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Request Files

You can define files to be uploaded by the client using File.

Info

To receive uploaded files, first install python-multipart.

E.g. pip install python-multipart.

This is because uploaded files are sent as "form data".

Import File

Import File and UploadFile from fastapi:

from fastapi import FastAPI, File, UploadFile

app = FastAPI()


@app.post("/files/")
async def create_file(file: bytes = File(...)):
    return {"file_size": len(file)}


@app.post("/uploadfile/")
async def create_upload_file(file: UploadFile):
    return {"filename": file.filename}

Define File Parameters

Create file parameters the same way you would for Body or Form:

from fastapi import FastAPI, File, UploadFile

app = FastAPI()


@app.post("/files/")
async def create_file(file: bytes = File(...)):
    return {"file_size": len(file)}


@app.post("/uploadfile/")
async def create_upload_file(file: UploadFile):
    return {"filename": file.filename}

Info

File is a class that inherits directly from Form.

But remember that when you import Query, Path, File and others from fastapi, those are actually functions that return special classes.

Tip

To declare File bodies, you need to use File, because otherwise the parameters would be interpreted as query parameters or body (JSON) parameters.

The files will be uploaded as "form data".

If you declare the type of your path operation function parameter as bytes, FastAPI will read the file for you and you will receive the contents as bytes.

Have in mind that this means that the whole contents will be stored in memory. This will work well for small files.

But there are several cases in which you might benefit from using UploadFile.

File Parameters with UploadFile

Define a file parameter with a type of UploadFile:

from fastapi import FastAPI, File, UploadFile

app = FastAPI()


@app.post("/files/")
async def create_file(file: bytes = File(...)):
    return {"file_size": len(file)}


@app.post("/uploadfile/")
async def create_upload_file(file: UploadFile):
    return {"filename": file.filename}

Using UploadFile has several advantages over bytes:

  • You don't have to use File() in the default value of the parameter.
  • It uses a "spooled" file:
    • A file stored in memory up to a maximum size limit, and after passing this limit it will be stored in disk.
  • This means that it will work well for large files like images, videos, large binaries, etc. without consuming all the memory.
  • You can get metadata from the uploaded file.
  • It has a file-like async interface.
  • It exposes an actual Python SpooledTemporaryFile object that you can pass directly to other libraries that expect a file-like object.

UploadFile

UploadFile has the following attributes:

  • filename: A str with the original file name that was uploaded (e.g. myimage.jpg).
  • content_type: A str with the content type (MIME type / media type) (e.g. image/jpeg).
  • file: A SpooledTemporaryFile (a file-like object). This is the actual Python file that you can pass directly to other functions or libraries that expect a "file-like" object.

UploadFile has the following async methods. They all call the corresponding file methods underneath (using the internal SpooledTemporaryFile).

  • write(data): Writes data (str or bytes) to the file.
  • read(size): Reads size (int) bytes/characters of the file.
  • seek(offset): Goes to the byte position offset (int) in the file.
    • E.g., await myfile.seek(0) would go to the start of the file.
    • This is especially useful if you run await myfile.read() once and then need to read the contents again.
  • close(): Closes the file.

As all these methods are async methods, you need to "await" them.

For example, inside of an async path operation function you can get the contents with:

contents = await myfile.read()

If you are inside of a normal def path operation function, you can access the UploadFile.file directly, for example:

contents = myfile.file.read()

async Technical Details

When you use the async methods, FastAPI runs the file methods in a threadpool and awaits for them.

Starlette Technical Details

FastAPI's UploadFile inherits directly from Starlette's UploadFile, but adds some necessary parts to make it compatible with Pydantic and the other parts of FastAPI.

What is "Form Data"

The way HTML forms (<form></form>) sends the data to the server normally uses a "special" encoding for that data, it's different from JSON.

FastAPI will make sure to read that data from the right place instead of JSON.

Technical Details

Data from forms is normally encoded using the "media type" application/x-www-form-urlencoded when it doesn't include files.

But when the form includes files, it is encoded as multipart/form-data. If you use File, FastAPI will know it has to get the files from the correct part of the body.

If you want to read more about these encodings and form fields, head to the MDN web docs for POST.

Warning

You can declare multiple File and Form parameters in a path operation, but you can't also declare Body fields that you expect to receive as JSON, as the request will have the body encoded using multipart/form-data instead of application/json.

This is not a limitation of FastAPI, it's part of the HTTP protocol.

