Answers for "how to save upload file in fastapi"

0

fastapi upload file save

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
Posted by: Guest on June-09-2021
0

how to save upload file in fastapi

from fastapi import FastAPI, File, UploadFile

app = FastAPI()


@app.post("/upload-file/")
async def create_upload_file(uploaded_file: UploadFile = File(...)):
    file_location = f"files/{uploaded_file.filename}"
    with open(file_location, "wb+") as file_object:
        file_object.write(uploaded_file.file.read())
    return {"info": f"file '{uploaded_file.filename}' saved at '{file_location}'"}
Posted by: Guest on May-08-2022

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