Question

我正在尝试将一个文件的内容复制到另一个文件，并且出现错误，我该怎么办？

for file in os.listdir('offer'):
    if '.css' in file:
        print(file)
        with open(f'offer/{file}', 'r+') as f:
            with open('offer/id.css', 'w+') as style_file:
                shutil.copyfile(f'offer/{f}', f'offer/{style_file}')

Traceback (most recent call last):
  File "C:/Users/Katerina/Desktop/python/test_attempt.py", line 50, in <module>
    shutil.copyfile(f'offer/{f}', f'offer/{style_file}')
  File "C:\Users\Katerina\AppData\Local\Programs\Python\Python37-32\lib\shutil.py", line 120, in copyfile
    with open(src, 'rb') as fsrc:
OSError: [Errno 22] Invalid argument: "offer/<_io.TextIOWrapper name='offer/id.css' mode='r+' encoding='cp1251'>"

Answer 1

我想您正在寻找类似的东西

with open('offer/id.css', 'w') as style_file:
    for file in os.listdir('offer'):
        if '.css' in file:
            #print(file)
            with open(f'offer/{file}', 'r+') as f:
                for line in f:
                    style_file.write(line)

shutil.copyfile的目的有所不同；它实际上并不允许您访问或修改两个文件的内容，而只是将一个文件复制到另一个文件。

Answer 2

我知道这不是您的问题，但是当我在运行[Errno 22] Invalid Argument时遇到一般的shutil.copy错误时，问题是该应用程序帐户的目标文件夹的权限已更改。我只需要将文件夹的Write Permissions还原到运行该应用程序的帐户即可。

Errno 22无效参数Python

2 个答案: