When passing the result to bash, I can't create a variable. Why?

Question

bash

There are search criteria, which are then passed to Bash for further processing: creating a file name and converting via FFmpeg.

There is an example file structure.

When I use the find command to find the files I need with their paths and simply ask them to output them, the utility works just perfect.

The result of the find command.

But if you redirect this execution to Bash to, for example, create a variable, edit its name and then pass it to FFmpeg for conversion, something like this happens:

Execution result redirected to Bash.

find ~/Desktop/Serial -type f \( -name "*.mkv" -o -name "*.avi" \) -exec sh -c 'path={}; echo $path' \;

If you simply ask to withdraw a variable from the result of execution, then it works out adequately:

Just echo.

Please tell me how to solve this problem. Thank you.

Answer 1

This problem is solved by simply quoting the assigned variable name.