Security.insertProviderAt(BouncyCastleProvider(), 1)
          Security.insertProviderAt(BouncyCastleJsseProvider(), 2)
        } else if (platformSystemProperty == CORRETTO_PROPERTY) {
          AmazonCorrettoCryptoProvider.install()

          AmazonCorrettoCryptoProvider.INSTANCE.assertHealthy()
        }

        Platform.resetForTests()

## Simple HTTP Basic Auth

* Import `HTTPBasic` and `HTTPBasicCredentials`.
* Create a "`security` scheme" using `HTTPBasic`.
* Use that `security` with a dependency in your *path operation*.
* It returns an object of type `HTTPBasicCredentials`:
    * It contains the `username` and `password` sent.

{* ../../docs_src/security/tutorial006_an_py39.py hl[4,8,12] *}

import android.os.Build;
import androidx.test.ext.junit.runners.AndroidJUnit4;

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.security.cert.Certificate;
import java.security.cert.CertificateException;
import java.security.cert.CertificateFactory;
import java.security.cert.X509Certificate;

import okhttp3.OkHttpClient;
import okhttp3.Protocol;
import okhttp3.Request;
import okhttp3.Response;

먼저 `OAuth2PasswordRequestForm`을 가져와 `/token`에 대한 *경로 작동*에서 `Depends`의 의존성으로 사용합니다.

//// tab | 파이썬 3.7 이상

```Python hl_lines="4  76"
{!> ../../docs_src/security/tutorial003.py!}
```

////

//// tab | 파이썬 3.10 이상

```Python hl_lines="2  74"
{!> ../../docs_src/security/tutorial003_py310.py!}
```

////

`OAuth2PasswordRequestForm`은 다음을 사용하여 폼 본문을 선언하는 클래스 의존성입니다:

* `username`.
* `password`.

============

Keeping the project small and stable limits our ability to accept new contributors. We are not
seeking new committers at this time, but some small contributions are welcome.

If you've found a security problem, please follow our [bug bounty][security] program.

If you've found a bug, please contribute a failing test case so we can study and fix it.

If you have a new feature idea, please build it in an external library. There are

============

Keeping the project small and stable limits our ability to accept new contributors. We are not
seeking new committers at this time, but some small contributions are welcome.

If you've found a security problem, please follow our [bug bounty][security] program.

If you've found a bug, please contribute a failing test case so we can study and fix it.

If you have a new feature idea, please build it in an external library. There are

- [Documentation Bugs and Feature Requests](https://github.com/istio/istio.io/issues)
- [Community and Governance Issues](https://github.com/istio/community/issues)

For security vulnerabilities, please don't report a bug (which is public) and instead follow

from typing import Optional

from dirty_equals import IsDict
from fastapi import Depends, FastAPI, Security
from fastapi.security import OAuth2, OAuth2PasswordRequestFormStrict
from fastapi.testclient import TestClient
from pydantic import BaseModel

app = FastAPI()

reusable_oauth2 = OAuth2(
    flows={
        "password": {
            "tokenUrl": "token",
            "scopes": {"read:users": "Read the users", "write:users": "Create users"},

* Importieren Sie `HTTPBasic` und `HTTPBasicCredentials`.
* Erstellen Sie mit `HTTPBasic` ein „`security`-Schema“.
* Verwenden Sie dieses `security` mit einer Abhängigkeit in Ihrer *Pfadoperation*.
* Diese gibt ein Objekt vom Typ `HTTPBasicCredentials` zurück:
    * Es enthält den gesendeten `username` und das gesendete `password`.

{* ../../docs_src/security/tutorial006_an_py39.py hl[4,8,12] *}

```Python hl_lines="25"
{!../../docs_src/security/tutorial002.py!}
```

## 获取用户

`get_current_user` 使用创建的（伪）工具函数，该函数接收 `str` 类型的令牌，并返回 Pydantic 的 `User` 模型：

```Python hl_lines="19-22  26-27"
{!../../docs_src/security/tutorial002.py!}
```

## 注入当前用户

在*路径操作* 的 `Depends` 中使用 `get_current_user`：

```Python hl_lines="31"
{!../../docs_src/security/tutorial002.py!}
```