Upload file using fastapi

Question

I am using FastAPI to upload a file according to the official documentation, as shown below:

@app.post("/create_file/")
async def create_file(file: UploadFile = File(...)):
      file2store = await file.read()
      # some code to store the BytesIO(file2store) to the other database

When I send a request using Python requests library, as shown below:

f = open(".../file.txt", 'rb')
files = {"file": (f.name, f, "multipart/form-data")}
requests.post(url="SERVER_URL/create_file", files=files)

the file2store variable is always empty. Sometimes (rarely seen), it can get the file bytes, but almost all the time it is empty, so I can't restore the file on the other database.

I also tried the bytes rather than UploadFile, but I get the same results. Is there something wrong in my code, or is the way I use FastAPI to upload a file wrong?

Answer 1

The below examples use the .file attribute of the UploadFile object to get the actual Python file (i.e., SpooledTemporaryFile), which allows you to call its methods, such as .read() and .close(), without having to await them. It is important, however, to define your endpoint with def in this case—otherwise, such operations would block the server until they are completed, if the endpoint was defined with async def. In FastAPI, a normal def endpoint is run in an external threadpool that is then awaited, instead of being called directly (as it would block the server).

If you have to define your endpoint with async def—as you might need to await for some other coroutines inside your route—then you should rather use asynchronous reading and writing of the contents, as demonstrated in this answer. Moreover, if you need to send additional data (such as JSON data) together with uploading the file(s), please have a look at this answer. I would also suggest you have a look at this answer, which explains the difference between def and async def endpoints.

Upload Single File

app.py

from fastapi import File, UploadFile

@app.post("/upload")
def upload(file: UploadFile = File(...)):
    try:
        contents = file.file.read()
        with open(file.filename, 'wb') as f:
            f.write(contents)
    except Exception:
        return {"message": "There was an error uploading the file"}
    finally:
        file.file.close()

    return {"message": f"Successfully uploaded {file.filename}"}

Read the File in chunks

As described in this answer, if the file is too big to fit into memory—for instance, if you have 8GB of RAM, you can’t load a 50GB file (not to mention that the available RAM will always be less than the total amount installed on your machine, as other applications will be using some of the RAM)—you should rather load the file into memory in chunks and process the data one chunk at a time. This method, however, may take longer to complete, depending on the chunk size you choose—in the example below, the chunk size is 1024 * 1024 bytes (i.e., 1MB). You can adjust the chunk size as desired.

from fastapi import File, UploadFile
        
@app.post("/upload")
def upload(file: UploadFile = File(...)):
    try:
        with open(file.filename, 'wb') as f:
            while contents := file.file.read(1024 * 1024):
                f.write(contents)
    except Exception:
        return {"message": "There was an error uploading the file"}
    finally:
        file.file.close()

    return {"message": f"Successfully uploaded {file.filename}"}

Another option would be to use shutil.copyfileobj(), which is used to copy the contents of a file-like object to another file-like object (have a look at this answer too). By default, the data is read in chunks with the default buffer (chunk) size being 1MB (i.e., 1024 * 1024 bytes) for Windows and 64KB for other platforms, as shown in the source code here. You can specify the buffer size by passing the optional length parameter. Note: If negative length value is passed, the entire contents of the file will be read instead—see f.read() as well, which .copyfileobj() uses under the hood (as can be seen in the source code here).

from fastapi import File, UploadFile
import shutil
        
@app.post("/upload")
def upload(file: UploadFile = File(...)):
    try:
        with open(file.filename, 'wb') as f:
            shutil.copyfileobj(file.file, f)
    except Exception:
        return {"message": "There was an error uploading the file"}
    finally:
        file.file.close()
        
    return {"message": f"Successfully uploaded {file.filename}"}

test.py

import requests

url = 'http://127.0.0.1:8000/upload'
file = {'file': open('images/1.png', 'rb')}
resp = requests.post(url=url, files=file) 
print(resp.json())

You may also want to have a look at this answer, which demonstrates another approach to upload a large file in chunks, using the .stream() method, which results in considerably minimising the time required to upload the file(s).

Upload Multiple (List of) Files

app.py

from fastapi import File, UploadFile
from typing import List

@app.post("/upload")
def upload(files: List[UploadFile] = File(...)):
    for file in files:
        try:
            contents = file.file.read()
            with open(file.filename, 'wb') as f:
                f.write(contents)
        except Exception:
            return {"message": "There was an error uploading the file(s)"}
        finally:
            file.file.close()

    return {"message": f"Successfuly uploaded {[file.filename for file in files]}"}    

Read the Files in chunks

As described earlier in this answer, if you expect some rather large file(s) and don't have enough RAM to accommodate all the data from the beginning to the end, you should rather load the file into memory in chunks, thus processing the data one chunk at a time (Note: adjust the chunk size as desired, below that is 1024 * 1024 bytes).

from fastapi import File, UploadFile
from typing import List

@app.post("/upload")
def upload(files: List[UploadFile] = File(...)):
    for file in files:
        try:
            with open(file.filename, 'wb') as f:
                while contents := file.file.read(1024 * 1024):
                    f.write(contents)
        except Exception:
            return {"message": "There was an error uploading the file(s)"}
        finally:
            file.file.close()
            
    return {"message": f"Successfuly uploaded {[file.filename for file in files]}"}   

or, using shutil.copyfileobj():

from fastapi import File, UploadFile
from typing import List
import shutil

@app.post("/upload")
def upload(files: List[UploadFile] = File(...)):
    for file in files:
        try:
            with open(file.filename, 'wb') as f:
                shutil.copyfileobj(file.file, f)
        except Exception:
            return {"message": "There was an error uploading the file(s)"}
        finally:
            file.file.close()

    return {"message": f"Successfuly uploaded {[file.filename for file in files]}"}  

test.py

import requests

url = 'http://127.0.0.1:8000/upload'
files = [('files', open('images/1.png', 'rb')), ('files', open('images/2.png', 'rb'))]
resp = requests.post(url=url, files=files) 
print(resp.json())

Answer 2

@app.post("/create_file/")
async def image(image: UploadFile = File(...)):
    print(image.file)
    # print('../'+os.path.isdir(os.getcwd()+"images"),"*************")
    try:
        os.mkdir("images")
        print(os.getcwd())
    except Exception as e:
        print(e) 
    file_name = os.getcwd()+"/images/"+image.filename.replace(" ", "-")
    with open(file_name,'wb+') as f:
        f.write(image.file.read())
        f.close()
   file = jsonable_encoder({"imagePath":file_name})
   new_image = await add_image(file)
   return {"filename": new_image}