Сбой этого кода Python:
from openpyxl import Workbook
wb = Workbook()
sheet = wb.create_sheet('*nice*')
Исключение:
/home/foo/local/bin/python /home/foo/src/invalid.py
Traceback (most recent call last):
File "/home/foo/src/invalid.py", line 5, in <module>
sheet = wb.create_sheet('*nice')
File "/home/foo/local/lib/python2.7/site-packages/openpyxl/workbook/workbook.py", line 158, in create_sheet
new_ws = Worksheet(parent=self, title=title)
File "/home/foo/local/lib/python2.7/site-packages/openpyxl/worksheet/worksheet.py", line 121, in __init__
_WorkbookChild.__init__(self, parent, title)
File "/home/foo/local/lib/python2.7/site-packages/openpyxl/workbook/child.py", line 50, in __init__
self.title = title or self._default_title
File "/home/foo/local/lib/python2.7/site-packages/openpyxl/workbook/child.py", line 93, in title
raise ValueError(msg)
ValueError: Invalid character * found in sheet title
Как создать лист, содержащий *
в заголовке?