[QUESTION] Is this the correct way to save an uploaded file ?

Hello, i am trying to save an uploaded file to disk, the following code works correctly but i wonder if it is the correct way to do that.

def parse(file: UploadFile = File(...)):

    extension = os.path.splitext(file.filename)[1]
    _, path = tempfile.mkstemp(prefix='parser_', suffix=extension)

    with open(path, 'wb') as f:
        f.write(file.file.read())

Basically i need to get the path of the temp file. I also have tried with .rollover() but file.file.name does not return the path (only the file descriptor)

[dmontagu]

@classywhetten FastAPI has almost no custom logic related to UploadFile – most of it is coming from starlette. So if you want a “save” wrapper or a better example of usage, it probably makes sense to ask in the starlette repo.

@damianoporta This line:

_, path = tempfile.mkstemp(prefix='parser_', suffix=extension)

causes the file with location path to actually be opened by python, which is what @vitalik was referring to when he said you need to close the tempfile handle (which is the thing named _ in your code).

---

Here are some utility functions that the people in this thread might find useful (at least as a starting point):

import shutil
from pathlib import Path
from tempfile import NamedTemporaryFile
from typing import Callable

from fastapi import UploadFile


def save_upload_file(upload_file: UploadFile, destination: Path) -> None:
    try:
        with destination.open("wb") as buffer:
            shutil.copyfileobj(upload_file.file, buffer)
    finally:
        upload_file.file.close()


def save_upload_file_tmp(upload_file: UploadFile) -> Path:
    try:
        suffix = Path(upload_file.filename).suffix
        with NamedTemporaryFile(delete=False, suffix=suffix) as tmp:
            shutil.copyfileobj(upload_file.file, tmp)
            tmp_path = Path(tmp.name)
    finally:
        upload_file.file.close()
    return tmp_path


def handle_upload_file(
    upload_file: UploadFile, handler: Callable[[Path], None]
) -> None:
    tmp_path = save_upload_file_tmp(upload_file)
    try:
        handler(tmp_path)  # Do something with the saved temp file
    finally:
        tmp_path.unlink()  # Delete the temp file

(I haven’t tested the above functions, just adapted them from slightly more complex functions in my own code.)

Note: you’d want to use the above functions inside of def endpoints, not async def, since they make use of blocking APIs.

I don’t know whether it handles out-of-memory issues reasonably well or not, but that is not an issue in my application – maximum request size is capped elsewhere, and the volume of concurrent requests I’m currently handling is such that OOM isn’t really an issue.

[classywhetten]

I’ve been digging through Flask’s documentation, and their file.save function is a wrapper around shutils.copyfileobj() in the standard library.

Here’s my working code:

@app.post("/files/upload")
def create_file(file: UploadFile = File(...)):
    global upload_folder
    file_object = file.file
    #create empty file to copy the file_object to
    upload_folder = open(os.path.join(upload_folder, file.filename), 'wb+')
    shutil.copyfileobj(file_object, upload_folder)
    upload_folder.close()
    return {"filename": file.filename}

I would much appreciate a wrapper around this be built into fastapi