Errno 22 无效参数 Python

Question

我正在尝试将一个文件的内容复制到另一个文件，但出现错误，我做错了什么？

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

我猜你正在寻找类似

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的目的不同；它实际上不允许您访问或修改任一文件的内容，它只是将一个文件复制到另一个文件。

Answer 2

我知道这不是你的问题，但是当我在运行 shutil.copy 时遇到一般 [Errno 22] Invalid Argument 错误时，问题是应用程序帐户的目标文件夹的权限已经改变。我只需要将文件夹的 Write Permissions 恢复到应用运行所在的帐户。

Errno 22 无效参数 Python

Errno 22 Invalid Argument Python

python

file

shutil

file-copying