Well,im trying to build a web applicattion in Java that starts a thread each time someone make a http request in the address.
This is a good pratice? Could work?? Advantages or disadvantages? Work is below and use a spring example,just added the thread that i want
ps: This is running in tomcat
> HomeController.java
#Controller public class HomeController {
private static final Logger logger = LoggerFactory.getLogger(HomeController.class);
/**
* Simply selects the home view to render by returning its name.
*/
#RequestMapping(value = "/", method = RequestMethod.GET)
public String home(Locale locale, Model model) {
logger.info("Welcome home! The client locale is {}.", locale);
Date date = new Date();
DateFormat dateFormat = DateFormat.getDateTimeInstance(DateFormat.LONG, DateFormat.LONG, locale);
String formattedDate = dateFormat.format(date);
model.addAttribute("serverTime", formattedDate );
new Teste().start();
return "home";
}}class Teste extends Thread{
#Override
public void run() {
while(true){
System.out.println("im in thread");
try {
Thread.sleep(5000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
> 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/n/javaee/web-app_2_5.xsd">
<!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
> servlet-context.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd
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">
<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->
<!-- Enables the Spring MVC #Controller programming model -->
<annotation-driven />
<!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<!-- Resolves views selected for rendering by #Controllers to .jsp resources in the /WEB-INF/views directory -->
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>
<context:component-scan base-package="com.example.threadTestes" />
</beans:beans>
threading in a managed environment is generally a bad idea. why not use some sort of abstraction like JMS to start a background handler every time someone sends a request ? that way you can control the number of active threads (jms pool size)
Starting a thread for every request is terrible idea in any application, managed or unmanaged, and #1 reason for that is that amount of memory OS can use to allocate a stack for new thread is limited; and not only it's limited, it is shared between all processes on the box (though your java process may be allowed to use only part of that). Now imagine those threads won't terminate at all, or won't terminate fast enough as new requests keep coming in - OS will eventually run out of space for new threads; depending on the OS and it's settings, result may vary from rather fascinating OutOfMemoryError in new Thread(), to the whole box becoming apparently unresponsive, because your java process consumed all available space and no new threads can be created at all. So: 1) for handling requests use at least thread pool (Executor); JMS is an option if you need persistence, but again you handle requests in the pool - it's just fed from JMS 2) for some sort of maintainence tasks you can launch a separate thread during app startup. In both cases clean up after yourself and stop additional threads during shutdown.
Related
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.
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 your web.xml deployment descriptor,
<path-info> is the part usually used by Spring to perform its internal routing (unless alwaysUseFullPath is set on the HandlerMapping).
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.
In my project using Spring MVC and JSON, I configure my web.xml file following as:
Web.xml
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
My Controller:
#Controller
public class HomeController {
#RequestMapping(value = "/", method = RequestMethod.GET)
public String home(Locale locale, Model model) {
return "home";
}
#RequestMapping(value = "/json", method = RequestMethod.GET, produces = "application/json")
public #ResponseBody List<Person> personsReturn(#RequestParam("name") String name, #RequestParam("age") int age) {
Person p1 = new Person("dat1", 11);
Person p2 = new Person("dat2", 22);
Person p3 = new Person("dat3", 33);
Person p4 = new Person("dat4", 44);
List<Person> lists = new ArrayList<Person>();
if (name.equals("dat")) {
lists.add(p1);
lists.add(p2);
} else {
lists.add(p3);
lists.add(p4);
}
return lists;
}
}
My Servlet-Context.xml:
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd
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">
<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->
<!-- Enables the Spring MVC #Controller programming model -->
<annotation-driven />
<!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<!-- Resolves views selected for rendering by #Controllers to .jsp resources in the /WEB-INF/views directory -->
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>
<context:component-scan base-package="com.luongkhanh.restfulspringmvc" />
</beans:beans>
I can get data from JSON by URL:
http://localhost:9999/restfulspringmvc/json?name=dat1&age=11
Result JSON:
[{"name":"dat3","age":33},{"name":"dat4","age":44}]
But If I want to get the image by URL at location, it's unsuccessful:
http://localhost:9999/restfulspringmvc/images/avatar.jpg
Logger in eclipse alert an exception:
ARN : org.springframework.web.servlet.PageNotFound - No mapping found for HTTP request with URI [/restfulspringmvc/images/avatar.jpg] in DispatcherServlet with name 'appServlet'
How to configure to get image from URL by Browser? Thank you so much !!
