How to configure Swagger UI's URL with Apache CXF?

Update your MyApp class to include the Swagger UI configuration.

import org.apache.cxf.jaxrs.openapi.OpenApiFeature;
import org.apache.cxf.jaxrs.swagger.ui.SwaggerUiConfig;

public class MyApp extends Application {
    // ...

    @Override
    public Set<Object> getSingletons() {
        final OpenApiFeature feature = new OpenApiFeature();

        // Set the Swagger UI configuration
        SwaggerUiConfig uiConfig = new SwaggerUiConfig()
                .url("/webappname/doc/openapi.json") // Set the URL of your swagger.json file
                .queryConfigEnabled(true);
        feature.setSwaggerUiConfig(uiConfig);

        singletons.add(feature);
        return singletons;
    }
}

In your web.xml, update the servlet and servlet-mapping to map the URL of your Swagger UI.

<servlet>
    <display-name>REST API Documentation</display-name>
    <servlet-name>RestAPIDoc</servlet-name>
    <servlet-class>org.apache.cxf.jaxrs.servlet.CXFNonSpringJaxrsServlet</servlet-class>
    <init-param>
        <param-name>jaxrs.serviceClasses</param-name>
        <param-value>com.abc.xyz.AgentManagerService</param-value>
    </init-param>
    <init-param>
        <param-name>jaxrs.features</param-name>
        <param-value>org.apache.cxf.jaxrs.openapi.OpenApiFeature</param-value>
    </init-param>
    <init-param>
        <param-name>jaxrs.providers</param-name>
        <param-value>
            org.apache.cxf.jaxrs.provider.MultipartProvider,
            com.fasterxml.jackson.jaxrs.json.JacksonJsonProvider
        </param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>RestAPIDoc</servlet-name>
    <url-pattern>/doc/*</url-pattern>
</servlet-mapping>

Make sure your swagger.json file is available at the specified URL (/webappname/doc/openapi.json) and contains the correct API definition for your application.

I hope it helps.