I tried to create a Spring Rest Controller, based on this example i create a controller like this.
DeveloperRestController.java
@RestController
public class DeveloperRestController {
@RequestMapping("/developer/list")
public Developer index() {
Developer developer = new Developer("Developername", "developer@yahoo.com");
return developer;
}
}
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<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>/</url-pattern>
</servlet-mapping>
<filter>
<filter-name>sitemesh</filter-name>
<filter-class>org.sitemesh.config.ConfigurableSiteMeshFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>sitemesh</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener> <!--Here we specify about the DispatcherServlet class in the Web Deployment Descriptor-->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/dispatcher-servlet.xml</param-value>
</context-param>
</web-app>
dispatcher-servlet.xml
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
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-4.2.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-4.2.xsd">
<context:component-scan base-package="com.developerdata.controller" />
<context:annotation-config />
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/jsp/" />
<property name="suffix" value=".jsp" />
</bean>
</beans>
But it shows 404 page not found, seems that spring tried to load a template...
Result:
Error 404 /WEB-INF/jsp/developer/list.jsp
what should i do?
Seems like you have a configuration issue. The sample you are basing yours on is spring boot based. So it handles the configuration for you. To get yours working you will need to add jackson to the classpath. If you are using maven then:
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.6.2</version>
</dependency>
Then you need to alter your spring config to include:
<mvc:annotation-driven />
From the spring documentation:
The above registers a RequestMappingHandlerMapping, a RequestMappingHandlerAdapter, and an ExceptionHandlerExceptionResolver (among others) in support of processing requests with annotated controller methods using annotations such as @RequestMapping, @ExceptionHandler, and others.
This also then enables the MappingJackson2HttpMessageConverter if jackson 2 is in your classpath.
References:
http://docs.spring.io/spring/docs/current/spring-framework-reference/htmlsingle/#mvc-config-enable
https://spring.io/guides/gs/rest-service/
http://www.mkyong.com/spring-mvc/spring-3-mvc-and-json-example/ (This one seems more appropriate for you to get started based on your problem)