from typing import Optional

from fastapi import FastAPI, File
from fastapi.testclient import TestClient

app = FastAPI()


@app.post("/files")
async def upload_files(files: Optional[list[bytes]] = File(None)):
    if files is None:
        return {"files_count": 0}
    return {"files_count": len(files), "sizes": [len(f) for f in files]}


def test_optional_bytes_list():
    client = TestClient(app)
    response = client.post(

# `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

    path2 = tmp_path / "test2.txt"
    path2.write_bytes(b"<file content2>")

    client = TestClient(app)
    with path.open("rb") as file, path2.open("rb") as file2:
        response = client.post(
            "/uploadfiles/",
            files=(
                ("files", ("test.txt", file)),
                ("files", ("test2.txt", file2)),
            ),
        )
    assert response.status_code == 200, response.text

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

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

@app.post("/optional-list-uploadfile")
async def read_optional_list_uploadfile(
    p: Annotated[Optional[list[UploadFile]], 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):

    path2 = tmp_path / "test2.txt"
    path2.write_bytes(b"<file content2>")

    client = TestClient(app)
    with path.open("rb") as file, path2.open("rb") as file2:
        response = client.post(
            "/uploadfiles/",
            files=(
                ("files", ("test.txt", file)),
                ("files", ("test2.txt", file2)),
            ),
        )
    assert response.status_code == 200, response.text

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_py39.py hl[14] *}

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

Але в деяких випадках Вам може знадобитися `UploadFile`.

## Параметри файлу з `UploadFile`

Визначте параметр файлу з типом `UploadFile`:

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

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

* Вам не потрібно використовувати `File()` у значенні за замовчуванням параметра.

    return {"file_size": len(p) if p else None}


@app.post("/optional-uploadfile", operation_id="optional_uploadfile")
async def read_optional_uploadfile(p: Annotated[Optional[UploadFile], File()] = None):
    return {"file_size": p.size if p else None}


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