在Komodo中运行openpyxl python脚本时出现PermissionError [errno 13]

Question

我在Windows 7上的Komodo edit 9和python
3.4中使用openpyxl脚本时遇到问题。我复制了一些openpyxl代码以进行学习，但无法从Komodo中执行。我收到权限错误13。我检查了路径，并且存在python34。当我使用IDLE或命令提示符时，将运行相同的脚本。我的Komodo命令当前为：％（python3）-u％F关于什么可能导致此问题的任何想法？代码和错误包括在下面

from openpyxl import Workbook
from openpyxl.compat import range
from openpyxl.cell import get_column_letter

wb = Workbook()

dest_filename = 'empty_book.xlsx'

ws1 = wb.active
ws1.title = "range names"

for row in range(1, 40):
    ws1.append(range(600))

ws2 = wb.create_sheet(title="Pi")

ws2['F5'] = 3.14

ws3 = wb.create_sheet(title="Data")
for row in range(10, 20):
    for col in range(27, 54):
        _ = ws3.cell(column=col, row=row, value="%s" % get_column_letter(col))
print(ws3['AA10'].value)
wb.save(filename = dest_filename)

-------开始错误-----------

AA
Traceback (most recent call last):
  File "C:\Users\PF15043\Desktop\Scripts\Ggizmo\excelReader.py", line 26, in <module>
    wb.save(filename = dest_filename)
  File "C:\Python34\lib\site-packages\openpyxl-2.3.0b1-py3.4.egg\openpyxl\workbook\workbook.py", line 254, in save
    save_workbook(self, filename)
  File "C:\Python34\lib\site-packages\openpyxl-2.3.0b1-py3.4.egg\openpyxl\writer\excel.py", line 195, in save_workbook
    writer.save(filename, as_template=as_template)
  File "C:\Python34\lib\site-packages\openpyxl-2.3.0b1-py3.4.egg\openpyxl\writer\excel.py", line 177, in save
    archive = ZipFile(filename, 'w', ZIP_DEFLATED, allowZip64=True)
  File "C:\Python34\lib\zipfile.py", line 923, in __init__
    self.fp = io.open(file, modeDict[mode])
PermissionError: [Errno 13] Permission denied: 'empty_book.xlsx'

Answer 1

这只是来自操作系统的错误，告诉您您无权在尝试创建的文件中创建文件。您应该指定要创建的文件的完整路径。