Using iTunes/Airplay for streaming live audio/video on local network?
I'm wondering if I can use iTunes to stream live audio and/or video to other computers, Airport Express, or Apple TVs on a local wifi network. I'm willing to jailbreak or hack anything if necessary.
So this is what I am hoping to do:
Camera-->Streaming Computer-->Airplay-->Local Wi-fi network-->Devices recieving audio/video
This device should be doing UPnP by default so you will not need most of the info on this page.
What the Link does have is the default access info and pictures.
In the Wan page turn Off the DOS or SPI firewall. (it will have one or the other)
Check in the Box to Disable.
Save Settings.
If this device is behind a Cable modem then also Allow it to Respond to Internet Pings on the same page.
(this a Check the Box to Do)
Save settings (Again if you did them above)
7:30 PM Monday; April 28, 2008
Similar Messages
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i have to transmit Captured Audio and video ..
here my steps
a)capture audio .. create a datasource for capturing device
b)capture video.. create a datasource for capturing device
c)merge both the datasource to get merged datasource
Still this i finished
now i want to transmit the media using merged datasource.. is it possible using AVtransmit2
if yes, how to do this?
help menow i want to transmit the media using merged datasource.. is it possible using AVtransmit2Why do you want to transmit the merged datasource? The RTP protocol requires that audio and video streams be transmitted separately, so you'd essentially be merging the video & audio only to have them sent in separate RTP streams, and received as separate RTP streams...
if yes, how to do this?You'd transmit a merged datasource just like any other datasource... -
Live Audio / Video Streaming Very Basic
I need to Stream Live Audio and Video, i went through a web site, and it gives a code,
[http://www.cs.odu.edu/~cs778/spring04/lectures/jmfsolutions/examplesindex.html#transmitboth]
in this, i dont no how to run the application, because we should have 2 port addresses each for Audio and Video.
how do u compile this and capture the file from else where using JMStudio.
Please help me, i am an absolute beginner to JMF.Please don't take this to be offensive, but if you're not able to figure this out on your own, you have absolutely no business playing around with something as advanced as JMF. It's not a question of intelligence or insult, but simply stated, if you don't understand the concept of a URL, and you don't have the ability to extrapolate beyond the exact command-line input you've been given... you need to go and learn the basics, because you lack the grasp of the fundamentals required for advanced programming.
With that in mind, the following is the answer to your question. If you can't understand it, it means that you lack the fundamentals necessary for JMF programmming, and you need to invest a significant amont of time aquiring those. My explanation should be quite clear to anyone with the proper Java programming fundamentals.
AVTransmit2 is sample code that can broadcast a single media source (live audio or live video or audio file or video file or video/audio file). It does not have the capability to broadcast more than once source, which is required for live audio and video support. It is designed to take in a single media locator, and broastcast it.
To meet your specifications, you will need to modify the main method so it is capable of passing in multiple media locators, and thus creating multiple instances of the AVTransmit2 class. To do this, you will either need to modify the command-line argument logic so it supports parsing multiple source arguments simultaniously.
Or, you could just rip out the command-line stuff and hard-code it for live audio and video. That's the "easy" way.
The default media locator for audio capture is javasound://0 and the default media locator for video capture (under Windows) is vfw://0 -
I live in Iran
Yesterday I wanted to use iTunes store for windows ; when i have open iTunes it gives me the error 0x80090318 .
I have search the web alot and i have check lots of Iranian and English website but have not find any solution for
this problem . Today I have find a software called "Hide my IP" i have opened up that
when this software was opened , itunes have said "iTunes can't verify the identify of the server "init.itunes.apple.com"".
then i have click on "continue" then iTunes said "iTunes could not connect to the iTunes Store. An unknown error occurred (310)."
And theres a matter that i think that have a connection with this topic and thats , that when i want to go to the lots
of websites like Google Yahoo! Mail and lots of other website my both browser ( FireFox and IE ) will say" This connection is untrusted " .
I think the solution of my second problem is the solution of my first problem too !Tried all the rest on this site did not work for me
Follow the link below & this one sorted it for me, it was a conflict in programs using Winsock & I got rid of my old program,
Sorted.
http://support.apple.com/kb/TS4123
Tom Ireland -
NO audio using Apple TV for streaming materials, and YES audio for Mac stored ones; what gives?
