I've created a custom filter in Spring Cloud Gateway by extending the class with "LoggingGlobalPreFilter"
I have the following routes.
return builder.routes().route(r -> r.path("/first/**").uri("http://localhost:8222"))
.route(r -> r.path("/seconf/**").uri("http://localhost:8333")).build();
Not sure how to add custom filter there. All the articles on the internet are talking about configuring filters in .yml file.
Thanks!
#Component
public class CustomFilter implements GlobalFilter, Ordered {
}
Related
Is it possible to use entity graph in Jpa Retrieving Outbound Gateway?
#Bean
public IntegrationFlow findContract() {
return f -> f
.handle(Jpa.retrievingGateway(this.entityManagerFactory)
.entityClass(Contract.class)
//.parameterExpression("contractNumber", "payload"))
// set entityGraph name
;
}
I would like to use approach similar to Spring Data Jpa
public interface ContractRepository extends JpaRepository<Contract, Long> {
#EntityGraph(value = "contract.documents", type = EntityGraphType.LOAD)
Contract findByContractNumber(String contractNumber);
}
This is just not implemented in the Spring Integration.
Feel free to raise an appropriate JIRA for such an improvement. Of course, Contribution is welcome.
Meanwhile as a workaround you can use a #ServiceActivator (.handle() in Java DSL) to call that ContractRepository.findByContractNumber() method.
Is there anyway that I can generate some metadata to add to the service when it registers.
We are moving from Eureka to Consul and I need to add a UUID value to the registered metadata when a service starts. So that later I can get this metadata value when I retrieve the service instances by name.
Some background: We were using this excellent front end UI from https://github.com/VanRoy/spring-cloud-dashboard. It is set to use the Eureka model for services in which you have an Application with a name. Each application will have multiple instances each with an instance id.
So with the eureka model there is a 2 level service description whereas the spring cloud model is a flat one where n instances each of which have a service id.
The flat model won't work with the UI that I referenced above since there is no distinction between application name and instance id which is the spring model these are the same.
So if I generate my own instance id and handle it through metadata then I can preserve some of the behaviour without rewriting the ui.
See the documentation on metadata and tags in spring cloud consul. Consul doesn't support metadata on service discovery yet, but spring cloud has a metadata abstraction (just a map of strings). In consul tags created with key=value style are parsed into that metadata map.
For example in, application.yml:
spring:
cloud:
consul:
discovery:
tags: foo=bar, baz
The above configuration will result in a map with foo→bar and baz→baz.
Based on Spencer's answer I added an EnvironmentPostProcessor to my code.
It works and I am able to add the metadata tag I want programmatically but it is a complement to the "tags: foo=bar, baz" element so it overrides that one. I will probably figure a way around it in the next day or so but I thougth I would add what I did for other who look at this answer and say, so what did you do?
first add a class as follows:
#Slf4j
public class MetaDataEnvProcessor implements EnvironmentPostProcessor, Ordered {
// Before ConfigFileApplicationListener
private int order = ConfigFileApplicationListener.DEFAULT_ORDER - 1;
private UUID instanceId = UUID.randomUUID();
#Override
public int getOrder() {
return this.order;
}
#Override
public void postProcessEnvironment(ConfigurableEnvironment environment,
SpringApplication application) {
LinkedHashMap<String, Object> map = new LinkedHashMap<>();
map.put("spring.cloud.consul.discovery.tags", "instanceId="+instanceId.toString());
MapPropertySource propertySource = new MapPropertySource("springCloudConsulTags", map);
environment.getPropertySources().addLast(propertySource);
}
}
then add a spring.factories in resources/META-INF with eht following line to add this processor
org.springframework.boot.env.EnvironmentPostProcessor=com.example.consul.MetaDataEnvProcessor
This works fine except for the override of what is in your application.yml file for tags
By default, Spring Data REST will expose search resources to urls under {resource_name}/search. For example, findBySubject_Id in the following code will be exposed to {baseUrl}/questions/search/findBySubject_Id?subjectId={subjectId}.
public interface QuestionRepository extends PagingAndSortingRepository<Question, String> {
Page<Question> findBySubject_Id(#Param("subjectId")String subjectId, Pageable pageable);
}
For compatiblility reasons, I need the exposing link to be {baseUrl}/questions?subjectId={subjectId}. Is there any way to do it?
