Question

setup.py

from distutils.core import setup
import py2exe

setup(console=['program.py'])

The error

Traceback (most recent call last):
File "program.py", line 427, in <module>
File "program.py", line 242, in __init__
WindowsError: [Error 267] The directory name is invalid: 'C:\\Users\\Bob\applications\\Program\\test\\v0.6\\dist\\library.zip/*.*'

The directory name refers to a zip file called library which is located in the dist folder and created during the compilation.

Line 240 – 246 of program.py

 file_list = []
 root_dir = sys.path[0]
 for path in os.listdir(root_dir):
    full_path = os.path.join(root_dir, path).lower()
    if os.path.isfile(full_path) and full_path.endswith('txt'):
        # create list of (filename, dir) tuples
        file_list.append((path.lower(), full_path))

Line 427 of program.py

gui = GuiTk(win)

Any ideas what causes the problem? I use Windows 7 64Bit and PortablePython 2.7.2.1 to create the executable. There are no other errors during the compilation procedure.

Answer 1

You are trying to list the items in sys.path(). From docs:

sys.path A list of strings that specifies the search path for
modules. Initialized from the environment variable PYTHONPATH, plus an
installation-dependent default.

As initialized upon program startup, the first item of this list,
path[0], is the directory containing the script that was used to
invoke the Python interpreter. If the script directory is not
available (e.g. if the interpreter is invoked interactively or if the
script is read from standard input), path[0] is the empty string,
which directs Python to search modules in the current directory first.
Notice that the script directory is inserted before the entries
inserted as a result of PYTHONPATH.

In the case of a py2exe executable like yours, sys.path is a list containing the path for library.zip (the archive that holds all the pure source modules py2exe find in your installation that could be needed for your executable to work).
But you can not use a zip archive for the path for os.listdir

>>> import os
>>> d = 'C:\\test.zip'
>>> os.listdir(d)
Traceback (most recent call last):
  File "<input>", line 1, in <module>
WindowsError: [Error 267] El nombre del directorio no es válido: 'C:\\test.zip/*.*'
>>> 

Probably you are not looking for sys.path but for the “current dir” as the name of your variable indicates.
If this is the case, then os.getcwd will do the job