Hello Cravate,
The following article provdies some useful troubleshooting that can help resolve audio issues with your Apple TV.
Apple TV (2nd and 3rd generation): Troubleshooting audio issues
http://support.apple.com/kb/TS5150
Cheers,
Allen -
Please, help with Live Audio/Video example from jmf solutions
Hello,
I�m desperate looking for a solution for a particular problem.
I�m trying to feed JMF with an AudioInputStream generated via Java Sound, so that I can send it via RTP. The problem is that I don�t know how to properly create a DataSource from an InputStream. I know the example Live Audio/Video Data from the jmf solutions focuses on something similar.
The problem is that I don�t know exactly how it works, os, the question is, how can I modify that example in order to use it and try to create a proper DataSource from the AudioInputStream, and then try to send it via RTP?
I think that I manage to create a DataSource and pass it to the class AVTransmit2 from the jmf examples, and from that DataSource create a processor, which creates successfully, and then find a corresponding format and try to send it, but when i try to send it or play it I get garbage sound, so I�m not really sure whether I create the DataSource correctly or not, as I�ve made some changes on the Live Audio/Video Data from the jmf solutions to construct a livestream from the audioinputstream. Actually, I don�t understand where in the code does it construct the DataSource from the livestream, from an inputStream, because there�s not constructor like this DataSource(InputStream) neither nothing similar.
Please help me as I�m getting very stuck with this, I would really appreciate your help,
thanks for your time, bye.import javax.media.*;
import javax.media.format.*;
import javax.media.protocol.*;
import java.io.IOException;
import javax.sound.sampled.AudioInputStream;
public class LiveAudioStream implements PushBufferStream, Runnable {
protected ContentDescriptor cd = new ContentDescriptor(ContentDescriptor.RAW);
protected int maxDataLength;
protected int vez = 0;
protected AudioInputStream data;
public AudioInputStream audioStream;
protected byte[] audioBuffer;
protected javax.media.format.AudioFormat audioFormat;
protected boolean started;
protected Thread thread;
protected float frameRate = 20f;
protected BufferTransferHandler transferHandler;
protected Control [] controls = new Control[0];
public LiveAudioStream(byte[] audioBuf) {
audioBuffer = audioBuf;
audioFormat = new AudioFormat(AudioFormat.ULAW,
8000.0,
8,
1,
Format.NOT_SPECIFIED,
AudioFormat.SIGNED,
8,
Format.NOT_SPECIFIED,
Format.byteArray);
maxDataLength = 40764;
thread = new Thread(this);
* SourceStream
public ContentDescriptor getContentDescriptor() {
return cd;
public long getContentLength() {
return LENGTH_UNKNOWN;
public boolean endOfStream() {
return false;
* PushBufferStream
int seqNo = 0;
double freq = 2.0;
public Format getFormat() {
return audioFormat;
public void read(Buffer buffer) throws IOException {
synchronized (this) {
Object outdata = buffer.getData();
if (outdata == null || !(outdata.getClass() == Format.byteArray) ||
((byte[])outdata).length < maxDataLength) {
outdata = new byte[maxDataLength];
buffer.setData(audioBuffer);
buffer.setFormat( audioFormat );
buffer.setTimeStamp( 1000000000 / 8 );
buffer.setSequenceNumber( seqNo );
buffer.setLength(maxDataLength);
buffer.setFlags(0);
buffer.setHeader( null );
seqNo++;
public void setTransferHandler(BufferTransferHandler transferHandler) {
synchronized (this) {
this.transferHandler = transferHandler;
notifyAll();
void start(boolean started) {
synchronized ( this ) {
this.started = started;
if (started && !thread.isAlive()) {
thread = new Thread(this);
thread.start();
notifyAll();
* Runnable
public void run() {
while (started) {
synchronized (this) {
while (transferHandler == null && started) {
try {
wait(1000);
} catch (InterruptedException ie) {
} // while
if (started && transferHandler != null) {
transferHandler.transferData(this);
try {
Thread.currentThread().sleep( 10 );
} catch (InterruptedException ise) {
} // while (started)
} // run
// Controls
