I want to expose service with Spring framework (not with spring boot). Then i can use the service to feed a dashboard. Charts in the dashboard need data with json format. My question is similar to this topic but with more question about code.[question]: Expose Service Layer directly in spring mvc
I first did the model, repository to access database. I am using Hibernate and MySQL. I run my application with a class containing the main method. Then i tried to add a rest controller to access the method findAll. But when i deployed the application on Tomcat, i only get the message 404 not found.
This is my first controller
@RestController
@RequestMapping("/fruit")
public class FruitController {
@Autowired
private IFruitRepository fruitRepo = new FruitRepository();
@RequestMapping( value = "/all", method = RequestMethod.GET )
public @ResponseBody List<Port> getFruit() {
List<Fruit> res = fruitRepo.findAll();
return res;
}
}
this is the interface
public interface IFruitRepository {
Boolean create(Fruit p);
Fruit findById(int id);
List<Fruit> findAll();
Fruit update(Fruit f);
boolean delete(int id);
}
this is the implementation of findAll method
public List<Fruit> findAll(){
List<Fruit> à_retourner = new ArrayList<>();
try (SessionFactory factory = HibernateUtil.getSessionFactory()) {
Session session = factory.openSession();
Query query = session.createQuery("from Fruit");
à_retourner = query.getResultList();
} catch (Exception e) {
System.out.println("exception _ findAll _ Fruit : " + e);
}
return à_retourner;
}
EDIT: web .xml
<?xml version="1.0" encoding="UTF-8"?> <web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd" version="4.0"> <context-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/applicationContext.xml</param-value> </context-param> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <servlet> <servlet-name>dispatcher</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>dispatcher</servlet-name> <url-pattern>*.form</url-pattern> </servlet-mapping> </web-app>
dispacher-servlet.xml
<?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd"> </beans>
applicationcontext.xml
<?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd"> </beans>
Should i add servlet , dispacher servlet , application context to find the resource through URI ?
I don't know what is exactly the url you are using to test the service but if you are trying to invoke /fruit/all, it won't work because the servlet dispatcher is configured to handle request that ends with .form. To make it work you should change the url-pattern of the servlet dispatcher to something like /fruit/*
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