CODEWITHSUNDEEP
python
Edison
Edison

Reputation: 4291

MetaData of downloaded zipped file

url='http://www.test.com/test.zip'
z = zipfile.ZipFile(BytesIO(urllib.urlopen(url).read()))
z.extractall(path='D:')

I am writing above code to download a zipped file from a url and have downloaded and extracted all files from it to a specified drive and it is working fine. Is there a way I can get meta data of all files extracted from z for example. filenames,file sizes and file extenstions etc?

Upvotes: 0

Views: 698

Answers (2)

SivanBH
SivanBH

Reputation: 392

I don't know of a built-in way to do that using the zipfile module, however it is easily done using os.path:

import os

EXTRACT_PATH = "D:"

z= zipfile.ZipFile(BytesIO(urllib.urlopen(url).read()))
z.extractall(path=EXTRACT_PATH)
extracted_files = [os.path.join(EXTRACT_PATH, filename) for filename in z.namelist()]
for extracted_file in extracted_files:
    # All metadata operations here, such as:
    print os.path.getsize(extracted_file)

Upvotes: 1

SuperBiasedMan
SuperBiasedMan

Reputation: 9969

Zipfile objects actually have built in tools for this that you can use without even extracting anything. infolist returns a list of ZipInfo objects that you can read certain information out of, including full file name and uncompressed size.

import os

url='http://www.test.com/test.zip'
z = zipfile.ZipFile(BytesIO(urllib.urlopen(url).read()))
info = z.infolist()
data = []
for obj in info:
    name = os.path.splitext(obj.filename)
    data.append(name[0], name[1], obj.file_size)

I also used os.path.splitext just to separate out the file's name from its extension as you did ask for file type separately from the name.

Upvotes: 1

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