public Object [] getControls() {
return controls;
public Object getControl(String controlType) {
try {
Class cls = Class.forName(controlType);
Object cs[] = getControls();
for (int i = 0; i < cs.length; i++) {
if (cls.isInstance(cs))
return cs[i];
return null;
} catch (Exception e) { // no such controlType or such control
return null;
and the other one, the DataSource,
import javax.media.Time;
import javax.media.protocol.*;
import java.io.IOException;
import java.io.InputStream;
import javax.sound.sampled.AudioInputStream;
public class CustomDataSource extends PushBufferDataSource {
protected Object [] controls = new Object[0];
protected boolean started = false;
protected String contentType = "raw";
protected boolean connected = false;
protected Time duration = DURATION_UNKNOWN;
protected LiveAudioStream [] streams = null;
protected LiveAudioStream stream = null;
public CustomDataSource(LiveAudioStream ls) {
streams = new LiveAudioStream[1];
stream = streams[0]= ls;
public String getContentType() {
if (!connected){
System.err.println("Error: DataSource not connected");
return null;
return contentType;
public byte[] getData() {
return stream.audioBuffer;
public void connect() throws IOException {
if (connected)
return;
connected = true;
public void disconnect() {
try {
if (started)
stop();
} catch (IOException e) {}
connected = false;
public void start() throws IOException {
// we need to throw error if connect() has not been called
if (!connected)
throw new java.lang.Error("DataSource must be connected before it can be started");
if (started)
return;
started = true;
stream.start(true);
public void stop() throws IOException {
if ((!connected) || (!started))
return;
started = false;
stream.start(false);
public Object [] getControls() {
return controls;
public Object getControl(String controlType) {
try {
Class cls = Class.forName(controlType);
Object cs[] = getControls();
for (int i = 0; i < cs.length; i++) {
if (cls.isInstance(cs))
return cs[i];
return null;
} catch (Exception e) { // no such controlType or such control
return null;
public Time getDuration() {
return duration;
public PushBufferStream [] getStreams() {
return streams;
hope this helps -
Can i make Live Audio/Video application between 2 users
Hello,
I am new to FLASH and FLEX.
I want to know if i can make a Live Audio/Video application(
using Microphone and Camera) for a website by using FMS. If yes
then should i use FMSS or FMIS. I will be using FLEX Buidler IDE .
Any one who has made this type of application or link to a
tutorial.
What i would like to make is like an application of Webcam
with 2 users can see/view each other and also talk on web site. And
alos how can i embed this application in java(EE) project.
I would be very thankful if you people can guide me in this
problem.
Hopefully i have explained my probelm.
Regards,
SaadYes you can make a Live A/V app with FMS! that is exactly
what it was designed for. You would need FMIS as that is the
interactive version that enables live capabilities. -
Problem with running example 'Generating Live Audio/Video Data'
Hello,
Concerning the example 'Generating Live Audio/Video Data', I'm having trouble with the run instructions.
http://java.sun.com/javase/technologies/desktop/media/jmf/2.1.1/solutions/LiveData.html
How does JMFRegistry know about the location of jmfsample?
How is 'live' resolved as a URL?
2.Register the package prefix for the new data source using JMFRegistry
- Run JMFRegistry
- In the Protocol Prefix List section add "jmfsample" and hit Commit.
4.Select File->Open URL and enter "live:"
Much thanks,
BenI'm getting the following error message: "Could not create player for live"Implies you've either not registered the "live:" protocol prefix in the JMF Registry, or, it couldn't load the class you registered for it...or it might be erroring out inside the actual live protocol, I'm not sure what that would look like, but a System.err.println statement in the constructor of both of those classes might be a good idea.
I added the output of javac (DataSource.class and LiveStream.class) to a directory on the classpath.