You have configured your DispatcherServlet to use the url-pattern / This overrides the default servlet mapping of the container.
The default servlet mapping is used to load any resources or requests without an explicit mapping in web.xml. Typically static resources are loaded using this. If you override it by specifying it as a pattern for your servlet in web.xml then you have to take care of loading the resources yourself
Spring MVC helps by providing resource handling via
You have to specify the mapping which matches your static resources url pattern and the location where your resources reside physically
You did specify
<resources mapping="/resources/**" location="/resources/" />
but your image does not reside in a resources folder under webapp Maven folder. Instead its in images folder. Hence you have to add another resources tag like this
<resources mapping="/images/**" location="/images/" />
NB The mapping attribute and the location attribute does not have to correspond. For example you can have a url like
http://localhost:9999/restfulspringmvc/images/avatar.jpg
and your images are in /static-resources/images under webapp the your mapping will be
<resources mapping="/images/**" location="/static-resources/images/" />
You can also have multiple resources specification in the same application context
I am trying to set up a webapp that get the body of a POST request and then create a database. I created a controller for POST and GET requests on a vFabric server using spring SVM. I started with the spring SVM web aplication template. Im using spring tool suite.
I am new to spring so I'm still grasping the concepts of bean and controllers.
my servlet configuration files goes like this:
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd
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">
<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->
<!-- Enables the Spring MVC #Controller programming model -->
<annotation-driven />
<!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<!-- Resolves views selected for rendering by #Controllers to .jsp resources in the /WEB-INF/views directory -->
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>
<context:component-scan base-package="com.spring.mongodb" />
</beans:beans>
Its basically a configuration I borrowed so I assume this is the problem.
the web.xml configuration goes like this:
<?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">
<!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
and my main controller is this:
#Controller
public class HomeController {
private static final Logger logger = LoggerFactory.getLogger(HomeController.class);
private static final Log log = LogFactory.getLog(HomeController.class);
private String bodyflow;
Coleccion Col;
#RequestMapping(value = "/", method = RequestMethod.POST)
public String getPOST(Locale locale, Model model, #RequestBody String body){
bodyflow = "char";
return "home";
}
#RequestMapping(value = "/", method = RequestMethod.GET)
public String home(Locale locale, Model model) {
logger.info("Welcome home! The client locale is {}.", locale);
Date date = new Date();
DateFormat dateFormat = DateFormat.getDateTimeInstance(DateFormat.LONG, DateFormat.LONG, locale);
String formattedDate = dateFormat.format(date);
model.addAttribute("serverTime", formattedDate );
model.addAttribute("postRequest", bodyflow );
return "home";
}
}
The problem with this code is that whereas the GET request is receive and executed by the method "home", the method getPOST does not execute, I assume it does not get the POST request and it I send to my server from the simple REST client google chrome. I assume is a configuration problem, do I need to add a diferent configuration to GET and POST?
You will notice that in the getPOST method I only set a variable, this is because I was testing if the problem was the model File being diferent in each method, but I realized the method isnt even running.
if you know of a better way of doing this please post it.
Thanks!
I'm pretty new to the Spring framework and web applications, though I'm experienced with Java. When I run my site on a local tomcat server the URL is: http://localhost:8080/myApp/
Now a request mapping delegates me to my home page with:
#RequestMapping(value = "/", method = RequestMethod.GET)
public String someMethod(Model model) { ...
return "index"; }
Now within a file index.xhtml I link to another page with link
but when I want to link back to the index page I have to use link. I searched for a solution and found:
<spring:url value='/apps' var="apps_url" />
link
But spring:url always resolves to http://localhost:8080/myApp/ - the page that I'm currently on. In addition, when I just use a link like this: link, it always resolves to http://localhost:8080/otherSite and not http://localhost:8080/myApp/otherSite like I expected. How can I get my link to work? Is http://localhost:8080/myApp implicitly defined as my context or can/should it be changed to http://localhost:8080/?
Also, is there any connection between the URL on local tomcat server and the URL the web application will have when it's published?
