Download file using fastapi

Download file using fastapi

I see the functions for uploading in an API, but I don't see how to download. Am I missing something? I want to create an API for a file download site. Is there a different API I should be using?

from typing import List
from fastapi import FastAPI, Query

app = FastAPI()
PATH "some/path"

@app.get("/shows/")
    def get_items(q: List[str] = Query(None)):
        '''
        Pass path to function.
        Returns folders and files.
        '''
    
        results = {}
    
        query_items = {"q": q}
        entry = PATH + "/".join(query_items["q"]) + "/"

        dirs = os.listdir(entry)
        results["folders"] = [val for val in dirs if os.path.isdir(entry+val)]
        results["files"] = [val for val in dirs if os.path.isfile(entry+val)]
        results["path_vars"] = query_items["q"]
    
        return results

Here is the sample bit of code for python to fetch files and dirs for a path, you can return the path as a list with a new entry in a loop to go deeper into a file tree. Passing a file name should trigger a download function, but I cant seem to get a download func going.

---

I figured it out,

from starlette.responses import FileResponse
@app.get("/shows/")
    def get_items(q: List[str] = Query(None)):
    '''
    Pass path to function.
    Returns folders and files.
    '''

    results = {}

    query_items = {"q": q}
    if query_items["q"]:
        entry = PATH + "/".join(query_items["q"])
    else:
        entry = PATH

    if os.path.isfile(entry):
        return download(entry)

    dirs = os.listdir(entry + "/")
    results["folders"] = [
        val for val in dirs if os.path.isdir(entry + "/"+val)]
    results["files"] = [val for val in dirs if os.path.isfile(entry + "/"+val)]
    results["path_vars"] = query_items["q"]

    return results

def download(file_path):
    """
    Download file for given path.
    """
    if os.path.isfile(file_path):
        return FileResponse(file_path)
        # return FileResponse(path=file_path)
    return None

I added this part

from starlette.responses import FileResponse
if os.path.isfile(entry):
    return download(entry)

Allows you to host static file. But for some reason all files download as "download" .extension. If you know how to ensure original file name, let me know.

---

This worked For me

from starlette.responses import FileResponse

return FileResponse(file_location, media_type='application/octet-stream',filename=file_name)

This will download the file with filename

---

Since we're talking about FastAPI, the proper way to return a file response is covered in their documentation, code snippet below:

from fastapi import FastAPI
from fastapi.responses import FileResponse

file_path = "large-video-file.mp4"
app = FastAPI()

@app.get("/")
def main():
    return FileResponse(path=file_path, filename=file_path, media_type='text/mp4')

---

FastAPI uses Starlette's FileResponse class so there are two ways to import FileResponse on your API code. But of course importing from FastAPI would be a better choice. You can follow the approach below to enable your API endpoints support file download.

Do not forget to add aiofiles to your dependency list. A basic requirements.txt file should look like (versions of modules might change in time, version 0.63.0 of fastapi strictly use starlette 0.13.6)

uvicorn==0.13.4
fastapi==0.63.0
starlette==0.13.6
aiofiles==0.6.0

And the API code

import os

from fastapi import FastAPI
from fastapi.responses import FileResponse


app = FastAPI()


@app.get("/")
async def main():
    file_name = "FILE NAME"
    # DEPENDS ON WHERE YOUR FILE LOCATES
    file_path = os.getcwd() + "/" + file_name
    return FileResponse(path=file_path, media_type='application/octet-stream', filename=file_name)

---

from fastapi import FastAPI from fastapi.responses import FileResponse import uvicorn import os

app = FastAPI()

@app.get("/download-file") def download_file(file_name: str): folder_path = r"C:\Users\HP\Desktop\excel files" file_location = f'{folder_path}{os.sep}{file_name}.xlsx'#os.sep is used to seperate with a
return FileResponse(file_location, media_type='application/octet-stream', filename=file_name)

uvicorn.run(app, port=9105)