return multiple pandas Dataframes as single Django zip files (one file per DF) HttpResponse

Question

I trying to figure out a way to return multiple dataframes from a Django view as Zip HttpResponse(zip_file, content_type="application/x-zip-compressed")

I tried this:

import zipfile
import datetime
def makeZipfiles(df_list):
    i = 0
    with zipfile.ZipFile('some_zip.zip', 'w') as csv_zip:
        for dfl in df_list:
            csv_zip.writestr(f"file_{str(i)}.csv", dfl.to_csv(index=False))
            i = i + 1

    return csv_zip

and in the view, I have the following:

zip_file = makeZipfiles(df_list)
response = HttpResponse(zip_file, content_type="application/x-zip-compressed")
return response

but when I try to look at the zip file in the download folder, I get an error that the 'archive is either unknown format or damaged'. The exported file is 1KB size and when I open in notepad I see this content only

"<zipfile.ZipFile [closed]>"

Please advise if what I am trying to do is feasible and if so, please provide a sample code.

Thank you

Answer 1

I haven't tried this out myself yet, but it seems to fit your demand quite well.

https://georepublic.info/en/blog/2019/zip-files-in-django-admin-and-python/

The author describes in detail how to get .csv files from the different dataframes and zip them in the end to one file for download.

His final code was the following:

file_list = [
. . .,
UserInfoResource()
]

def getfiles():
    fileSet = {}
    date_now = dt.datetime.now().strftime('%Y%m%d%H%M')
    for file in file_list:
        dataset = file.export()
        dataset = dataset.csv
        name = file._meta.model.__name__ + date_now
        fileSet[name] = dataset

    return fileSet

def download_zip(request):
    files = getfiles()
    zip_filename = 'Survey_Data' + dt.datetime.now().strftime('%Y%m%d%H%M') + '.zip'
zip_buffer = io.BytesIO()
    with zipfile.ZipFile(zip_buffer, "a", zipfile.ZIP_DEFLATED, False) as zip_file:
        for k, file in files.items():
            zip_file.writestr(k + '.csv', file)
    zip_buffer.seek(0)
    resp = HttpResponse(zip_buffer, content_type='application/zip')
    resp['Content-Disposition'] = 'attachment; filename = %s' % zip_filename
    return resp

I will update this answer if I manage to apply this on my own website. (Let us know if you found a better answer in the meantime, thanks ;) )