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


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`
    attribute to access the raw standard Python file (blocking, not async), useful and

@app.post("/optional-list-uploadfile")
async def read_optional_list_uploadfile(
    p: Annotated[list[UploadFile] | None, File()] = None,
):
    return {"file_size": [file.size for file in p] if p else None}


@pytest.mark.parametrize(
    "path",
    [
        "/optional-list-bytes",
        "/optional-list-uploadfile",
    ],
)
def test_optional_list_schema(path: str):

String buildNumber = System.getenv('BUILD_NUMBER')

if (buildNumber) {
  File uploadFile = file("build/${buildNumber}.tar.bz2")
  project.gradle.buildFinished { result ->
    println "build complete, generating: $uploadFile"
    if (uploadFile.exists()) {
      project.delete(uploadFile)
    }

    try {
      ant.tar(destfile: uploadFile, compression: "bzip2", longfile: "gnu") {
        fileset(dir: projectDir) {

Але є кілька випадків, у яких вам може бути корисно використовувати `UploadFile`.

## Параметри файлу з `UploadFile` { #file-parameters-with-uploadfile }

Визначте параметр файлу з типом `UploadFile`:

{* ../../docs_src/request_files/tutorial001_an_py310.py hl[14] *}

Використання `UploadFile` має кілька переваг перед `bytes`:

Aber es gibt viele Fälle, in denen Sie davon profitieren, `UploadFile` zu verwenden.

## Datei-Parameter mit `UploadFile` { #file-parameters-with-uploadfile }

Definieren Sie einen Datei-Parameter mit dem Typ `UploadFile`:

{* ../../docs_src/request_files/tutorial001_an_py310.py hl[14] *}

`UploadFile` zu verwenden, hat mehrere Vorzüge gegenüber `bytes`:

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"),
):

from typing import Annotated

from fastapi import FastAPI, File, UploadFile

app = FastAPI()


@app.post("/files/")
async def create_file(file: Annotated[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:

@app.post("/required-uploadfile", operation_id="required_uploadfile")
async def read_required_uploadfile(p: Annotated[UploadFile, File()]):
    return {"file_size": p.size}


@pytest.mark.parametrize(
    "path",
    [
        "/required-bytes",
        "/required-uploadfile",
    ],
)
def test_required_schema(path: str):
    openapi = app.openapi()
    body_model_name = get_body_model_name(openapi, path)