key="labels.storage_file"/></label>
                                                        <div class="form-inline col-sm-9">
                                                            <input type="file" name="uploadFile" id="uploadFile"  class="form-control-file"/>

# `UploadFile` class

You can define *path operation function* parameters to be of the type `UploadFile` to receive files from the request.

You can import it directly from `fastapi`:

```python
from fastapi import UploadFile
```

::: fastapi.UploadFile
    options:
        members:
            - file
            - filename
            - size
            - headers
            - content_type
            - read

from starlette.datastructures import UploadFile as StarletteUploadFile
from typing_extensions import Annotated, Doc


class UploadFile(StarletteUploadFile):
    """
    A file uploaded in a request.

    Define it as a *path operation function* (or dependency) parameter.

    If you are using a regular `def` function, you can use the `upload_file.file`

from fastapi import FastAPI, File, UploadFile
from typing_extensions import Annotated

app = FastAPI()


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


@app.post("/uploadfile/")
async def create_upload_file(
    file: Annotated[UploadFile, File(description="A file read as UploadFile")],
):

from typing import Annotated

from fastapi import FastAPI, File, UploadFile

app = FastAPI()


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


@app.post("/uploadfile/")
async def create_upload_file(file: 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("/")

from typing import Annotated

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

app = FastAPI()


@app.post("/files/")
async def create_files(files: Annotated[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]}

Mas há muitos casos em que você pode se beneficiar do uso de `UploadFile`.

## Parâmetros de Arquivo com `UploadFile`

Defina um parâmetro de arquivo com um tipo de `UploadFile`:

{* ../../docs_src/request_files/tutorial001_an_py39.py hl[14] *}

Utilizar `UploadFile` tem várias vantagens sobre `bytes`:

* Você não precisa utilizar o `File()` no valor padrão do parâmetro.

    response = client.post("/files/")
    assert response.status_code == 200, response.text
    assert response.json() == {"message": "No file sent"}


def test_post_uploadfile_no_body():
    response = client.post("/uploadfile/")
    assert response.status_code == 200, response.text
    assert response.json() == {"message": "No upload file sent"}


def test_post_file(tmp_path):
    path = tmp_path / "test.txt"

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`:

//// tab | Python 3.9+

```Python hl_lines="14"
{!> ../../docs_src/request_files/tutorial001_an_py39.py!}
```

////