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I have been developing an application using Visual Studio Code and running it via Visual Studio Code's built-in terminal. My application uses texture loading via the arcade library.

Example:

texture_a = arcade.load_texture("Version 1\Images\texture_a.png")
  • The script that loads the files is "Version1\Main_Files\Graphics.py"
  • The script that I run is "Version1\Main.py"
  • I am not sure if that is useful. I am fairly experienced with Python, but I don't really understand the nuances of paths.

My application works fine/ loads and displays the textures when using Visual Studio Code, but when I try to run the Python file from the command line, I get a "Cannot locate resource" error. Why does the path work when running from Visual Studio Code terminal, but doesn't work from the normal terminal?

I am using Windows 11.

righteousness_wealth
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Wills
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    This is usually a problem with the so-called working directory. VS Code may run your script with a different working directory than the path you start the script from in the terminal. Try importing `os` at the start of your script and adding something like `print(os.getcwd())`, and compare what you see in both cases. – Grismar Aug 20 '23 at 22:43
  • This *sounds* like a duplicate. – Peter Mortensen Aug 20 '23 at 22:44
  • @Grismar Thank you. They are different: one is in C:\Users\myName, and the other is is W:\ Code Files\ ApplicationName. How does this effect how python interprets the path? How might I be able to fix it? – Wills Aug 20 '23 at 22:47
  • Is `"Version 1\Images\texture_a.png"` a file path? Is this code you wrote or from a library? That path is relative to the current working directory, and thats different in your two cases. Its common to have a standard directory where you put the files you need. If they are resources for the program, maybe in the same place as .py files. Or perhaps in your Documents folder. Maybe its something you pass into the program from the command line. – tdelaney Aug 20 '23 at 23:03
  • Thank you. That is very helpful. I can definitely make the paths absolute, but I host my project on git hub and would like for the app to work on different devices. Is there a way to change the directory from within my program. – Wills Aug 20 '23 at 23:12
  • See the answer I just posted - that also shows how to change the directory from within your script. Keep in mind that other devices may have different directory structures. You may want to make the working directory configurable at least. – Grismar Aug 20 '23 at 23:12

1 Answers1

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This is usually a problem with the so-called working directory. VS Code may run your script with a different working directory than the path you start the script from in the terminal. Try importing os at the start of your script and adding something like print(os.getcwd()), and compare what you see in both cases.

In your case, you see C:\Users\myName and W:\Code Files\ApplicationName. What does that mean?

Any program (across OSes even) has a working directory. All relative paths are relative to that working directory. For example, if you open('test.txt') it will be opened in that directory. Only if you specify an absolute path (like open('C:\Temp\test.txt')) does the working directory not matter.

In your case, if you need to open Version 1\Images\texture_a.png, you have some options:

  1. Accept that this is relative to where the user of your script starts the script; this can be useful, for example if your script was designed to operate on some files in a directory of a user's choosing.

  2. Provide an absolute path, if you need that image to be in a specific location, regardless of where the script is, or where it was started. For example C:\Users\myName\Version 1\Images\texture_a.png instead of relying on that being the working directory.

  3. Change the working directory to where the script should be running, so you can keep using the same relative paths: os.chdir('C:\Users\myName') before opening any files, so that this will be the working directory (assuming you can be sure it exists).

You can also find out where the script itself is (with something like pathlib.Path(__file__).parent) and use that as a prefix to your paths, or as the working directory for files you open.

Grismar
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  • For those that find it useful reading this later, I implemented this for the general case with: ' `import os / import pathlib / PATH = pathlib.Path(__file__).parent.parent.parent / os.chdir(PATH) ` – Wills Aug 21 '23 at 00:17