python program doesn't work when opened by run

Question

I'm using Windows 7 x64, when I open my program by Windows - run it does not work properly. It stars, but the commands does not work the way they do, when I double click it.

/run cmd /c start "" "C:\Python27\Scripts\bot.bat"
/run cmd /c start python "C:\Python27\Scripts\bot.py"
/run python "C:\Python27\Scripts\bot.py"

I tried these and all of them, failed. While a simple double click on the .bat file or the .py work.

The bat file just calls for the python file

@echo off
start "" "C:\Python27\Scripts\bot.py"

The error when I open it by Windows - Run is

[Errno 2] No such file or directory: 'list.txt'

list.txt is inside Scripts folder and when opened by double click it always worked.

Update

I open the files for read using

g = open("list.txt","r")

and again for write:

g = open("list.txt","w")

I've tried James solution and it worked, but since I have many methods using these, I will get a lot of work as it is not just find and replace, it envolves indentation and also the names of lists changes according which method.

This is an error internally being raised in your application. most likely the result of "assuming" path(s). Don't make assumptions! — James Mills, May 05 '15 at 02:47
With regards to your update; You'll have to work out how to best "refactor" your code. My first suggestion would be to make more use of functions and in general make your code modular and reusable. — James Mills, May 05 '15 at 03:26
because `with open` has generated different kinds of errors that I does not recognize, like `'file' object has no attribute '__getitem__'` when I'm trying call .readlines() . I might change `g = open("C:\Python27\Scripts\list.txt","r").readlines()` to solve it faster. — Commentator, May 05 '15 at 03:56

James Mills · Answer 1 · 2015-05-05T03:22:09.967

You'll want to do something like this in your application:

import os
import sys

with open(os.path.join(os.path.dirname(sys.argv[0]), "lists.txt"), "r") as f:
    # do something with lists.txt

This removes the assumption that lists.txt will be in the current directory or similar.

Note: that sys.argv[0] should be the "full absolute path" to the program being executed and hopefully C:\Python27\Scripts\bot.py on your system.

Update: Alternative to using sys.argv[0] (Thank you Alex Taylor) as a means of "determining your entrypoint's directory" you could also use __file__ which is a "global" in Python module(s) that is the "full path" to that module. The only caveat here is that this won't work if your "package" is zipped or otherwise an importable archive. See: __file__

score 2 · Accepted Answer · edited May 23 '17 at 11:51

Similar to James' answer, but using the __file__ macro as the way of getting the currently executing script:

import os.path

with open(os.path.join(os.path.dirname(os.path.abspath(__file__)), 'list.txt'), 'r') as list_file:
    list_data = list_file.read()

The issue is that the working directory is set to the location you double-clicked from, but launching from the command line in the ways you have provided does not. Opening a command prompt to the location of the script and launching from there would also work since the file would be in the working directory.

The __file___ macro is generally considered to be the best way of determining a python script location.

python program doesn't work when opened by run

Update

2 Answers2