Uploading a zip file in python flask without form

Question

I'm trying to upload a zip file to my server using python flask request and then unzip it using zipfile module.

Here is my code:

@app.route('/uoload', methods=['POST'])
def upload ():

    data = request.data
    current_path = os.getcwd()
    filename = "file.zip"
    with open(os.path.join(upload_path, filename), 'w') as file:
        file.write(data)
    try:
        with zipfile.ZipFile(os.path.join(current_path + filename)) as zf:
            zf.extractall(os.path.join(upload_path))
    except BadZipfile as e:
        print e
        return "", 406

But it seems like the uploaded file is damaged somehow. Because when i'm trying to unzip it, BadzipFile exception occurs and it says : "Bad magic number for file header" .

shifloni shifloni · Accepted Answer · 2015-11-30T22:01:55

The problem seems to be that you are using open to create a zip archive. When you use open, python will create write the data to the file and will name it like you wanted. That doesn't make it a zip archive. That is why it fails to extract data from the file.

If you want to create a zip archive use:

import zipfile

zf = zipfile.ZipFile('file.zip', mode='w')
zf.write('add_this_file_to_zip_archive.txt')
zf.close()

Uploading a zip file in python flask without form

1 Answers