How to extract a specific file from an archive donwloaded from internet using only memory

I'm looking for a way to extract a specific file (knowing his name) from an archive containing multiple ones, without writing any file on the hard drive.

I tried to use both StringIO and zipfile, but I only get the entire archive, or the same error from Zipfile (open require another argument than a StringIo object)

Needed behaviour:

archive.zip #containing ex_file1.ext, ex_file2.ext, target.ext
extracted_file #the targeted unzipped file

archive.zip = getFileFromUrl("file_url")
extracted_file = extractFromArchive(archive.zip, target.ext)


What I've tried so far:

import zipfile, requests

data = requests.get("file_url")
zfile = StringIO.StringIO(zipfile.ZipFile(data.content))
needed_file = zfile.open("Needed file name", "r").read()


2 Answers
2

There is a builtin library, zipfile, made for working with zip archives.

https://docs.python.org/2/library/zipfile.html

You can list the files in an archive:

ZipFile.namelist()


and extract a subset:

ZipFile.extract(member[, path[, pwd]])


EDIT:
This question has in-memory zip info. TLDR, Zipfile does work with in-memory file-like objects.

Python in-memory zip library

I finally found why I didn't succeed to do it after few hours of testing :

I was bufferring the zipfile object instead of buffering the file itself and then open it as a Zipfile object, which raised a type error.

Here is the way to do :

import zipfile, requests

data = requests.get(url) # Getting the archive from the url
zfile = zipfile.ZipFile(StringIO.StringIO(data.content)) # Opening it in an emulated file
filenames = zfile.namelist() # Listing all files
for name in filesnames:
if name == "Needed file name": # Verify the file is present
needed_file = zfile.open(name, "r").read() # Getting the needed file content
break


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