Hey all,
I'm starting to develop a new application and I'm researching using Camel's
REST DSL vs. Spring Boot and it's REST support. The last application I wrote
used Apache Camel + Spring Boot, but I only used the external properties
support from Boot. For this new application, I have a skeleton of the previous
project. The new project needs to do REST (XML-based). Looking at a HelloWorld
example in Spring MVC vs. Camel, I'm leaning towards using Spring MVC.
Camel
----
@Component
public class HelloWorldRoute extends RouteBuilder {
@Override
public void configure() throws Exception {
// servlet is configured in Application.java
restConfiguration().component("servlet").bindingMode(RestBindingMode.json);
rest("/say")
.get("/hello")
.to("direct:talk");
from("direct:talk")
.process(exchange -> {
HelloWorld hw =
HelloWorld.builder().message("Howdy World!").build();
exchange.getIn().setBody(hw);
});
}
}
Spring MVC
----
@RestController
@Api("Hello")
public class HelloWorldController {
@RequestMapping(value = "/api/hello", method = RequestMethod.GET)
public HelloWorld sayHello() {
return HelloWorld.builder().message("Hello World").build();
}
}
I was able to get Swagger working for both services, albeit with different
endpoints. One issue I found is that the "base.path" has to be configured or
calling the methods by clicking on buttons in Swagger UI doesn't work. With
Spring Boot, I'm able to configure Swagger with the following:
@Configuration
@EnableSwagger
public class SwaggerConfig implements EnvironmentAware {
public static final String DEFAULT_INCLUDE_PATTERN = "/api/.*";
private RelaxedPropertyResolver propertyResolver;
@Override
public void setEnvironment(Environment environment) {
this.propertyResolver = new
RelaxedPropertyResolver(environment, "swagger.");
}
/**
* Swagger Spring MVC configuration
*/
@Bean
public SwaggerSpringMvcPlugin
swaggerSpringMvcPlugin(SpringSwaggerConfig springSwaggerConfig) {
return new SwaggerSpringMvcPlugin(springSwaggerConfig)
.apiInfo(apiInfo())
.genericModelSubstitutes(ResponseEntity.class)
.includePatterns(DEFAULT_INCLUDE_PATTERN);
}
/**
* API Info as it appears on the swagger-ui page
*/
private ApiInfo apiInfo() {
return new ApiInfo(
propertyResolver.getProperty("title"),
propertyResolver.getProperty("description"),
propertyResolver.getProperty("termsOfServiceUrl"),
propertyResolver.getProperty("contact"),
propertyResolver.getProperty("license"),
propertyResolver.getProperty("licenseUrl"));
}
}
With Camel, it's a bit less code, but if I don't override the "base.path", it
defaults to localhost:8080. The Spring MVC Swagger implementation figures out
the correct base path on its own.
/**
* Swagger Camel Configuration
*/
@Bean
public ServletRegistrationBean swaggerServlet() {
ServletRegistrationBean swagger = new
ServletRegistrationBean(new SpringRestSwaggerApiDeclarationServlet(),
"/swagger/*");
Map<String, String> params = new HashMap<>();
params.put("base.path", "https://localhost:8443/rest";);
params.put("api.title", propertyResolver.getProperty("title"));
params.put("api.description",
propertyResolver.getProperty("description"));
params.put("api.termsOfServiceUrl",
propertyResolver.getProperty("termsOfServiceUrl"));
params.put("api.version",
propertyResolver.getProperty("version"));
params.put("api.license",
propertyResolver.getProperty("license"));
params.put("api.licenseUrl",
propertyResolver.getProperty("licenseUrl"));
swagger.setInitParameters(params);
return swagger;
}
Is it possible to improve the SpringRestSwaggerApiDeclarationServlet so it gets
the path from CamelServlet and it doesn't have to be hardcoded?
Thanks,
Matt