C:\Program Files\JMF2.1.1e\lib\jmfsample\media\protocol\liveEh, that looks a little questionable to me. I'm not 100% sure that the JRE will automaticlly descend into the package subdirectories like that, looking for classfiles, for every folder on the path. I am, of course, fully open to the idea that it does and I just never thought about it...but I guess I just thought it only did that for JAR files, not CLASS files. Regardless, I'd recommend:
1) Make sure you've registered the protocol prefix "live:" correctly in JMF Registry
2) Try to run it with the 2 compiled class files in the same folder as your project
3) Try to run it with the 2 compiled class files in the lib directory, if that's on the classpath
4) Try to run it with the 2 compiled class files installed in the JRE as an extension (google for how to do this because I don't remember off the top of my head)
5) Reinstall JMF and see if that helps -
I'm trying to stream live desktop video to a remote client. I can stream for about 3 seconds then it gives me a memory error saying that there isn't enough memory... Is there a way to flush the memory instead of trying to retain it? Becuase i'm sure if I could just flush the data after I transmit it, the memory error would go away. I'm using the datasource.java and livestream.java from the screen grabber example. Oh by the way i'm transmitting at 30 fps thats why it takes so little time for memory to run out. Anyway thanks in advance, Dana Garcia
Edited by: Dana_Garcia on Sep 10, 2009 8:17 AM
Edited by: Dana_Garcia on Sep 10, 2009 8:18 AMThat's stange, I wouldn't expect that to happen if you're streaming.
Can you post your code? No need to include the 2 sample files... -
Bandwidth required for Lync 2013 audio video call
Hi,
what is the required bandwidth required for lync 2013 audio video call from out side of company network. considering users will use owa integration with exchange 2013 for audio video call.
i tried with lync bandwidth calculator but could not figure it out.
actually i have some remote site where users have connectivity of 160 kbps only.There are a number of variables such as call type and video resolution, but I would suggest using this table as a guide:
http://technet.microsoft.com/en-us/library/jj688118.aspx
For example for capacity purposes with a Lync peer-to-peer call you're looking at 57Kbps (86 with FEC)
If this helped you please click "Vote As Helpful" if it answered your question please click "Mark As Answer" | Blog
www.lynced.com.au | Twitter
@imlynced -
I am unable to use iTunes credit for game credit purchase. A box comes up with " an error has occurred Please try later" any advice greatly appreciate. Thanks
mgpainter wrote :
prince2012 wrote:
Hi mgpainter
Welcome to the Community
Sorry I am not a RIM employee, But still regarding your problem have a look at this RIM Knowledge Base to resolve the error .Try it and see if it resolves your problem : KB25882 .
Prince
Click " Like " if you want to Thank someone.
If Problem Resolves mark the post(s) as " Solution ", so that other can make use of it.
Thanks for the welcome and the reply!
I looked at KB25882 and my settings are good, so that didn't help.
I had this problem over a year ago, and someone at RIM found a bit set that shouldn't have been set and fixed it (that is my terminology for what happened).
I am really looking for a way to contact RIM directly. I hate going through the mill with my carrier to get it escalated to RIM for resolution...... <sigh>
Hi mgpainter
Most easiest way to Contact RIM is through your Carrier but as you don't prefer that than there is another way of support called Incident Based Support but just want to inform you that its not free.If you want have a try BlackBerry Incident Based Support .
Good Luck
Prince
Click " Like " if you want to Thank someone.
If Problem Resolves mark the post(s) as " Solution ", so that other can make use of it.
Click " Like " if you want to Thank someone.
If Problem Resolves mark the post(s) as " Solution ", so that other can make use of it. -
I have been using Itunes Apps for some time. Today I signed to Itunes to get Apps updated. Apple said I must update my account. I did that with a new password. Itunes said invalid support. Now what can I do?
What do you mean "invalid support"? What EXACTLY did it say? Did it say you had an unsupported SIM after updating the firmware on your phone?
-
Our computer crashed that we mainly used iTunes on for upgrades, new purchases, etc. We've downloaded iTunes onto my laptop but none of my apps are in the library. We've made this laptop an authorized computer, but it's still not working. Why is this?
See Recover your iTunes library from your iPod or iOS device.
tt2 -
I downloaded Itunes 10.7 and my audio/video does't work can anybody help me
Could you walk us through the problem in some more detail, please?
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Can you still use your iPhone for apps, music, and videos after it is disconnected from your carrier?
Using an iPhone without a wireless service plan
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