VSC fails to connect to (any) python kernel.

After the upgrade to the latest VSC version on Linux Ubuntu 20.04 I cannot run any jupyter notebook. When I run any cell, using any environment (that used to work till the previous version), VSC seems to fail connecting to the python kernel, a dialog shows up and says:

Connecting to kernel: base (Python 3.9.5): Activating Python Environment 'Python 3.9.5 ('base')'

It never disappears.

Environment data

• VS Code version:

Version: 1.69.0
Commit: 92d25e35d9bf1a6b16f7d0758f25d48ace11e5b9
Date: 2022-07-07T08:29:47.439Z
Electron: 18.3.5
Chromium: 100.0.4896.160
Node.js: 16.13.2
V8: 10.0.139.17-electron.0
OS: Linux x64 5.13.0-52-generic

• Jupyter Extension version (available under the Extensions sidebar):

v2022.6.1001902341

• Python Extension version (available under the Extensions sidebar):

v2022.10.0

• OS (Windows | Mac | Linux distro) and version:

Linux Ubuntu 20.04

• Python and/or Anaconda version:

conda 4.11.0

• Type of virtual environment used (N/A | venv | virtualenv | conda | …):

conda

• Jupyter server running: Local | Remote | N/A

local

Expected behaviour

Cell is run.

Actual behaviour

Cell is never run, for every conda environment and kernel I can select, I read:

Connecting to kernel: base (Python 3.9.5): Activating Python Environment 'Python 3.9.5 ('base')'

Steps to reproduce:

In my case it is enough to open any jupyter notebook, select any environemnt and run any cell.

[amitbha]

maybe because Jupyter extension is loaded before the Python extension is ready. I reload the jupyter extension, it works

1. disable jupyter extension
2. restart vscode
3. open the .ipynb file
4. enable jupyter extension
5. run all

[DonJayamanne]

Closing this issue as its been over 4 weeks, since the information was requested. We’ll be happy to reopen the issue when the requested information has been provided.