Here are some of my application files:
servlet-context.xml:
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:tx="http://www.springframework.org/schema/tx"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.1.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->
<!-- Enables the Spring MVC #Controller programming model -->
<annotation-driven />
<context:component-scan base-package="myApp" />
<tx:annotation-driven transaction-manager="transactionManager"/>
<!-- Handles HTTP GET requests for /resources/** and /css/** by efficiently
serving up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<resources mapping="/css/**" location="/css/" />
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".xhtml" />
</beans:bean>
</beans:beans>
excerpt from web.xml:
<!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
It is good practice to put all your links as follows
Other site
${pageContext.servletContext.contextPath} always gives your application root, when you are developing use http://localhost:8080/myApp, then your application root is /myapp, but when you want to place your application in production generally your application root will be /, using ${pageContext.servletContext.contextPath} before links you ensure it will work in both cases
When facing same problem I use c liblary:
<%# taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>
Other site
This should do the trick:
<spring:url value="/some_link" context="myApp" var="apps_url" />
link
apps_url should be equal to: http://localhost:8080/myApp/some_link
Tomcat can have many web apps executing at the same time. Consider the scenario:
tomcat is running on 8080
tomcat web apps folder has three apps running in it
appA.war
appB.war
ROOT.war
The URL http://localhost:8080/ will be get the request to tomcat and tomcat must decide which of three apps should service the request since the URL does not identify the app the request will be routed by tomcat to the ROOT.war which is the root web app in tomcat.
When you have a JSP in appA.war with link in it then the web browser that parses the output will figure that the request should be submitted to root of the server since the href started with / therefore the browser puts togethec the url http://localhost:8080/otherSite when the request get to tomcat tomcat assumes that /otherSite is refering to a web app called otherSite.war but of course that is not there.
if you use link the browser will assume this is a relative URL and so it will send the request to http://localhost:8080/appA/otherSite because the page that generate the url was served out of the context http://localhost:8080/appA
So in general you need to make sure that the URLs from a JSP page are relative or that you add the context name to the generated url, this is what <spring:url /> and <c:url /> tags do.
is there any connection between the url on local tomcat server and the
url the webapplication will have when its published?
the URL for the context is the same as the name of the war file in the tomcat webapps folder unless you change that in the server.xml or in the context.xml file.
This question already has answers here:
Why does Spring MVC respond with a 404 and report "No mapping found for HTTP request with URI [...] in DispatcherServlet"?
(13 answers)
Closed 6 years ago.
I'm absolutely newbie with Java and Spring and I want to learn from example.
I'm using an out-of-the-box configuration / installation
Mac OSX
Springsource Tool Suite as IDE
Spring 2.8.1.RELEASE
vfabric-tc-server-developer-2.6.1.RELEASE
I've tried to generate a new project based on "Spring Template Project". Then I've chosen the "Spring MVC Project". The sample project is generated. After that, without modifying anything, I've tried to execute de "home.jsp" page via "Run As". The Web Server starts and finally I received the error in the console Tab.
No mapping found for HTTP request with URI [/myproject/] in
DispatcherServlet with name 'appServlet'
And this other output in these web pages:
http://localhost:8080/myproject/WEB-INF/views/home.jsp
http://localhost:8080/myproject
Here you can see an image on how my project is structured (auto generated for STS):
What is wrong?
Here you can see the content of the web.xml file:
<?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">
<!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
The root-context.xml file has this content.
<?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-3.0.xsd">
<!-- Root Context: defines shared resources visible to all other web components -->
</beans>
And finally the servlet-context.xml has this content.
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->
<!-- Enables the Spring MVC #Controller programming model -->
<annotation-driven />
<!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<!-- Resolves views selected for rendering by #Controllers to .jsp resources in the /WEB-INF/views directory -->
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>
<context:component-scan base-package="com.mycompany.myapp" />
</beans:beans>
Does anybody have an idea to solve it?
Spring's convention is to assume that the <servlet-name> in the web.xml for the DispatcherServlet matches the beginning of the Spring servlet context XML file. Rename servlet-context.xml to appServlet-context.xml and see if that helps.
Everything under WEB-INF is not accessible from the outside, and the point of an MVC framework is to invoke a controller before dispatching to the view, so invoking the JSP directly should not be done anyway.
And you probably don't have any Spring controller mapped to the root URL, so obviously, there is nothing at the URL http://localhost:8080/myproject/.
Add a Controller to your application that returns a ModelAndView instance with name "home". Then configure some handler mapping to that Controller and try to access your web app with a URL similar to this one: http://localhost:8080/myproject/home.do. More information can be found here and here.