求教遍历文件夹

lemon_wb

1. import os
   
2. 
   
3. import openpyxl
   
4. from openpyxl import workbook
   
5. from openpyxl import load_workbook
   
6. 
   
7. filepath = 'D:\\单户施工'
   
8. filelist = os.listdir(filepath)
   
9. files=[]
   
10. 
    
11. for i in range(len(filelist)):
    
12.     child = os.path.join('%s\\%s' % (filepath, filelist[i]))
    
13.     wb = openpyxl.load_workbook(child)

复制代码

=======================================
执行出现Traceback (most recent call last):
  File "d:/Users/test/PycharmProjects/pythonProject/main.py", line 13, in <module>
    wb = openpyxl.load_workbook(child)

    wb = openpyxl.load_workbook(child) 这个语句那里有问题？

Twilight6

检查下目录下是不是没有其他目录，若有就要在 for 循环多加个 if 判断是否为文件夹，进行跳过

另外就是注意要把此时打开此文件下的 excel 关闭

lemon_wb

Twilight6 发表于 2021-9-22 20:14
检查下目录下是不是没有其他目录，若有就要在 for 循环多加个 if 判断是否为文件夹，进行跳过

另外 ...

最后一级目录了，下面没再有目录了。也没有打开excel

Twilight6

lemon_wb 发表于 2021-9-22 20:18
最后一级目录了，下面没再有目录了。也没有打开excel

那发下你的完整报错，最后一句报错关键内容你没有拷贝上

lemon_wb

Twilight6 发表于 2021-9-22 20:20
那发下你的完整报错，最后一句报错关键内容你没有拷贝上

D:\Users\test\PycharmProjects\pythonProject1\venv\Scripts\python.exe D:/Users/test/PycharmProjects/pythonProject1/new.py
Traceback (most recent call last):
  File "D:/Users/test/PycharmProjects/pythonProject1/new.py", line 13, in <module>
    wb = openpyxl.load_workbook(child)
  File "D:\Users\test\PycharmProjects\pythonProject1\venv\lib\site-packages\openpyxl\reader\excel.py", line 318, in load_workbook
    data_only, keep_links)
  File "D:\Users\test\PycharmProjects\pythonProject1\venv\lib\site-packages\openpyxl\reader\excel.py", line 126, in __init__
    self.archive = _validate_archive(fn)
  File "D:\Users\test\PycharmProjects\pythonProject1\venv\lib\site-packages\openpyxl\reader\excel.py", line 96, in _validate_archive
    raise InvalidFileException(msg)
openpyxl.utils.exceptions.InvalidFileException: openpyxl does not support  file format, please check you can open it with Excel first. Supported formats are: .xlsx,.xlsm,.xltx,.xltm

Process finished with exit code 1

lemon_wb

是不支持xls类型，是吗？那想打开
xls类型的怎么办呢？

Twilight6

lemon_wb 发表于 2021-9-22 21:19
是不支持xls类型，是吗？那想打开
xls类型的怎么办呢？

在 for 循环里面加个 print('%s\\%s' % (filepath, filelist[i])) 看看路径

lemon_wb

Twilight6 发表于 2021-9-22 21:25
在 for 循环里面加个 print('%s\\%s' % (filepath, filelist)) 看看路径

1. import os
   
2. 
   
3. import openpyxl
   
4. from openpyxl import workbook
   
5. from openpyxl import load_workbook
   
6. 
   
7. filepath = 'D:\\test\\2021年1月\\202001历下单户结算'
   
8. filelist = os.listdir(filepath)
   
9. files=[]
   
10. 
    
11. for i in range(len(filelist)):
    
12.     child = os.path.join('%s\\%s' % (filepath, filelist[i]))
    
13.     print('%s\\%s' % (filepath, filelist))
    
14.     wb = openpyxl.load_workbook(child)

复制代码

返回
D:\Users\test\PycharmProjects\pythonProject1\venv\Scripts\python.exe D:/Users/test/PycharmProjects/pythonProject1/new.py
Traceback (most recent call last):
  File "D:/Users/test/PycharmProjects/pythonProject1/new.py", line 14, in <module>
    wb = openpyxl.load_workbook(child)
  File "D:\Users\test\PycharmProjects\pythonProject1\venv\lib\site-packages\openpyxl\reader\excel.py", line 318, in load_workbook
    data_only, keep_links)
  File "D:\Users\test\PycharmProjects\pythonProject1\venv\lib\site-packages\openpyxl\reader\excel.py", line 126, in __init__
    self.archive = _validate_archive(fn)
  File "D:\Users\test\PycharmProjects\pythonProject1\venv\lib\site-packages\openpyxl\reader\excel.py", line 96, in _validate_archive
    raise InvalidFileException(msg)
openpyxl.utils.exceptions.InvalidFileException: openpyxl does not support the old .xls file format, please use xlrd to read this file, or convert it to the more recent .xlsx file format.


Process finished with exit code 1

Twilight6

lemon_wb 发表于 2021-9-22 21:28
返回
D:%users\test\PycharmProjects\pythonProject1\venv\Scripts\python.exe D:/Users/test/Pycha ...

好吧  这不怎么懂了，另外我之前代码的 [ i ] 被吞了