room does not work

[andresbelt]

how the room works?

if it's a different room the video is shown

example:

room : 1 presenter show video

room: 2 you see the video of room 1 Call not accepted for the following reason: Another user is currently acting as presenter. Try again later ...

[ayushjn20]

A room contains single presenter/broadcaster and multiple viewers. One can create one's own room and start broadcasting, and others can join the room to view it; or create a room and wait as a viewer for someone to start broadcasting as a presenter

[alienmaster9]

all rooms getting same video

[ayushjn20]

After running the code I realised that there is no event that fires the name of the room from client to the server, and thus the server takes the room parameter undefined in all the cases.You can do this: Instead of a form field, use window.prompt in the first line of js, and ask the user to input a room name, send the name to the server as 'joinRoom' event using socket.emit and then continue. Also, on the server side, insert a check statement for room name.

[dflm25]

Same error, all rooms getting same video someone has a solution for this issue

[AkashNagina]

Now able to generate multiple presenters by creating random rooms but how viewer can see streaming of specific room?

[AkashNagina]

Worked fine, Here we made some changes: 1-Followed approach of @ayushjn20 2-server.js: data.room to data, data.username to data.

[ghost]

Hi ayushjn20, instead of a form field, I am taken window.prompt() And I am sending my room name to the server using socket.emit, it is worked fine for me thank you for your Solution