On 4/14/21, Quentin Bock <[email protected]> wrote:
>
> this is the only part of the code that causes the error
>
> file = open('Egils Saga 1-15.txt', "r")
Here's an app_abspath() function to resolve a filename against the
directory of the main script:
import os
import sys
def get_main_file():
if hasattr(sys, 'frozen'):
return sys.executable
main = getattr(sys.modules.get('__main__'), '__file__', '')
return os.path.abspath(main) if main else ''
def get_main_dir():
return os.path.dirname(get_main_file()) or os.getcwd()
def app_abspath(filename):
return os.path.join(get_main_dir(), filename)
file = open(app_abspath('Egils Saga 1-15.txt'), 'r')
In the frozen script case, sys.executable is the main 'script'. For a
"-c" command, there is no main file, so it uses the current working
directory.
Using the variable name "file" is fine so long as compatibility with
Python 2 isn't required. In Python 3, "file" is not a reserved keyword
and not the name of a builtin function or type.
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