Zip 文件并保存到特定文件夹

Question

this might be a silly question but I'm struggling a lot finding solution to it.

So I have a file in the given folder:

Output\20190101_0100\20190101_0100.csv

Now I want to zip the file and save it to same location. So here's my try:

zipfile.ZipFile('Output/20190101_0100/20190101_0100_11.zip', mode='w', compression=zipfile.ZIP_DEFLATED).write('Output/20190101_0100/20190101_0100_11.csv')

But it's making a folder insider zip folder and saving it, as shown below:

Output\20190101_0100\20190101_0100_11.zip\Output\20190101_0100\20190101_0100_11.csv

Can someone tell me how can I save my file directly in the same location or location mentioned below:

Output\20190101_0100\20190101_0100_11.zip\20190101_0100_11.csv

Answer 1

Rephrasing of question

The question is slightly confusing because Output\20190101_0100\20190101_0100_11.zip\Output\20190101_0100\20190101_0100_11.csv won't be a file, but rather Output\20190101_0100\20190101_0100_11.csv will be a file within the zip file Output\20190101_0100\20190101_0100_11.zip (if I am not mistaken)

Just to restate your problem (if I understood it correctly):

• You have a file Output\20190101_0100\20190101_0100.csv (a file 20190101_0100.csv in the Output -> 20190101_0100 sub directory)
• You want to create the zip file Output/20190101_0100/20190101_0100_11.zip ( 20190101_0100_11.zip in the Output -> 20190101_0100.zip directory)
• You want to add the aforementioned CSV file Output\20190101_0100\20190101_0100.csv but without the leading path, ie as 20190101_0100_11.csv rather than Output\20190101_0100\20190101_0100.csv .

Or to not get confused with too many similar directories, let's simplify it as:

• You have a file test.csv in the sub directory sub-folder
• You want to create the zip file test.zip
• You want to add the aforementioned CSV file test.csv but without the leading path, ie as test.csv rather than sub-folder/test.csv .

Answer

From the ZipFile.write documentation:

Write the file named filename to the archive, giving it the archive name arcname (by default, this will be the same as filename , but without a drive letter and with leading path separators removed).

That means that arcname will default to the passed in filename (it doesn't have a drive letter or leading path separator).

If you want to remove the sub folder part, just pass in arcname as well. eg:

import zipfile

with zipfile.ZipFile('path-to-zip/test.zip', 'w') as zf:
    zf.write('sub-folder/test.csv', arcname='test.csv')

Answer 2

You could try using a raw path:

zipfile.ZipFile('Output/20190101_0100/20190101_0100_11.zip', mode='w', compression=zipfile.ZIP_DEFLATED).write(r'C:\...\Output\20190101_0100\20190101_0100_11.csv')