title: Pass a file to Python via the Right-Click 'Open With' context menu in windows, to then open in another program
body: <p>I use a piece of software that when you close, saves your current configuration, however, next time I open the software by clicking on a file associated with it, it tries to run that file through the same configuration, which I don't want. I have therefore created a python script to first open up the app data for that program, reset the configuration back to default, and then open the program again. </p>
  
  <p>This works fine if I just try to load the program without a file, but I would like to be able to click on a file associated with that program (usually I can just double click on the file to open it in the program), then use the 'open with' function in windows to select my script, and then open the file in the program once my script has cleared out the last configuration.
  Currently, if I do this, it clears the configuration and opens the program but with no file loaded.</p>
  
  <p>Essentially I am asking how I can pass the file to python, and then get python to pass that file to the third party application.</p>
  
  <p>I am using the 'os.startfile('link to .exe') function to open the program, but how do I get the file path into python as an argument/string by clicking on the file and opening it with my script?</p>
  
  <pre><code>path = 'path/to/file/selected'  # passed to python from selecting the file in windows explorer before starting script
  
  # execute some code
  
  os.startfile('path')
  </code></pre>
  
  <p>I'm a bit of a beginner when it comes to python so any help would be appreciated!</p>
  
  <p>Using python 3.6 on windows 10</p>