android apk is available, most probably you can find it code on website or you can de-compile it.
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地球回转人心会变
3楼-- · 2019-03-13 22:45

actually SIP clients can talk peer-to-peer, they just need to know their IP addresses and UDP ports where they listen to SIP messages.

You can play around with normal SIP clients on two comuters (X-Lite for Windows, Twinkle for Linux, and some others exist too) and try establishing a call between them without server registration. It's quite possible.

Also you can run a minimalistic SIP server somewhere in local LAN. For example, FreeSWITCH can be minimized to a quite tiny footprint.

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▲ chillily
4楼-- · 2019-03-13 22:48

OK so if you are looking for some peer-2-peer communications, I think wifi is the way to go (better distance and speeds). If you can develop for only newer versions of Android then WI-FI Direct is the way to go, but that will only work on Android 4.0 and above

In order to have something run on below 4.0 you are going to have to go with a 3rd party library. I know Qualcomm has a library called alljoyn, but not sure how good it is.

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别忘想泡老子
5楼-- · 2019-03-13 22:49

Of course it is possible! Why you would need the internet? As long as you are both connected to the same network that is fine! Below is the java and xml for a working app.

On start up it will provide you with your own local port, for example "52022".. this is random every time and unfortunately that can't be helped. We then enter the IP address of the other phone and THEIR randomly generated port number and press connect. They do exactly the same and BAM you're connected :) This test app obviously requires you to be close by to exchange port numbers, but in my proper app I was easily able to request each port number before connecting. Hope this helps!

public class MainActivity extends Activity {

AudioGroup m_AudioGroup;
AudioStream m_AudioStream;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
      StrictMode.setThreadPolicy(policy);
      try {   
          AudioManager audio =  (AudioManager) getSystemService(Context.AUDIO_SERVICE); 
          audio.setMode(AudioManager.MODE_IN_COMMUNICATION);
          m_AudioGroup = new AudioGroup();
          m_AudioGroup.setMode(AudioGroup.MODE_NORMAL);
          m_AudioStream = new AudioStream(InetAddress.getByAddress(getLocalIPAddress ()));
          int localPort = m_AudioStream.getLocalPort();
          m_AudioStream.setCodec(AudioCodec.PCMU);
          m_AudioStream.setMode(RtpStream.MODE_NORMAL);

          ((TextView)findViewById(R.id.lblLocalPort)).setText(String.valueOf(localPort));

          ((Button) findViewById(R.id.button1)).setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) {
                String remoteAddress = ((EditText)findViewById(R.id.editText2)).getText().toString();
                String remotePort = ((EditText)findViewById(R.id.editText1)).getText().toString();

                  try {
                    m_AudioStream.associate(InetAddress.getByName(remoteAddress), Integer.parseInt(remotePort));
                } catch (NumberFormatException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                } catch (UnknownHostException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
                  m_AudioStream.join(m_AudioGroup);
            }
        });

          ((Button) findViewById(R.id.button2)).setOnClickListener(new OnClickListener() {

                @Override
                public void onClick(View v) {
                      m_AudioStream.release();
                }
            });

      } catch (Exception e) {
       Log.e("----------------------", e.toString());
       e.printStackTrace();
      }
}

public static byte[] getLocalIPAddress () {
    byte ip[]=null;
       try {
           for (Enumeration<NetworkInterface> en = NetworkInterface.getNetworkInterfaces(); en.hasMoreElements();) {
               NetworkInterface intf = en.nextElement();
               for (Enumeration<InetAddress> enumIpAddr = intf.getInetAddresses(); enumIpAddr.hasMoreElements();) {
                   InetAddress inetAddress = enumIpAddr.nextElement();
                   if (!inetAddress.isLoopbackAddress()) {
                    ip= inetAddress.getAddress();
                   }
               }
           }
       } catch (SocketException ex) {
           Log.i("SocketException ", ex.toString());
       }
       return ip;

}

}

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:id="@+id/LinearLayout1"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:orientation="vertical"
android:paddingBottom="@dimen/activity_vertical_margin"
android:paddingLeft="@dimen/activity_horizontal_margin"
android:paddingRight="@dimen/activity_horizontal_margin"
android:paddingTop="@dimen/activity_vertical_margin"
tools:context=".MainActivity" >

<TextView
    android:id="@+id/lblLocalPort"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:text="@string/localPort" />

<EditText
    android:id="@+id/editText2"
    android:layout_width="match_parent"
    android:layout_height="wrap_content"
    android:layout_marginTop="20dp"
    android:ems="10"
    android:hint="@string/iPHint"
    android:inputType="phone" />

<EditText
    android:id="@+id/editText1"
    android:layout_width="match_parent"
    android:layout_height="wrap_content"
    android:layout_marginTop="20dp"
    android:ems="10"
    android:hint="@string/portHint"
    android:inputType="number" >

    <requestFocus />
</EditText>

<LinearLayout
    android:layout_width="match_parent"
    android:layout_height="wrap_content" 
    android:layout_marginTop="20dp">

    <Button
        android:id="@+id/button1"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:text="@string/connect" />

    <Button
        android:id="@+id/button2"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:text="@string/Disconnect" />
</LinearLayout>

EDIT: The IP address method stopped working at API 22, use below code:

private byte[] getLocalIPAddress() {   
    byte[] bytes = null;

    try {
        // get the string ip
        WifiManager wm = (WifiManager) getSystemService(WIFI_SERVICE);
        String ip = Formatter.formatIpAddress(wm.getConnectionInfo().getIpAddress());

        // convert to bytes
        InetAddress inetAddress = null;
        try {
            inetAddress = InetAddress.getByName(ip);
        } catch (UnknownHostException e) {
            e.printStackTrace();
        }

        bytes = new byte[0];
        if (inetAddress != null) {
            bytes = inetAddress.getAddress();
        }

    } catch (Exception e) {
        e.printStackTrace();
        Toast.makeText(this, R.string.phone_voip_incompatible, Toast.LENGTH_SHORT).show();
    }

    return bytes;
}
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Anthone
6楼-- · 2019-03-13 22:49

It is not possible because VOIP call Pass through internet and via sip server.

for Example . if you want to call outside from your country via VOIP dailer, you must need internet access because it is not possible to communicate via Bluetooth.

Thanks.

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