kandi X-RAY | voicechat Summary
kandi X-RAY | voicechat Summary
VoiceChat is a simple VoIP application written in Java that supports uPnP, conversations with multiple users and basic compression with comfort noise.
Top functions reviewed by kandi - BETA
- Initialize the components .
- Runs the loop
- Invoked when the JButton1 action is clicked .
- Get external IP address
- Add a message to the broadcast queue .
- Adds a new client connection .
- Returns true if the packet can be killed .
- Gets internal IP address .
- Print message dialog .
- Sets the id of the channel .
voicechat Key Features
voicechat Examples and Code Snippets
Trending Discussions on voicechat
I am maintaining a Push-to-talk VoIP app. When a PTT call is running the app create an audio session...
ANSWERAnswered 2021-Mar-16 at 15:25
I hope you are testing this on an actual device and not a simulator.
In the code, have you tried using this:
I would like to know how to kick/disconnect someone from the voice channel the user is in, or a specific channel. I learned how to kick someone and here is my code:...
ANSWERAnswered 2021-Apr-04 at 20:53
Working on making a bot for a personal discord server, wanting it to simply join general voicechat when command "Blyat" is detected, play mp3 file, when its done to leave. Code:...
ANSWERAnswered 2021-Mar-12 at 02:12
Your problem comes from an unidentified variable. Fortunatly that is very easy to fix. All you have to do is define
command before you call it.
For this we'll splice the message content into two parts. The command and any arguments that may be included like mentions or other words. We do that with:
i am having trouble while developing web application using spring boot
i am applying spring security and having errors i dont know how to solve the error im also new at spring boot and first time using spring security my credentials are not been accessed by the compiler and having issues
this is my dao class...
ANSWERAnswered 2021-Mar-03 at 10:39
You have a very strange return type in your
I have a swift project that uses the GoogleWebRTC pod.
When trying to negotiate the OPUS codec for audio calls i find that the peer connection is successfully setup, however i am experiencing one way audio. SRTP is being sent from my iPhone to the other party successfully, and SRTP is being sent from the other party to my iPhone, however my phone/app is not playing the incoming SRTP to the user. If i negotiate any other codec (G722 for example) then i get 2 way audio, it's just when i try to negotiate OPUS that i don't hear any incoming audio on my iPhone.
Couldn't see anything relevant in the logs, but looking for some pointers on how to troubleshoot this or what could potentially be the cause of this issue.
I'm using the google WebRTC iOS SDK.
Here is the code in my webrtc class where i initialize the audio session if that helps....
ANSWERAnswered 2021-Feb-19 at 13:45
For anybody else who stumbles across this, I wasn't using the audiosession provided by callkit in the didActivate method of the callprovider protocol.
Here's my amended configureAudioSession
I want my bot to send a message if im not in a voice channel when i type a command.
Heres my current code:...
ANSWERAnswered 2020-Nov-07 at 08:54
Member.voice will be None, you need to check that
Below is the revised code:
I am trying to mute a user in a certain voice channel without creating a role "Muted" for it. Here is my code:...
ANSWERAnswered 2020-Jun-10 at 21:04
https://discordpy.readthedocs.io/en/latest/api.html?highlight=discord%20utils#discord.Member.edit You can use .edit(mute = True) to mute a user you need.
I'm using AUAudioUnit to play audio that the app is streaming from the server. My code works fine in the foreground. But when I background the app, it won't play the audio. I got the following error.
[aurioc] AURemoteIO.cpp:1590:Start: AUIOClient_StartIO failed (561145187)
The error code 561145187 means
This error type usually occurs when an app starts a mixable recording from the background and it isn’t configured as an Inter-App Audio app.
This is how I set up the AVAudioSession in Swift:...
ANSWERAnswered 2020-Apr-21 at 15:32
To record audio in the background, you have to start the audio unit in the foreground. Then the audio unit will continue to run in the background. If you don't have any data to play, you can play silence.
If you are not recording audio, then don't use the playAndRecord session type. Use one of the play-only session types instead.
I'm trying install a tap on the output audio that is played on my app. I have no issue catching buffer from microphone input, but when it comes to catch sound that it goes trough the speaker or the earpiece or whatever the output device is, it does not succeed. Am I missing something?
In my example I'm trying to catch the audio buffer from an audio file that an AVPLayer is playing. But let's pretend I don't have access directly to the AVPlayer instance.
The goal is to perform Speech Recognition on an audio stream....
ANSWERAnswered 2020-May-06 at 14:47
I was facing the same problem and during 2 days of brainstorming found the following.
Apple says that For AVAudioOutputNode, tap format must be specified as nil. I'm not sure that it's important but in my case, that finally worked, format was nil. You need to start recording and don't forget to stop it.
Removing tap is really important, otherwise you will have file that you can't open.
Try to save the file with the same audio settings that you used in source file.
Here's my code that finally worked. It was partly taken from this question Saving Audio After Effect in iOS.
I have a TCP Server and Client in Java. The Server can send commands to the Client, the Client will then execute the command, for example: send an image to the Server.
Im sending the data with a bytearray and thats working.
But lets imagine, I want to send an image and a file separately. How would the Server supposed to know which is the right bytearray? Or if I want to make a VoiceChat (which needs to be sending bytearrays continously) and separately sending an image?
Thats my code send bytes:
ANSWERAnswered 2020-Apr-10 at 12:38
You need to design a "protocol" for the communication. A protocol defines what are the messages that can be exchanged and how they are represented in the lower level data stream.
A quick and easy protocol is where you first send the length of the data you are going to send, and then the data:
No vulnerabilities reported
Reuse Trending Solutions
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page