How to expose service with spring 5? - mysql

Question

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 ?

Answer 1

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/*