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I am trying to run pyvtt with its default configuration in a throwaway, private LXC guest (so no name, no DNS configuration whatsoever, just a local IP address). From inside the LXC:
root@bullseye-bl35UhyVrM:~# ip -br a sh dev eth0
eth0@if8 UP 192.168.155.120/24 fe80::7c91:7fff:fea4:201b/64
root@bullseye-bl35UhyVrM:~# getent hosts 192.168.155.120
root@bullseye-bl35UhyVrM:~#
I access the app as http://192.168.155.120:8080 from my local browser. However, when trying to log in, the app gives me ws://localhost:8080 as a websocket URL, which the browser can't connect to.
From my understanding of the code, the URL is generated as something like "localhost by default, unless I can find a better value"; this corner case does not seem to be covered.
The text was updated successfully, but these errors were encountered:
if you're running the VTT in your LAN, you need to tell the VTT which IP to use to generate such URLs (because it cannot tell whether you're going public on the internet, private on your LAN or super-private for localhost-testing).
Check your settings.json (defaults at ~/.local/share/pyvtt/) and use your LAN-IP as domain inside the hosting-sections.
Yes, thanks! I was hoping to get something a bit more automagic, maybe generated from the initial HTTP request, but I understood after posting that the websocket URL is global and not per-request.
Hello,
I am trying to run pyvtt with its default configuration in a throwaway, private LXC guest (so no name, no DNS configuration whatsoever, just a local IP address). From inside the LXC:
I access the app as
http://192.168.155.120:8080
from my local browser. However, when trying to log in, the app gives mews://localhost:8080
as a websocket URL, which the browser can't connect to.From my understanding of the code, the URL is generated as something like "localhost by default, unless I can find a better value"; this corner case does not seem to be covered.
The text was updated successfully, but these errors were encountered: