En Thu, 16 Oct 2008 00:16:13 -0300, MOS37F MOS37F <mos37f@gmail.c om>
escribió:
If you want some kind of answer:
- shorten the code to the minimum fragment that still shows the problem
- don't use non-standard modules
- try to ask a specific question: what input you used, what output you
were expecting, and what you actually got.
As a general comment, instead of 'E:\DIR\DIR\DIR \ShelfFile' use either:
'E:\\DIR\\DIR\\ DIR\\ShelfFile' or
r'E:\DIR\DIR\DI R\ShelfFile'
See http://www.python.org/doc/faq/general/#id54
and os.path.join in this case.
--
Gabriel Genellina
escribió:
import shelve
from ShelfFile import ABC
from zipfile import ZipFile, ZIP_DEFLATED
ShelfFilePath = 'E:\DIR\DIR\DIR \ShelfFile'
from ShelfFile import ABC
from zipfile import ZipFile, ZIP_DEFLATED
ShelfFilePath = 'E:\DIR\DIR\DIR \ShelfFile'
- shorten the code to the minimum fragment that still shows the problem
- don't use non-standard modules
- try to ask a specific question: what input you used, what output you
were expecting, and what you actually got.
As a general comment, instead of 'E:\DIR\DIR\DIR \ShelfFile' use either:
'E:\\DIR\\DIR\\ DIR\\ShelfFile' or
r'E:\DIR\DIR\DI R\ShelfFile'
See http://www.python.org/doc/faq/general/#id54
FilePath1 = RetrievedObject .DirectoryPath1 + '\\' + 'Filename.txt'
--
Gabriel Genellina