Please let me know where to fix the problem if possible.
I want to host a personal application on a virtual server.
The server has a window for configuring the python application.
In this window, you need to select the python application launch file, and also specify the entry point to the application.
The test run file for HomeRun.py has the following code:
import os import sys sys.path.insert(0, os.path.dirname(__file__)) def application(environ, start_response): start_response('200 OK', [('Content-Type', 'text/plain')]) message = 'It works!\n' version = 'Python %s\n' % sys.version.split() response = '\n'.join([message, version]) return [response.encode()]
If I specify it as a launch file, and also indicate that the entry point is an application (function), then when I launch the application from the link, I get the message “It works, and the python version”
But, how can I start the streamlit application in this case?
The start page of my application streamlit to be in the Home.py module.
On my home computer, I would need to run the command:
streamlit run Home.py
But I don’t understand how to run this application on the server, because it starts there through the “Run application” button.
I have already used streamlit many times, but all this experience was on the local computer.
I will be glad for your tips.