File paths within my Python script fail when launching via double-click (Windows 10)

Question

I have a python script that reads and writes to files that are located relative to it, in directories above and beside it. When I run my script via Cygwin using

python script.py

The program works perfectly. However, when I run it by navigating through the windows GUI to my file and double clicking, I get a blank cmd prompt and then my program runs fine until I reach the point where I need to access the other files, at which point it fails and gives me this message in the cmd prompt that opens itself:

../FFPRM.TXT                                                                                                             
../2025510296/FFPRM_000.TXT                                                                                             
Exception in Tkinter callback                                                                                           
Traceback (most recent call last):                                                                                        
  File "C:\Users\rbanks\AppData\Local\Programs\Python\Python35-32\lib\tkinter\__init__.py", line 1549, in __call__          
    return self.func(*args)                                                                                     
  File "C:\Users\rbanks\Desktop\TSAC\EXECUTABLE\T-SAC_GUI.py", line 705, in run_exe                                         
    invalid_entry, output_text = self.apply()                                                                              
  File "C:\Users\rbanks\Desktop\TSAC\EXECUTABLE\T-SAC_GUI.py", line 694, in apply                                           
    p = subprocess.Popen(['cp', output_file_path, output_file_path_id])                                                   
  File "C:\Users\rbanks\AppData\Local\Programs\Python\Python35-32\lib\subprocess.py", line 950, in __init__          
    restore_signals, start_new_session)                                                                                   
  File "C:\Users\rbanks\AppData\Local\Programs\Python\Python35-32\lib\subprocess.py", line 1220, in _execute_child          startupinfo)                                                                                                        
FileNotFoundError: [WinError 2] The system cannot find the file specified 

I am deploying this script as well as the directory structure as a zip for users to be able to unzip and use anywhere on their PC, so it is important for me to be able to run it with a simple double click and my relative file paths.

My first thought was the cmd prompt that was opening and executing my script was in a different environment, but when I run:

cd
pause

in a .cmd script, I get:

C:\Users\rbanks\Desktop\TSAC\EXECUTABLE>pause 

Which is the correct location.

I am not having any luck with Google, I assume because I can't seem to construct a sufficient search query. Could someone point me in the right direction please?

Answer 1

[edit] the other answer is correct(at least i suspect) but i will leave this here in the hopes that it helps the op in the future with path problems ... and doing something like this is just generally good practice

---

use

BASEPATH = os.path.abspath(os.path.dirname(__file__))

at the top of your script

the later

txt_file = os.path.join(BASEPATH,"my_file.txt")

or even

txt_file = os.path.join(BASEPATH,"..","my_file.txt")

this gives you the benefit of being able to do things like

if not os.path.exists(txt_file):
   print "Cannot find file: %r"%txt_file

which will likely give you a better idea about what your problem might actually be (if its simply path related at least)