OSError: [Errno 22] Ungültiges Argument in load_workbook

Question

Wenn ich openpexl Operating Excel verwende, löst der folgende Code den Fehler aus. Ich kann den Grund nicht herausfinden.

wb=load_workbook(r'C:\Users\Administrator\Desktop\11.xls') 

Fehlermeldung:

C:\Users\Administrator\AppData\Local\Programs\Python\Python36\python.exe 
C:/Users/Administrator/Desktop/data_process.py 
Traceback (most recent call last): 
File "C:/Users/Administrator/Desktop/data_process.py", line 3, in <module> 
wb = load_workbook(r'‪C:\Users\Administrator\Desktop\11.xls') 
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python36\lib\site-packages\openpyxl\reader\excel.py", line 171, in load_workbook 
archive = _validate_archive(filename) 
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python36\lib\site-packages\openpyxl\reader\excel.py", line 118, in _validate_archive 
archive = ZipFile(filename, 'r', ZIP_DEFLATED) 
File "C:\Users\Administrator\AppData\Local\Programs\Python\Python36\lib\zipfile.py", line 1082, in __init__ 
self.fp = io.open(file, filemode) 

OSError: [Errno 22] Invalid argument: '\u202aC:\\Users\\Administrator\\Desktop\\11.xls' 

Answer 1

Probieren Sie folgendes statt 1 '\' versuchen Doppel \

from openpyxl import load_workbook 
wb=load_workbook(r'C:\\Users\\Administrator\\Desktop\\11.xls') 

Lassen Sie mich mit, ob das funktioniert.