claim_prefix  (string)    JWT claim namespace prefix e.g. "customer1/"
scopes        (csv)       Comma separated list of OpenID scopes for server, defaults to advertised scopes from discovery document e.g. "email,admin"
comment       (sentence)  optionally add a comment to this setting
```

and ENV based options

```
mc admin config set myminio/ identity_openid --env

KEY:

Gradle uses pull requests for contributions. Fork [gradle/gradle](https://github.com/gradle/gradle) and clone your fork. Configure your Git username and email with:

    git config user.name 'First Last'
    git config user.email ******@****.***

### IntelliJ IDEA

#### Import Gradle into IntelliJ

To import Gradle into IntelliJ:

Ayrıca güvenlik gereksinimleri için herhangi bir model veya veri kullanabilirsiniz (bu örnekte bir Pydantic `User` modeli).

Ancak belirli bir data model, class ya da type kullanmak zorunda değilsiniz.

Modelinizde bir `id` ve `email` olsun, ama `username` olmasın mı istiyorsunuz? Elbette. Aynı araçları kullanabilirsiniz.

Sadece bir `str` mı istiyorsunuz? Ya da sadece bir `dict`? Veya doğrudan bir veritabanı class model instance'ı? Hepsi aynı şekilde çalışır.

                "Expected merged artifactId to be TARGET but was " + merged.getArtifactId());
    }

    @Test
    void mergeSameContributors() {
        Contributor contributor =
                Contributor.newBuilder().email("******@****.***").build();

        Model target =
                Model.newBuilder().contributors(Arrays.asList(contributor)).build();

        Model source =

And you can use any model or data for the security requirements (in this case, a Pydantic model `User`).

But you are not restricted to using some specific data model, class or type.

Do you want to have an `id` and `email` and not have any `username` in your model? Sure. You can use these same tools.

Do you want to just have a `str`? Or just a `dict`? Or a database class model instance directly? It all works the same way.

Sie sind jedoch nicht auf die Verwendung von bestimmten Datenmodellen, Klassen, oder Typen beschränkt.

Möchten Sie eine `id` und eine `email` und keinen `username` in Ihrem Modell haben? Kein Problem. Sie können dieselben Tools verwenden.

			Optional:    true,
			Type:        "string",
		},
		config.HelpKV{
			Key:         Scopes,
			Description: `Comma separated list of OpenID scopes for server, defaults to advertised scopes from discovery document e.g. "email,admin"` + defaultHelpPostfix(Scopes),
			Optional:    true,
			Type:        "csv",
		},
		config.HelpKV{
			Key:         Vendor,

                setAttributeValue(result, fessConfig.getLdapAttrGivenName(), o -> user.setGivenName(o.toString()));
                setAttributeValue(result, fessConfig.getLdapAttrMail(), o -> user.setMail(o.toString()));
                setAttributeValue(result, fessConfig.getLdapAttrEmployeeNumber(), o -> user.setEmployeeNumber(o.toString()));

Quando você instala o FastAPI com `pip install "fastapi[standard]"`, ele vem com o grupo `standard` de dependências opcionais:

Utilizado pelo Pydantic:

* [`email-validator`](https://github.com/JoshData/python-email-validator) - para validação de email.

Utilizado pelo Starlette:

* [`httpx`](https://www.python-httpx.org) - Obrigatório caso você queira utilizar o `TestClient`.

E você pode usar qualquer modelo ou dado para os requisitos de segurança (neste caso, um modelo Pydantic `User`).

Mas você não está restrito a usar um modelo de dados, classe ou tipo específico.

Você quer ter apenas um `id` e `email`, sem incluir nenhum `username` no modelo? Claro. Você pode usar essas mesmas ferramentas.