Hi I am just starting to learn Spring mvc, I use spring mvc #annotation and I have 3 servlets dispatchers (appservlet, admin, student):
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">
<!-- 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>admin</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/admin-servlet.xml</param-value>
</init-param>
</servlet>
<servlet>
<servlet-name>student</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
</servlet>
<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>
<servlet-mapping>
<servlet-name>admin</servlet-name>
<url-pattern>*.admin</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>student</servlet-name>
<url-pattern>*.student</url-pattern>
</servlet-mapping>
</web-app>
After a test in a method I want a redirect to another servlet dispatcher, either the .admin or a .student. How can I do that please?
#controller
#Controller
public class AuthentificationController {
#RequestMapping(value = "verifier", method = RequestMethod.POST)
public ModelAndView redir(#ModelAttribute("per") Personne per,
HttpSession session) {
if (p1.equals(per)) {
session.setAttribute("login", per);
ModelAndView M = new ModelAndView(".admin");
return M;
}
if (p2.equals(per)) {
ModelAndView M = new ModelAndView(".student");
return M;
} else {
ModelAndView M = new ModelAndView("home");
return M;
}
}
}
The problem is that ModelAndView returns (.jsp) but I want a redirection for another servlet-dispatcher !!
Here you are:
#RequestMapping(value = "verifier", method = RequestMethod.POST)
public String redir(#ModelAttribute("per") Personne per, HttpSession session) {
if (p1.equals(per)) {
session.setAttribute("login", per);
return "redirect:.admin";
} else if (p2.equals(per)) {
return "redirect:.student"
} else {
return "home";
}
}
Related
need your help,I am getting HTTP Status 404 when access this url
localhost:8080/SurakartaSmart/adm-ser/view-all
below is my controller
#Controller
#RequestMapping("/adm-ser")
public class adminController {
#Autowired
adminService serv;
#RequestMapping(value="/view-all", method=RequestMethod.GET)
public #ResponseBody List<adminModel>viewall()throws Exception{
List<adminModel>data = null;
try{
data = serv.viewall();
}catch(Exception e){
e.getMessage();
}
return data;
}
}
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" 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>surakarta-smar</display-name>
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-config.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
where lies the fault that I need to improve?
I have an app. Due to some issues, in my web.xml, I had to change the url pattern from "/" to "*.action", and added a welcome-file-list. This is my web.xml now:
<?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>
<welcome-file-list>
<welcome-file>home.jsp</welcome-file>
</welcome-file-list>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>*.action</url-pattern>
</servlet-mapping>
</web-app>
Before I did the change, I was able to reach home.jsp by simply typing localhost:8080/myapp in my browser. What should I type now to reach the home page?
Thank you.
NOTE: Im using Spring.
EDIT:
Forgot my HomeController:
#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 = "/")
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 );
return "home";
}
}
Your request should ends with ".action" try to sent localhost:8080/myapp/some.action
And change mapping in controller to "/some.action"
My dispatcher servlet mapping
<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/springconfig/dispatcher-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.html</url-pattern>
</servlet-mapping>
And the controller has handler like
#RequestMapping("moduleone")
public class ApplicationController {
#RequestMapping(value="Login.html",method=RequestMethod.GET)
public ModelAndView showLoginPage(){
ModelAndView mv=new ModelAndView("../moduleone/Login");
mv.addObject("loginForm", new LoginForm());
return mv;
}
#RequestMapping(value="Home.html", method = RequestMethod.GET)
public ModelAndView showHome(HttpServletRequest request) {
ModelAndView mv=new ModelAndView("Home");
mv.addObject("customerName",appCon.getFirstName() );
return mv;
}
}
Is it possible to handler request that are not mapped in controller
like
http://localhost:8090/Project/moduleone/invalidpage.html
http://localhost:8090/Project/moduleone/invalidurl/invalidpage
I have tried #RequestMapping(value="*",method=RequestMethod.GET) but doest work
As 404 (page not found) actually produces an exception on web container level, containers usually provide an exception handling mechanism, thus you can try exception (or so called error) handling, as shown below;
First create a controller
#Controller
public class PageNotFoundErrorController {
#RequestMapping(value="/pageNotFound.html")
public String handlePageNotFound() {
// do something
return "pageNotFound";
}
}
and configure web.xml in order to map the error to the controller written above;
<error-page>
<error-code>404</error-code>
<location>/pageNotFound.html</location>
</error-page>
you can also extend it by simply adding 403, 500 and other error-codes to web.xml and mapping them to any controller.
What is even more fascinating is that you can also map any exception (even the ones created by your code); here you can find a nice example about it http://www.mkyong.com/spring-mvc/spring-mvc-exception-handling-example/
I try the code block and if change your scenario a bit i can handle it.