Optional File Upload

You can make a file optional by using standard type annotations and setting a default value of None:

from typing import Optional

from fastapi import FastAPI, File, UploadFile

app = FastAPI()


@app.post("/files/")
async def create_file(file: Optional[bytes] = File(None)):
    if not file:
        return {"message": "No file sent"}
    else:
        return {"file_size": len(file)}


@app.post("/uploadfile/")
async def create_upload_file(file: Optional[UploadFile] = None):
    if not file:
        return {"message": "No upload file sent"}
    else:
        return {"filename": file.filename}

from fastapi import FastAPI, File, UploadFile

app = FastAPI()


@app.post("/files/")
async def create_file(file: bytes | None = File(None)):
    if not file:
        return {"message": "No file sent"}
    else:
        return {"file_size": len(file)}


@app.post("/uploadfile/")
async def create_upload_file(file: UploadFile | None = None):
    if not file:
        return {"message": "No upload file sent"}
    else:
        return {"filename": file.filename}

UploadFile with Additional Metadata

You can also use File() with UploadFile, for example, to set additional metadata:

from fastapi import FastAPI, File, UploadFile

app = FastAPI()


@app.post("/files/")
async def create_file(file: bytes = File(..., description="A file read as bytes")):
    return {"file_size": len(file)}


@app.post("/uploadfile/")
async def create_upload_file(
    file: UploadFile = File(..., description="A file read as UploadFile")
):
    return {"filename": file.filename}

Multiple File Uploads

It's possible to upload several files at the same time.

They would be associated to the same "form field" sent using "form data".

To use that, declare a list of bytes or UploadFile:

from typing import List

from fastapi import FastAPI, File, UploadFile
from fastapi.responses import HTMLResponse

app = FastAPI()


@app.post("/files/")
async def create_files(files: List[bytes] = File(...)):
    return {"file_sizes": [len(file) for file in files]}


@app.post("/uploadfiles/")
async def create_upload_files(files: List[UploadFile]):
    return {"filenames": [file.filename for file in files]}


@app.get("/")
async def main():
    content = """
<body>
<form action="/files/" enctype="multipart/form-data" method="post">
<input name="files" type="file" multiple>
<input type="submit">
</form>
<form action="/uploadfiles/" enctype="multipart/form-data" method="post">
<input name="files" type="file" multiple>
<input type="submit">
</form>
</body>
    """
    return HTMLResponse(content=content)

from fastapi import FastAPI, File, UploadFile
from fastapi.responses import HTMLResponse

app = FastAPI()


@app.post("/files/")
async def create_files(files: list[bytes] = File(...)):
    return {"file_sizes": [len(file) for file in files]}


@app.post("/uploadfiles/")
async def create_upload_files(files: list[UploadFile]):
    return {"filenames": [file.filename for file in files]}


@app.get("/")
async def main():
    content = """
<body>
<form action="/files/" enctype="multipart/form-data" method="post">
<input name="files" type="file" multiple>
<input type="submit">
</form>
<form action="/uploadfiles/" enctype="multipart/form-data" method="post">
<input name="files" type="file" multiple>
<input type="submit">
</form>
</body>
    """
    return HTMLResponse(content=content)

You will receive, as declared, a list of bytes or UploadFiles.

Technical Details

You could also use from starlette.responses import HTMLResponse.

FastAPI provides the same starlette.responses as fastapi.responses just as a convenience for you, the developer. But most of the available responses come directly from Starlette.

Multiple File Uploads with Additional Metadata

And the same way as before, you can use File() to set additional parameters, even for UploadFile:

from typing import List

from fastapi import FastAPI, File, UploadFile
from fastapi.responses import HTMLResponse

app = FastAPI()


@app.post("/files/")
async def create_files(
    files: List[bytes] = File(..., description="Multiple files as bytes")
):
    return {"file_sizes": [len(file) for file in files]}


@app.post("/uploadfiles/")
async def create_upload_files(
    files: List[UploadFile] = File(..., description="Multiple files as UploadFile")
):
    return {"filenames": [file.filename for file in files]}


@app.get("/")
async def main():
    content = """
<body>
<form action="/files/" enctype="multipart/form-data" method="post">
<input name="files" type="file" multiple>
<input type="submit">
</form>
<form action="/uploadfiles/" enctype="multipart/form-data" method="post">
<input name="files" type="file" multiple>
<input type="submit">
</form>
</body>
    """
    return HTMLResponse(content=content)

from fastapi import FastAPI, File, UploadFile
from fastapi.responses import HTMLResponse

app = FastAPI()


@app.post("/files/")
async def create_files(
    files: list[bytes] = File(..., description="Multiple files as bytes")
):
    return {"file_sizes": [len(file) for file in files]}


@app.post("/uploadfiles/")
async def create_upload_files(
    files: list[UploadFile] = File(..., description="Multiple files as UploadFile")
):
    return {"filenames": [file.filename for file in files]}


@app.get("/")
async def main():
    content = """
<body>
<form action="/files/" enctype="multipart/form-data" method="post">
<input name="files" type="file" multiple>
<input type="submit">
</form>
<form action="/uploadfiles/" enctype="multipart/form-data" method="post">
<input name="files" type="file" multiple>
<input type="submit">
</form>
</body>
    """
    return HTMLResponse(content=content)

Recap

Use File, bytes, and UploadFile to declare files to be uploaded in the request, sent as form data.