Return multiple files from fastapi

Question

Using fastapi, I can't figure out how to send multiple files as a response. For example, to send a single file, I'll use something like this

from fastapi import FastAPI, Response

app = FastAPI()

@app.get("/image_from_id/")
async def image_from_id(image_id: int):

    # Get image from the database
    img = ...
    return Response(content=img, media_type="application/png")

However, I'm not sure what it looks like to send a list of images. Ideally, I'd like to do something like this:

@app.get("/images_from_ids/")
async def image_from_id(image_ids: List[int]):

    # Get a list of images from the database
    images = ...
    return Response(content=images, media_type="multipart/form-data")

However, this returns the error

    def render(self, content: typing.Any) -> bytes:
        if content is None:
            return b""
        if isinstance(content, bytes):
            return content
>       return content.encode(self.charset)
E       AttributeError: 'list' object has no attribute 'encode'

Not sure, but if `content` is a type of `List` then loop content: `for c in content: c.encode() ...` — felipsmartins, Apr 11 '20 at 20:08
@felipsmartins the objects in the list are bytes already, running `img.encode()` on them doesn't work `'bytes' object has no attribute 'encode'` — Hooked, Apr 11 '20 at 20:24

shluvme · Answer 1 · 2022-09-18T01:33:46.953

Furthermore, you can create the zip on-the-fly and stream it back to the user using a StreamingResponse object:

import os
import zipfile
import StringIO
from fastapi.responses import StreamingResponse

zip_subdir = "/some_local_path/of_files_to_compress"

def zipfile(filenames):
    zip_io = BytesIO()
    with zipfile.ZipFile(zip_io, mode='w', compression=zipfile.ZIP_DEFLATED) as temp_zip:
        for fpath in filenames:
            # Calculate path for file in zip
            fdir, fname = os.path.split(fpath)
            zip_path = os.path.join(zip_subdir, fname)
            # Add file, at correct path
            temp_zip.write(fpath, zip_path)
    return StreamingResponse(
        iter([zip_io.getvalue()]), 
        media_type="application/x-zip-compressed", 
        headers = { "Content-Disposition": f"attachment; filename=images.zip"}
    )

please correct to `temp_zip.write(fpath, zip_path)` – Radoslaw Jan 27 '22 at 13:51 — Radoslaw, Jan 27 '22 at 13:51
also zip_subdir = "archive" here – Gaurav Hazra Sep 06 '22 at 19:26 — Gaurav Hazra, Sep 06 '22 at 19:26

kia · Accepted Answer · 2020-04-12T01:10:17.003

Zipping is the best option that will have same results on all browsers. you can zip files dynamically.

import os
import zipfile
import StringIO


def zipfiles(filenames):
    zip_subdir = "archive"
    zip_filename = "%s.zip" % zip_subdir

    # Open StringIO to grab in-memory ZIP contents
    s = StringIO.StringIO()
    # The zip compressor
    zf = zipfile.ZipFile(s, "w")

    for fpath in filenames:
        # Calculate path for file in zip
        fdir, fname = os.path.split(fpath)
        zip_path = os.path.join(zip_subdir, fname)

        # Add file, at correct path
        zf.write(fpath, zip_path)

    # Must close zip for all contents to be written
    zf.close()

    # Grab ZIP file from in-memory, make response with correct MIME-type
    resp = Response(s.getvalue(), mimetype = "application/x-zip-compressed")
    # ..and correct content-disposition
    resp['Content-Disposition'] = 'attachment; filename=%s' % zip_filename

    return resp


@app.get("/image_from_id/")
async def image_from_id(image_id: int):

    # Get image from the database
    img = ...
    return zipfiles(img)

As alternative you can use base64 encoding to embed an (very small) image into json response. but i don't recommend it.

You can also use MIME/multipart but keep in mind that i was created for email messages and/or POST transmission to the HTTP server. It was never intended to be received and parsed on the client side of a HTTP transaction. Some browsers support it, some others don't. (so i think you shouldn't use this either)

Hi @kia, and thank you for the answer! Stack Overflow discourages single link answers (and it may get downvoted as a result). The answer would be improved if you could provide a small example that shows how to use `aiofiles` in this particular context. — Hooked, Apr 11 '20 at 20:21
Hi @Hooked and @kia, unless I'm missing something, zipping files as in your example is a blocking IO operation that might mess the asynchronous event loop under the hood. Consider either making a synchronous handler (remove the `async` in the definition of `image_for_id`, or consider running the `zipfiles` function under the control of a `TheadPoolExecutor` as in this example : https://docs.python.org/3/library/asyncio-eventloop.html#asyncio.loop.run_in_executor — glenfant, May 19 '20 at 16:56
@glenfant Can you please tell me why **"a blocking IO operation might mess the asynchronous event loop under the hood"**? thank you. — kia, May 20 '20 at 00:02

vozman · Answer 3 · 2022-02-04T11:12:10.910

I've got some problems with @kia's answer on Python3 and latest fastapi so here is a fix that I got working it includes BytesIO instead of Stringio, fixes for response attribute and removal of top level archive folder

import os
import zipfile
import io


def zipfiles(filenames):
    zip_filename = "archive.zip"

    s = io.BytesIO()
    zf = zipfile.ZipFile(s, "w")

    for fpath in filenames:
        # Calculate path for file in zip
        fdir, fname = os.path.split(fpath)

        # Add file, at correct path
        zf.write(fpath, fname)

    # Must close zip for all contents to be written
    zf.close()

    # Grab ZIP file from in-memory, make response with correct MIME-type
    resp = Response(s.getvalue(), media_type="application/x-zip-compressed", headers={
        'Content-Disposition': f'attachment;filename={zip_filename}'
    })

    return resp

@app.get("/image_from_id/")
async def image_from_id(image_id: int):

    # Get image from the database
    img = ...
    return zipfiles(img)

Return multiple files from fastapi

3 Answers3