//This one is OK
http://localhost:8090/Project/moduleone/invalidpage.html
//add invalid.html not a folder it should be file
http://localhost:8090/Project/moduleone/invalidurl/invalidpage.html
HomeController.java
#RequestMapping(value = {"*/*.html","*.html"}, method = RequestMethod.GET)
public String test(HttpServletResponse response) throws IOException {
return new String("home");
}
dispatcher-servlet.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>TestSpringMVC</display-name>
<context-param>
<param-name>contextClass</param-name>
<param-value>
org.springframework.web.context.support.AnnotationConfigWebApplicationContext
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>SpringDispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/springconfig/dispatcher-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>SpringDispatcher</servlet-name>
<url-pattern>*.html</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>30</session-timeout>
</session-config>
</web-app>
I can handle both request with this way.
I think you should define an exception page for your second scenario.
Also you can read this issue
I have spring application, in which I use org.apache.cxf for soap and spring MVC for displayng some pages.
My web.xml contains two servlets :CXFServlet and mvc-dispatcher
<servlet>
<servlet-name>CXFServlet</servlet-name>
<servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>CXFServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>WEB-INF/servlet-context.xml</param-value>
</init-param>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/hello</url-pattern>
</servlet-mapping>
When I has been used #ResponseBody in my controller everything was fine.
#Controller
#RequestMapping("/hello")
#ResponseBody
public class HelloController {
#RequestMapping(method = RequestMethod.GET)
public String printWelcome() {
return "hello" ;
}
}
but then i was needed to use jsp I have to use the following
#Controller
#RequestMapping("/hello")
public class HelloController {
#RequestMapping(method = RequestMethod.GET)
public ModelAndView printWelcome(ModelMap model) {
model.addAttribute("message", "hello");
return new ModelAndView("hello") ;
}
}
and when I request http://localhost:8080/hello I get "No service was found" instead of "hello"
I found that if I delete following from web.xml
<servlet>
<servlet-name>CXFServlet</servlet-name>
<servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>CXFServlet</servlet-name>
<servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
my controller works fine.
The Servlet container you are using is matching CXFServlet instead of mvc-dispatcher for the URI http://localhost:8080/hello, resulting in your request being sent to CXFServlet, and the error message "No service was found" being returned by CXFServlet. To quote the Servlet 3.0 spec,
Versions of this specification prior to 2.5 made use of these mapping
techniques as a suggestion rather than a requirement, allowing servlet
containers to each have their different schemes for mapping client
requests to servlets.
http://download.oracle.com/otndocs/jcp/servlet-3.0-fr-eval-oth-JSpec/
You will likely need to configure you CXFServlet mapping to something else, e.g.
<servlet-mapping>
<servlet-name>CXFServlet</servlet-name>
<url-pattern>/services/*</url-pattern>
</servlet-mapping>
You might want to mention the container (Tomcat, Glassfish, etc.) that you are using, as there could also be a bug preventing this from working correctly.
I am using Apache CXF with Spring , please tell me how the CXFServlet reads the myapp-ws-context.xml
<web-app>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:myapp-ws-context.xml</param-value>
</context-param>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<servlet>
<display-name>CXF Servlet</display-name>
<servlet-name>CXFServlet</servlet-name>
<servlet-class>
org.apache.cxf.transport.servlet.CXFServlet
</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>CXFServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
Have you seen sources of org.apache.cxf.transport.servlet.CXFServlet (open source)?
Everything is more than explicit:
#Override
protected void loadBus(ServletConfig sc) {
ApplicationContext wac = WebApplicationContextUtils.
getWebApplicationContext(sc.getServletContext());
String configLocation = sc.getInitParameter("config-location");
if (configLocation == null) {
try {
InputStream is = sc.getServletContext().getResourceAsStream("/WEB-INF/cxf-servlet.xml");
if (is != null && is.available() > 0) {
is.close();
configLocation = "/WEB-INF/cxf-servlet.xml";
}
} catch (Exception ex) {
//ignore
}
}
if (configLocation != null) {
wac = createSpringContext(wac, sc, configLocation);
}
if (wac != null) {
setBus(wac.getBean("cxf", Bus.class));
} else {
setBus(BusFactory.newInstance().createBus());
}
}
Note that WebApplicationContextUtils is a Spring class that tries to find an application context in servlet context attribute named: org.springframework.web.context.WebApplicationContext.ROOT.
Actually your classpath:myapp-ws-context.xml is read by Spring, not CXF.
By adding below configuration in your web.xml, it would be read by Spring and the context would be loaded:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:myapp-ws-context.xml</param-value>
</context-param>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
But you could configure your Servlet/WebApp scope of Spring objects, like multipartResolver etc., to make the objects' scopes clearly, by enhancing your CXFServlet configuration like below:
<servlet>
<display-name>CXF Servlet</display-name>
<servlet-name>CXFServlet</servlet-name>
<servlet-class>
org.apache.cxf.transport.servlet.CXFServlet
</servlet-class>
<init-param>
<param-name>config-location</param-name>
<param-value>/WEB-INF/your-webapp-scope-spring-config.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
*Please take note that from your web-app context, you can access all objects in the context loaded from contextConfigLocation. *