server = server.clone();

            String password = server.getPassword();
            if (securityDispatcher.isAnyEncryptedString(password)) {
                try {
                    if (securityDispatcher.isLegacyEncryptedString(password)) {
                        problems.add(new DefaultSettingsProblem(
                                "Legacy/insecurely encrypted password detected for server " + server.getId(),

///

## Password hashing

"Hashing" means converting some content (a password in this case) into a sequence of bytes (just a string) that looks like gibberish.

Whenever you pass exactly the same content (exactly the same password) you get exactly the same gibberish.

But you cannot convert from the gibberish back to the password.

### Why use password hashing

     */
    public String getPassword() {
        return password;
    }

    /**
     * Set the user's password which is used when connecting to the repository.
     *
     * @param password password of the user
     */
    public void setPassword(String password) {
        this.password = password;
    }

    /**
     * Get the username used to access the repository.
     *

If it doesn't receive it, it returns an HTTP 401 "Unauthorized" error.

And returns a header `WWW-Authenticate` with a value of `Basic`, and an optional `realm` parameter.

That tells the browser to show the integrated prompt for a username and password.

Then, when you type that username and password, the browser sends them in the header automatically.

        this.host = host;
    }

    /**
     * Get user's password used to log in to proxy server.
     *
     * @return user's password at proxy host
     */
    public String getPassword() {
        return password;
    }

    /**
     * Set the user's password for the proxy server.
     *
     * @param password password to use to log in to a proxy server
     */

Um dies zu lösen, konvertieren wir zunächst den `username` und das `password` in UTF-8-codierte `bytes`.

Dann können wir `secrets.compare_digest()` verwenden, um sicherzustellen, dass `credentials.username` `"stanleyjobson"` und `credentials.password` `"swordfish"` ist.

{* ../../docs_src/security/tutorial007_an_py39.py hl[1,12:24] *}

Dies wäre das gleiche wie:

```Python

        return "Secure console password prompt";
    }

    @Override
    public Optional<String> configTemplate() {
        return Optional.empty();
    }

    @Override
    public String handle(String config) throws SecDispatcherException {
        if (NAME.equals(config)) {
            try {
                return prompter.promptForPassword("Enter the master password: ");
            } catch (PrompterException e) {

## Pegue o `username` (nome de usuário) e `password` (senha)

É utilizado o utils de segurança da **FastAPI** para obter o `username` e a `password`.

OAuth2 especifica que ao usar o "password flow" (fluxo de senha), que estamos usando, o cliente/usuário deve enviar os campos `username` e `password` como dados do formulário.

        with:
          username: ${{ secrets.DOCKERHUB_USERNAME }}
          password: ${{ secrets.DOCKERHUB_TOKEN }}
      -
        name: Login to GCR
        uses: docker/login-action@9780b0c442fbb1117ed29e0efdff1e18412f7567 # v3.3.0
        with:
          registry: gcr.io
          username: _json_key
          password: ${{ secrets.GCP_CREDS }}
      -
        name: Login to AR

    with pytest.raises(RuntimeError, match=multipart_incorrect_install_error):
        app = FastAPI()

        @app.post("/")
        async def root(username: str = Form(), password: str = Form()):
            return username  # pragma: nocover


def test_incorrect_multipart_installed_form_file(monkeypatch):
    monkeypatch.setattr("python_multipart.__version__", "0.0.12")