Docker: Permission denied: '/root/.cache'" about fossology HOT 3 CLOSED

Hi @GMishx,

I've looked into this issue. Currently we are using scancode-wrapper.cc file to construct the command to execute runscanonfiles.py. I checked the permissions of both scripts, and I noticed a change in the user and group for the .cache directory.

This indicates that the fossy user is attempting to create a directory that needs root access. I'm unsure about the ideal solution for this. Any insights would be greatly appreciated. Please take a look.

Thank you!

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Hi @xsuchy / @GMishx / @dvjsharma ,

Setting the following environment variable SCANCODE_CACHE='/home/fossy/.cache fixed the scan code issue for us.

or enforcing in the below.
https://github.com/fossology/fossology/blob/4.4.0/src/scancode/agent/scancode_wrapper.cc#L73

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Thank you @kannan-seenivasan.
I discovered a similar solution during my research. The Scancode toolkit already contains a related issue that closely mirrors ours, where the calling users are not the same (#2355 & #357). To tackle this, they have introduced options and environment variables. Please see CHANGELOG.rst.

I tested the solution by adding the SCANCODE_CACHE environment variable at the container level in the docker-compose.yml file, and it worked perfectly. There was no need to modify scancode_wrapper.cc since the issue did not exist in the local installation (which was also tested). However, even if we were to update it, I believe using the --cache-dir option would be a more suitable approach.

I've submitted a PR to update the docker-compose.yml, which will resolve this issue and #2667. I will make the necessary modifications as advised.

EDIT: It was discussed in community meeting to set the environment variable before calling the runscanonfiles.py file, not in the docker-compose.yml file. Additionally it was decided to store cache at /usr/local/share/fossology/scancode/agent/. I have updated the PR accordingly.

EDIT: It was suggested to set the environment variable inside runscanonfiles.py and not to hardcode the part, but get it inside the script itself (using __file__).

CC @GMishx

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