[Django]-How to read contents of zip file in memory on a file upload in python?

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6👍

✅

The first argument to zipfile.ZipFile() can be a file object rather than a pathname. I think the Django UploadedFile object supports this use, so you can read directly from that rather than having to copy into a file.

You can also open the file directly from the zip archive rather than extracting that into a file.

import json
import zipfile

@csrf_exempt
def get_zip(request):
    try:
        if request.method == "POST":
            try:
                client_file = request.FILES['file']
                # unzip the zip file to the same directory 
                with zipfile.ZipFile(client_file, 'r') as zip_ref:
                    first = zip_ref.infolist()[0]
                    with zip_ref.open(first, "r") as fo:
                        json_content = json.load(fo)
                doSomething(json_content)
                return HttpResponse(0)

            except Exception as e:
                return HttpResponse(1)
👤Barmar

6👍

From what I understand, what @jason is trying to say here is to first open a zipFile just like you have done here with zipfile.ZipFile(file_path + client_file.name, 'r') as zip_ref:.

class zipfile.ZipFile(file[, mode[, compression[, allowZip64]]])

  Open a ZIP file, where file can be either a path to a file (a string) or a file-like object.

And then use BytesIO read in the bytes of a file-like object. But from above you are reading in r mode and not rb mode. So change it as follows.

with open(filename, 'rb') as file_data:
    bytes_content = file_data.read()
    file_like_object = io.BytesIO(bytes_content)
    zipfile_ob = zipfile.ZipFile(file_like_object)

Now zipfile_ob can be accessed from memory.

👤AzyCrw4282

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