protected <RESULT extends Group> RESULT createEntity(final Map<String, Object> source, final Class<? extends RESULT> entityType) {
        try {
            final RESULT result = entityType.newInstance();
            result.setName(DfTypeUtil.toString(source.get("name")));
            result.setAttributes(source.entrySet().stream().filter(e -> !"name".equals(e.getKey()))

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.

그리고 보안 요구 사항에 대한 모든 모델 또는 데이터를 사용할 수 있습니다(이 경우 Pydantic 모델 `User`).

그러나 일부 특정 데이터 모델, 클래스 또는 타입을 사용하도록 제한되지 않습니다.

모델에 `id`와 `email`이 있고 `username`이 없길 원하십니까? 맞습니다. 이들은 동일한 도구를 사용할 수 있습니다.

`str`만 갖고 싶습니까? 아니면 그냥 `dict`를 갖고 싶습니까? 아니면 데이터베이스 클래스 모델 인스턴스를 직접 갖고 싶습니까? 그들은 모두 같은 방식으로 작동합니다.

if ! docker container inspect tf >/dev/null 2>&1 ; then
  # Pass all existing TFCI_ variables into the Docker container
  env_file=$(mktemp)
  env | grep ^TFCI_ > "$env_file"

  WORKING_DIR="$TFCI_GIT_DIR"
  if [[ `uname -s | grep -P '^MSYS_NT'` ]]; then
    env_file=$(cygpath -m $env_file)
    # Host dirs can only be mapped to an existing drive inside the container, so
    # T:\ is replaced with C:\.

            }
        }
        if (!found) {
            Plugin plugin = new Plugin();
            plugin.setPrefix(goalPrefix);
            plugin.setArtifactId(artifactId);
            plugin.setName(name);

            getMetadata().addPlugin(plugin);
        }
    }

    public Object getKey() {
        return groupId;
    }

    public boolean isSnapshot() {
        return false;
    }

To keep the Groovy DSL ergonomic, we generate methods as necessary from the non-Groovy equivalents.

Doing this provides the following specific benefits:
- **Reduce the API surface** - We no longer need to maintain two methods.
- **Consistency** - All languages have consistent access to the same APIs and ergonomics in the DSL.

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.

    }

  companion object {
    val isSupported: Boolean =
      try {
        // Trigger an early exception over a fatal error, prefer a RuntimeException over Error.
        Class.forName("org.bouncycastle.jsse.provider.BouncyCastleJsseProvider", false, javaClass.classLoader)

        true
      } catch (_: ClassNotFoundException) {
        false
      }

        FtpServer server = null;
        try {
            String username = "testuser";
            String password = "testpass";
            server = startFtpServer(FTP_PORT, username, password);
            Map<String, Object> params = new HashMap<String, Object>();
            FtpAuthentication auth = new FtpAuthentication();
            auth.setUsername(username);
            auth.setPassword(password);

    public HtmlTransformer htmlTransformer;

    @Override
    protected void setUp() throws Exception {
        super.setUp();
        htmlTransformer = new HtmlTransformer();
        htmlTransformer.setName("htmlTransformer");
        Map<String, String> featureMap = newHashMap();
        featureMap.put("http://xml.org/sax/features/namespaces", "false");
        htmlTransformer.setFeatureMap(featureMap);