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Downloading multiple files in Flask

I am trying to provide the client side the option of downloading some files in Flask. There can be multiple files or a single file available for the user/client to download.
However I am not able to understand how to provide the user the option to download multiple files.
Here is what I have tried so far:
#app.route('/download_files')
def download():
count=0
download_list=[]
for path in pathlib.Path("dir1/dir2").iterdir():
if path.is_file():
for i in names:
if pathlib.PurePosixPath(path).stem == i:
count += 1
download_list.append(path)
return send_file(download_list, as_attachment=True, mimetype="text/plain", download_name="Downloaded Files", attachment_filename="Generated Files")
This does not work properly even with a single file. The file type I am trying to download is text file with the extension .sql .
Will I somehow have to zip multiple files and then provide the download option? Please guide with my available options.

In order to offer several files together as a download, you only have the option of compressing them in an archive.
In my example, all files that match the specified pattern are listed and compressed in a zip archive. This is written to the memory and sent by the server.
from flask import Flask
from flask import send_file
from glob import glob
from io import BytesIO
from zipfile import ZipFile
import os
app = Flask(__name__)
#app.route('/download')
def download():
target = 'dir1/dir2'
stream = BytesIO()
with ZipFile(stream, 'w') as zf:
for file in glob(os.path.join(target, '*.sql')):
zf.write(file, os.path.basename(file))
stream.seek(0)
return send_file(
stream,
as_attachment=True,
download_name='archive.zip'
)

You haven't provided a code sample where you actually getting these files or this file. Minimum working example would be like this:
from flask import Flask, request
app = Flask(__name__)
#app.route('/download_files', methods=['POST'])
def download():
file = request.files['file'] # for one file
files = request.files.getlist("file[]") # if there're multiple files provided
if __name__ == "__main__":
app.run()
After what your file variable will be an object of werkzeug.FileStorage and files variable will be a list of these objects.
And to download all these files you can check this question.

Download CSV File in Flask Best Practice

Right now I have a flask app in which part of the functionality allows me to select a date range and see data from a sql database from that selected date range. I then can click a button and it exports this to a csv file which is just saved in the flask project directory. I want the user to be able to download this csv file. I want to know what the best practice for a user to download a dynamic csv file. Should I send_file() and then delete the file after user has downloaded since this data shouldn't be saved and the user won't be using that file again. Should the file be saved to the database and then deleted out of the db? Or can I just keep it within the flask directory? Please provide insight if possible, thank you so much.

#brunns pointed it in very right direction.
You don't have to save the file in your database or in your file structure or anywhere. It will get created in memory on user request.
I've done this with django for pdf and for csv files it'll work in the same way with flask too. Basics are all same.
for python3 use io.StringIO, for python2 use StringIO
from io import StringIO
import csv
from flask import make_response
#app.route('/download')
def post(self):
si = StringIO.StringIO()
cw = csv.writer(si)
cw.writerows(csvList)
output = make_response(si.getvalue())
output.headers["Content-Disposition"] = "attachment; filename=export.csv"
output.headers["Content-type"] = "text/csv"
return output
Courtesy: vectorfrog

Based on #xxbinxx's answer, used with pandas
from io import StringIO
import csv
from flask import make_response
#app.route('/download')
def download_csv(self, df: pd.DataFrame):
si = StringIO()
cw = csv.writer(si)
cw.writerows(df.columns.tolist())
cw.writerows(df.values.tolist())
output = make_response(si.getvalue())
output.headers["Content-Disposition"] = "attachment; filename=export.csv"
output.headers["Content-type"] = "text/csv"
return output

Download file from root directory using flask

I am generating a xlsx file which I would like to download once it is created. The file is created using a module called 'xlsxwriter'. It saves the file in my root directory, however I cant figure out how to access it via flask, so that it starts a download.
This is how I create the file:
workbook = xlsxwriter.Workbook('images.xlsx')
worksheet = workbook.add_worksheet()
worksheet.write(..someData..)
It saves the file in my root directory.
Now I am trying to access it in order to download it via flask:
app = Flask(__name__, static_url_path='')
#app.route('/download')
def download():
# do some stuff
return Response(
app.send_static_file('images.xlsx'),
mimetype="xlsx",
headers={"Content-disposition":
"attachment; filename=images.xlsx"})
However, I get a 404 Error. Is using send_static_file the correct way to go here?

I found a solution using 'send_file' instead. Providing the path to my file like so:
from flask import send_file
return send_file(pathToMyFile, as_attachment=True)

Download a file with Django

This might perhaps be a simple question, but I somehow just can not find the solution. Django offers a lot about uploading file, but how do I do to download a file.
Let's assume we have a button on HTML on uploads/something.txt as a file.
I tried with django.views.static.serve, however what this did it would open a file on webpage.
My question is simple: What is the best and most pythonic way for user of our website to download a file?

You need to read that file.
Serve it using HttpResponse along with proper content type.
Here's some sample code:
content = open("uploads/something.txt").read()
return HttpResponse(content, content_type='text/plain')
This should serve a text file.
But as you described, on some browser, it will not ask to download the file, rather, it would show it in the browser. If you want to show a download prompt, use this:
response = HttpResponse(open("uploads/something.txt", 'rb').read())
response['Content-Type'] = 'text/plain'
response['Content-Disposition'] = 'attachment; filename=DownloadedText.txt'
return response
However, please note that it might be a better idea to serve static contents or uploaded files via nginx or the reverse proxy of your choice. Sending large files through Django might not be the most optimum way of doing that.

import os
from django.conf import settings
from django.http import HttpResponse, Http404
def download(request, path):
file_path = os.path.join(settings.MEDIA_ROOT, path)
if os.path.exists(file_path):
with open(file_path, 'rb') as fh:
response = HttpResponse(fh.read(), content_type="application/vnd.ms-excel")
response['Content-Disposition'] = 'inline; filename=' + os.path.basename(file_path)
return response
raise Http404

Maybe a little late but here is my solution:
def render(self, value):
return format_html('<a href="/media/{0}" download>{0}</a>', value)

Help improve my file upload method (Pyramid framework)

Currently, I am using the following method for uploading files (via HTML form) in Pyramid.
if request.params.get('form.submitted'):
upload_directory = os.getcwd() + '/myapp/static/uploads/'
my_file = request.POST.get('thumbnail')
saved_file = str(upload_directory) + str(my_file.filename)
perm_file = open(saved_file, 'w')
shutil.copyfileobj(my_file.file, perm_file)
my_file.file.close()
perm_file.close()
I am just wondering, is this a good way of saving file uploads, are there any security concerns with my method? How else can I improve my method. Thanks.

You'll want to use something like werkzug's safe_join rather than just adding the upload directory to the given file name. An attacker could create a POST with a filename of ../../../some/important/path and cause this script to overwrite some file outside of your upload_directory.