在 Python 中将文件从一个位置复制到另一个位置
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Copy a file from one location to another in Python
提问by robster
I have a list called fileListcontaining thousands of filenames and sizes something like this:
我有一个名为fileList的列表，其中包含数千个文件名和大小，如下所示：
['/home/rob/Pictures/some/folder/picture one something.jpg', '143452']
['/home/rob/Pictures/some/other/folder/pictureBlah.jpg', '473642']
['/home/rob/Pictures/folder/blahblahpicture filename.jpg', '345345']
I want to copy the files using fileList[0] as the source but to another whole destination. Something like:
我想使用 fileList[0] 作为源将文件复制到另一个完整的目的地。就像是：
copyFile(fileList[0], destinationFolder)
and have it copy the file to that location.
并让它将文件复制到该位置。
When I try this like so:
当我这样尝试时：
for item in fileList:
copyfile(item[0], "/Users/username/Desktop/testPhotos")
I get an error like the following:
我收到如下错误：
with open(dst, 'wb') as fdst:
IsADirectoryError: [Errno 21] Is a directory: '/Users/username/Desktop/testPhotos'
What could be something I could look at to get this working? I'm using Python 3 on a Mac and on Linux.
我可以看看什么才能让它发挥作用？我在 Mac 和 Linux 上使用 Python 3。
回答by Psytho
You have to give a full nameof the destination file, not just a folder name.
您必须提供目标文件的全名，而不仅仅是文件夹名称。
You can get the file name using os.path.basename(path)and then build the destionation path usin os.path.join(path, *paths)
您可以使用获取文件名os.path.basename(path)，然后使用使用构建目标路径os.path.join(path, *paths)
for item in fileList:
filename = os.path.basename(item[0])
copyfile(item[0], os.path.join("/Users/username/Desktop/testPhotos", filename))
回答by Rakesh
Use os.path.basenameto get the file name and then use it in destination.
使用os.path.basename来获取文件名，然后在目的地使用它。
import os
from shutil import copyfile
for item in fileList:
copyfile(item[0], "/Users/username/Desktop/testPhotos/{}".format(os.path.basename(item[0])))
回答by Z Yoder
You could just use the shutil.copy() command:
你可以只使用 shutil.copy() 命令：
e.g.
例如
import shutil
for item in fileList:
shutil.copy(item[0], "/Users/username/Desktop/testPhotos")
[From the Python 3.6.1 documentation. I tried this and it works.]
[来自 Python 3.6.1 文档。我试过了，它有效。]
回答by Thom
You probably need to cut off the file name from the source string and put it behind the destination path, so it is a full file path.
您可能需要从源字符串中截取文件名并将其放在目标路径后面，因此它是完整的文件路径。
