How to open a excel file?

[xueguan li]

If my excel file is saved on path of " D:/python/test.xlsx",how can i open it.

my code is like below:

   import openpyxl;

   from openpyxl import load_workbook

   wb = load_workbook('D:\python\test.xlsx')

but,the running  result is error,and the indication is:

Traceback (most recent call last):

  File "<pyshell#9>", line 1, in <module>

    wb = load_workbook('D:\python\test.xlsx')

  File "C:\Python27\lib\site-packages\openpyxl-2.3.0b2-py2.7.egg\openpyxl\reader\excel.py", line 150, in load_workbook

    archive = _validate_archive(filename)

  File "C:\Python27\lib\site-packages\openpyxl-2.3.0b2-py2.7.egg\openpyxl\reader\excel.py", line 114, in _validate_archive

    archive = ZipFile(filename, 'r', ZIP_DEFLATED)

  File "C:\Python27\lib\zipfile.py", line 699, in __init__

    self.fp = open(file, modeDict[mode])

IOError: [Errno 22] invalid mode ('rb') or filename: 'D:\\python\test.xlsx'

What is the problem ,please give me help,Thanks!

[Charlie Clark]

Am .09.2015, 11:46 Uhr, schrieb xueguan li <lxg...@gmail.com>:

> If my excel file is saved on path of " D:/python/test.xlsx",how can i
> open it.

On Windows you have to be very careful with file names because the \
character is treated specially in Python. The best approach is always to
use the os.path.join function to assemble a path with the correct
separators for the operating system. A quick workaround is to use a Python
raw string. The following two statements are equivalent and should work:

wb = load_workbook('D:\\python\\test.xlsx')

wb = load_workbook(r'D:\python\test.xlsx')

Charlie
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