You can not override /search url. If you really need a link without search, you should override response handler for the entity via writing your own controller with annotation #RepositoryRestController
I am trying to use the new HystrixFeign support in Feign. Here is what my code looks like
route66Client =
HystrixFeign.builder()
.logger(new Slf4jLogger())
.encoder(new GsonEncoder())
.target(Route66Client.class, "http://route66/");
The Route66Client is defined as
public interface Route66Client {
#RequestLine("POST /route")
ApiResponse getRoute(
RouteRequest request);
}
When i try to run the code. I get UnknownHostException. As it is not able to resolve route66 for its hostname. Anyone knows what i could be missing ?
Further i had this working with regular Feign ( not HystrixFeign ). All i did was to annotate my Route66Client class with #FeignClient("...") and then injecting Route66Client in the calling class ( So no Feign.builder() was used )
But i couldn't find some #HystrixFeignClient annotation. So i went ahead and started using the HystrixFeign.builder(). But then when i did that the serviceName->Address resolution stopped working.
If you want the benefits of eureka, don't use the builder directly. Put #EnableFeignClients on an #Configuration class (usually your application). Then label Route66Client with #FeignClient("route66") and inject Route66Client. This is only available in Brixton's 2nd Milestone. See the documentation.
I'm new to Spring MVC and Hibernate. I'm trying to start a project by following tutorials but I have been running into problems as my project structure is not consistent with the tutorials I am reading.
I have downloaded the latest STS and I do see the option of creating an Spring MVC project. However it is based on Spring 3 and still uses XML files. From what I have read it looks like there is a way to do it without XML files since Spring 3. I prefer annotations over XML files greatly.
How can I create a Spring MVC 4 application that is based on annotations and relies on xml files minimally?
EDIT:
I want to create a web project
Here is a squeletal example of full java configuration. You will need :
a class extending AbstractAnnotationConfigDispatcherServletInitializer to replace the old web.xml file
one or more #Configuration annotaded class(es) to initialize the root context (replaces the old applicationContext.xml)
one or more #Configuration annotaded class(es) to initialize the DispatcherServlet context (replaces the old dispatcher-servlet.xml)
This is the web.xml :
public class WebAppConf extends AbstractAnnotationConfigDispatcherServletInitializer {
#Override
protected Class<?>[] getRootConfigClasses() {
// declare root context configuration classes
return new Class<?>[]{ RootConf.class };
}
#Override
protected Class<?>[] getServletConfigClasses() {
// declare servlet context configuration classes
return new Class<?>[]{ ServletConf.class };
}
#Override
protected String[] getServletMappings() {
// mapping of DispatcherServlet
return new String[]{"/"};
}
#Override
protected void customizeRegistration(Dynamic registration) {
// additional configuration, here for MultipartConfig
super.customizeRegistration(registration);
MultipartConfigElement multipartConf = new MultipartConfigElement("", 200000L, -1L, 0);
registration.setMultipartConfig(multipartConf);
}
}
RootConf will declare business model, service and dao beans and is not shown here.
ServletConf declares the controllers and servlet configuration :
#Configuration
#EnableWebMvc
// declare where to find annotated controllers
#ComponentScan({"org.example.web"})
public class ServletConf extends WebMvcConfigurerAdapter {
#Bean
MultipartResolver multipartResolver() {
return new StandardServletMultipartResolver();
}
#Bean
ViewResolver internalViewResolver() {
// the view resolver bean ...
InternalResourceViewResolver resolver = new InternalResourceViewResolver();
resolver.setPrefix("/WEB-INF/jsp/");
resolver.setSuffix(".jsp");
return resolver;
}
}
As said above, it is squeletal, but it comes from a working minimal example so you should be able to start with that and extend it at will. In my example, the above three classes live in a org.example.config package that will never be scanned for autodetecting other configuration classes or annotated beans.
Hope it helps ...
I know this doesn't answer your question fully, but hopefully the links will be useful.
WebApplicationInitializer - A 100% code based approach to configuration
as well as AnnotationConfigWebApplicationContext
Also, if you have the time, reading the relevant sections of Spring's MVC chapter of their documentation is helpful.
I Wish that this link will be helpful for you Spring security with annotation Mkyong
The latest versions of STS integrate the Spring guides from https://spring.io/guides directly, try the "Import Spring Getting Started Content" wizard. There are good guides included for creating a Spring Boot based web service, for example, among many others.