CODEWITHSUNDEEP
javaeclipsespringjsptomcat
Michael
Michael

Reputation: 2128

Spring project in Eclipse - 404

I try to run the app on Tomcat but keep getting this error "The requested resource (/testapp/) is not available." - what might be wrong? I guess my XML setup is incorrect but not sure how to fix it.

dispatcher-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:mvc="http://www.springframework.org/schema/mvc"
    xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
        http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
        http://www.springframework.org/schema/mvc
        http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
        http://www.springframework.org/schema/context 
        http://www.springframework.org/schema/context/spring-context-3.0.xsd">

    <context:component-scan base-package="controller" />

    <mvc:view-controller path="/" view-name="HelloWorldPage" />

    <bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix" value="/WEB-INF/view/" />
        <property name="suffix" value=".jsp" />
    </bean>

</beans>

web.xml

<?xml version="1.0" encoding="utf-8"?>

<web-app xmlns="http://java.sun.com/xml/ns/javaee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
        http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" version="2.5">

    <display-name>Spring Web MVC Application</display-name>

    <servlet>
        <servlet-name>mvc-dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>mvc-dispatcher</servlet-name>
        <url-pattern>*.htm</url-pattern>
    </servlet-mapping>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/dispatcher-servlet.xml</param-value>
    </context-param>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

</web-app>

HelloWorldController.java - using Spring annotations.

package controller;

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.servlet.ModelAndView;

@Controller
@RequestMapping("/welcome")
public class HelloWorldController {

    @RequestMapping(method = RequestMethod.GET)
    public ModelAndView helloWorld() {

        ModelAndView model = new ModelAndView("HelloWorldPage");
        model.addObject("msg", "hello world");

        return model;
    }
}

Upvotes: 1

Views: 1897

Answers (2)

Peter Szanto
Peter Szanto

Reputation: 7722

in your web.xml you map spring to *.htm but your controller is mapped to

@RequestMapping("/welcome")

you should map to

@RequestMapping("/welcome.htm")

Upvotes: 2

Cristiano Fontes
Cristiano Fontes

Reputation: 5088

In case you are not getting any exception on eclipse console, I think you are probably not hitting the right contextRoot of your project, mostly it's the exact name of your application, but sometimes it's different...

If you are not typing the app name wrongly try this.

Click on the project right mouse button click on properties.

Then search for Context.

you are going to find a section called ContextRoot

Use the string in that box to mount your Contextroot like this

localhost:8080/stringFound/

and hit enter in the browser.

Upvotes: 0

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