Question

我需要重命名大约2000个文件，因为它们目前在文件名的末尾有每个客户的帐号，但我被要求将其更改为帐户名。

通过从文件名末尾切换帐号并在Excel电子表格中查找以查找帐户名称，我可以毫无问题地获取帐户名称。

# open excel spreadsheet containing account names against numbers

xl = Dispatch("Excel.Application")
wb = xl.Workbooks.Add(r"C:\path\to\accountnumbers.xlsx")
ws = wb.Worksheets(1)
row = 1
col = 1
empty = False

# get all the filenames of the files inside the sales pack folder

while not empty:
    val = ws.Cells(row,col).value
    val2 = ws.Cells(row,col+1).value
    for path, subdirs, files in os.walk(r"C:\path\to\Sales Packs"):
        for filename in files:
            accNo = filename[11:len(filename)-5]
            if accNo == val:
                accName = val2
                os.rename(filename, filename[0:11]+accName+".xlsx")  
    row += 1
    if val == None:
        empty = True

xl.Quit()
pythoncom.CoUninitialize()

当我运行时，我收到以下错误：

Traceback (most recent call last):
  File "C:/Users/Ryan/Documents/filenamechange.py", line 29, in <module>
    os.rename(filename, filename[0:11]+accName+".xlsx")
FileNotFoundError: [WinError 2] The system cannot find the file specified: 'Sales_Pack_66 Books Ltd.xlsx'

有谁知道我哪里出错了？

编辑：

代码现在正在运作。

# open excel spreadsheet containing account names against numbers

xl = Dispatch("Excel.Application")
wb = xl.Workbooks.Add(r"C:\path\to\accountnumbers.xlsx")
ws = wb.Worksheets(1)
row = 1
col = 1
empty = False

# get all the filenames of the files inside the sales pack folder

while not empty:
    val = ws.Cells(row,col).value
    val2 = ws.Cells(row,col+1).value
    for path, subdirs, files in os.walk(r"C:\path\to\Sales Packs"):
        for filename in files:
            accNo = filename[11:len(filename)-5]
            if accNo == val:
                accName = val2
                filename = os.path.join(path, filename)
                os.rename(filename, path+r"\Sales Pack_"+accName+".xlsx")
    row += 1
    if val == None:
        empty = True

xl.Quit()
pythoncom.CoUninitialize()

Answer 1

我认为你应该使用完整的路径：

filename = os.path.join("C:\path\to\Sales Packs", filename)
os.rename(filename, filename[0:11]+accName+".xlsx")

根据文件夹中已有的文件名重命名文件夹中的所有文件

1 个答案: