Issue
I am new to spring mvc and tomcat. I have developed a demo spring mvc project and trying to deploy it on tomcat 9 through eclipse. Server starts successfully but when i try to access the url from browser i get 404 with below error message on screen. :
Message The requested resource [/spring-mvc-demo/] is not available
Description The origin server did not find a current representation for the target resource or is not willing to disclose that one exists.
Below are my code details :
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app 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_3_1.xsd"
id="WebApp_ID" version="3.1">
<display-name>spring-mvc-demo</display-name>
<absolute-ordering />
<!-- Spring MVC Configs -->
<!-- Step 1: Configure Spring MVC Dispatcher Servlet -->
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-mvc-demo-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<!-- Step 2: Set up URL mapping for Spring MVC Dispatcher Servlet -->
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/spring-mvc-demo</url-pattern>
</servlet-mapping>
</web-app>
spring-mvc-demo-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"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context.xsd
http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc.xsd">
<!-- Step 3: Add support for component scanning -->
<context:component-scan base-package="main.webapp" />
<!-- Step 4: Add support for conversion, formatting and validation support -->
<mvc:annotation-driven/>
<!-- Step 5: Define Spring MVC view resolver -->
<bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/view/" />
<property name="suffix" value=".jsp" />
</bean>
</beans>
HomeController.java
package main.webapp.springdemo.controller;
import javax.annotation.PostConstruct;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
@Controller
public class HomeController {
@PostConstruct
public void init() {
System.out.println("HomeController bean getting intiated");
}
@RequestMapping("/")
public String getMainMenu() {
return "main-menu";
}
}
I tried running this app on tomcat 9 through eclipse as well as manually but in both the scenarios i got same error.
Solution
Assuming the application deploys without errors (check your logs), the URI path you are using to access it is almost certainly wrong. In a servlet environment the URI path decomposes as:
<context-path><servlet-path><path-info>
where:
<context-path>
is the prefix to your application. In the Eclipse server configuration page this is called just "path" and defaults to/<project-name>
,<servlet-path>
is configured through the<servlet-mapping>
element in yourweb.xml
deployment descriptor,<path-info>
is the part usually used by Spring to perform its internal routing (unlessalwaysUseFullPath
is set on theHandlerMapping
).
Therefore (theoretically) you should try accessing:
http://localhost:8080/projectName/spring-mvc-demo/
There is however another problem: your servlet mapping is an exact mapping, that does not match anything else beyond /projectName/spring-mvc-demo
. You should replace it with a prefix mapping (see this question for an overview of servlet mappings):
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/spring-mvc-demo/*</url-pattern>
</servlet-mapping>
If you want to shorten your URL, the DispatcherServlet
is usually mapped as default servlet /
. Be aware not to use the catchall prefix mapping /*
, which would override the mapping of the JSP servlet.
Remark: There are some cases when Spring does not use the part after <servlet-path>
for its routing, e.g. when you use an exact or extension mapping, the whole path after <context-path>
is used.
Therefore if you use:
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/spring-mvc-demo</url-pattern>
</servlet-mapping>
you should specify:
@RequestMapping("/spring-mvc-demo")
public String getMainMenu() {
...
}
and use the URL http://localhost:8080/appName/spring-mvc-demo
.
Answered By - Piotr P. Karwasz
Answer Checked By - Mildred Charles (JavaFixing Admin)