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Spring MVC webapp not loading on browser

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.

No mapping found for HTTP request with URI in web.xml

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

java spring resource folder: FileNotFoundException

I cannot access files in resource directory. What is the problem?
in 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.test" />
</beans:beans>
in 'HomeController.java'
#RequestMapping(value = "/", method = RequestMethod.GET)
public String home(Locale locale, Model model) {
File file = new File("/resources/sample/text.txt");
if (file.exists()) {
logger.info("file found.");
} else {
logger.info("file not found.");
}
return "home";
}
in web.xml (default)
<?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>

Oups, that is not the correct way to access resources in a web application.
First, your resource paths are relative to the ServetContext path, so you should not try to use them as file system absolute paths. But in addition, you cannot be sure that they exist as true files.
The servlet specifications allow a servlet container not to explode a war file, prodided it is able to correctly serve resources through ServletContext.getResourceAsStream(java.lang.String path)
That means that you should only use resources as read-only through ServletContext.getResourceAsStream(java.lang.String path) and never try to use them as true files.

Threads inside webapplication Java

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.

No mapping found for HTTP request with URI [/myproject/] in DispatcherServlet with name 'appServlet' [duplicate]

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.