Can't Open SLEAP Project After Crashing During Saving

[PrestonSheng]

Hello,

I had a SLEAP project crash while it was saving, and I haven't been able to open it again. Here is the error in my command line:

C:/Users/Abdus-Saboor Lab/Desktop/SLEAP Videos for PRESTON/Project_Fixed2.slp doesn't match ext for json or json.zip
Traceback (most recent call last):
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\gui\commands.py", line 287, in openProject
self.execute(OpenProject, filename=filename, first_open=first_open)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\gui\commands.py", line 241, in execute
command().execute(context=self, params=kwargs)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\gui\commands.py", line 137, in execute
self.do_with_signal(context, params)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\gui\commands.py", line 161, in do_with_signal
cls.do_action(context, params)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\gui\commands.py", line 725, in do_action
context.loadProjectFile(filename)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\gui\commands.py", line 272, in loadProjectFile
self.execute(LoadProjectFile, filename=filename)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\gui\commands.py", line 241, in execute
command().execute(context=self, params=kwargs)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\gui\commands.py", line 135, in execute
okay = self.ask(context, params)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\gui\commands.py", line 696, in ask
labels = Labels.load_file(filename, video_search=gui_video_callback)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\io\dataset.py", line 1969, in load_file
filename, for_object="labels", video_search=video_search, *args, **kwargs
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\io\format\main.py", line 113, in read
return disp.read(filename, *args, **kwargs)
File "C:\ProgramData\miniconda3\envs\sleap\lib\site-packages\sleap\io\format\dispatch.py", line 59, in read
raise TypeError("No file format adaptor could read this file.")
TypeError: No file format adaptor could read this file.

Is there anything that I can do to fix this? Thank you all so much!

[gitttt-1234]

Hi @PrestonSheng !

Which sleap version are you using? Could you share your slp file here (with associated videos or you can export it as a pkg.slp from the gui and share it instead of uploading all videos) so that we could debug better from our side?

Thanks,

Divya

[PrestonSheng]

Hi Divya,

I can send in my SLEAP file, but there are many videos and would be difficult to share. Currently, it won't open in the GUI at all unfortunately, so I don't even get to the part where it accesses the videos in the project. Uploading now though!

I am using SLEAP version 1.4.1a2.

Preston

[gitttt-1234]

Hi @PrestonSheng !

Apologies for the delay. I tried opening the file directly using the h5 package, but it couldn’t be read — this isn’t an issue with the SLEAP frontend. It appears the file may have become corrupted, possibly during transfer from another device. If you have access to an earlier or original version of the file, that would be helpful.

Thanks,

Divya