Jamovi 2.6.26 & 2.6.44 Fail to launch on Azure VM

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Quint
Posts: 11
Joined: Wed Sep 14, 2022 1:53 pm

Jamovi 2.6.26 & 2.6.44 Fail to launch on Azure VM

Post by Quint »

Hello,

I'm installing Jamovi 2.6.26 & 2.6.44 onto Azure VMs and the most resent versions fail to launch.
The starting Jamovi splash screen comes up but animation just keeps going and doesn't launch the program.

If I install version 2.3.28, it works fine.

The debug output is:

C:\Program Files\jamovi 2.6.44.0\bin>
jamovi.exe -debug
**********************************************
C:\Program Files\jamovi 2.6.44.0\bin>
C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\utils\latexify.py:394: SyntaxWarning: invalid escape sequence '\S'
wrps = [wrps[0] + '\\colorbox{' + re.findall('background-color:(\S*?)[;"]', cspc)[0] + '}{', wrps[1] + '}']
C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\utils\latexify.py:395: SyntaxWarning: invalid escape sequence '\S'
cspc = re.sub('background-color:\S*?[;"]', '"', cspc)
C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\utils\latexify.py:397: SyntaxWarning: invalid escape sequence '\S'
wrps = [wrps[0] + '\\textcolor{' + re.findall('color:(\S*?)[;"]', cspc)[0] + '}{', wrps[1] + '}']
C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\utils\latexify.py:398: SyntaxWarning: invalid escape sequence '\S'
cspc = re.sub('="color:\S*?[;"]', '=""', cspc)
C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\utils\latexify.py:399: SyntaxWarning: invalid escape sequence '\S'
cspc = wrps[0] + re.findall('<span style="*?>([\S\s]*?)<\/span>', cspc)[0] + wrps[1]

C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\__main__.py:121: DeprecationWarning: There is no current event loop
loop = get_event_loop()

jamovi
version: 0.0.0
cli: 0.0.0

Task exception was never retrieved
future: <Task finished name='Task-2' coro=<Server._run() done, defined at C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\server.py:720> exception=FileExistsError(17, 'Cannot create a file when that file already exists')>
Traceback (most recent call last):
File "C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\server.py", line 735, in _run
await self._session.start()
File "C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\session.py", line 104, in start
await self._modules.read()
File "C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\modules\modules.py", line 225, in read
await self._reread_installed()
File "C:\Program Files\jamovi 2.6.44.0\Resources\server\jamovi\server\modules\modules.py", line 339, in _reread_installed
os.makedirs(user_module_path, exist_ok=True)
File "<frozen os>", line 225, in makedirs
FileExistsError: [WinError 183] Cannot create a file when that file already exists: 'C:/Users/Myusername/AppData/Roaming/jamovi\\modules'
**********************************************

If I delete that module's folder (in the line above) and launch the program, it works. If I run the program again, I get the message saying cannot create a file when that file already exists.

These are shared VMs in Azure for student lab usage. We are using FSlogix profiles which are basically roaming profiles.

Thoughts on how to fix this?
User avatar
jonathon
Posts: 2856
Joined: Fri Jan 27, 2017 10:04 am

Re: Jamovi 2.6.26 & 2.6.44 Fail to launch on Azure VM

Post by jonathon »

hi,

this is quite unusual. according to the error (and i'm not saying the error makes any sense) there is a *file* (rather than a directory) called 'modules' at that location.

jonathon
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