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.
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.
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.