问题
I have the standard Spring 4.x files in Netbeans 8.0.2. I added a controller, RegisterController.java in the controllers-package. I also added one model, a User, with some basic information about the user. Next I made 2 .jsp files, Registration.jsp and RegistrationSuccess.jsp.
Here is my RegisterController:
@Controller
@RequestMapping(value="/register")
public class RegisterController {
@RequestMapping(method=RequestMethod.GET)
public String viewRegistration(Model model){
User user = new User();
model.addAttribute("userForm", user);
List<String> professionList = new ArrayList<>();
professionList.add("Developer");
professionList.add("Designer");
professionList.add("IT Manager");
model.addAttribute("professionList", professionList);
return "Registration";
}
@RequestMapping(method=RequestMethod.POST)
public String processRegistration(@ModelAttribute("userForm") User user, Map<String, Object> Model){
System.out.println("username: " + user.getUsername());
System.out.println("password: " + user.getPassword());
System.out.println("email: " + user.getEmail());
System.out.println("birth date: " + user.getBirthDate());
System.out.println("profession: " + user.getProfession());
return "RegistrationSuccess";
}
}
Now, going to myProject/register results in a 404. I am confused how Spring manages the routing though. There is a web.xml looking like this:
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee" 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">
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/applicationContext.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.htm</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>
30
</session-timeout>
</session-config>
<welcome-file-list>
<welcome-file>redirect.jsp</welcome-file>
</welcome-file-list>
</web-app>
What I think this means is every url with *.htm goes to the dispatcher-servlet:
<?xml version='1.0' encoding='UTF-8' ?>
<!-- was: <?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:p="http://www.springframework.org/schema/p"
xmlns:aop="http://www.springframework.org/schema/aop"
xmlns:tx="http://www.springframework.org/schema/tx"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-4.0.xsd
http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-4.0.xsd">
<bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/>
<!--
Most controllers will use the ControllerClassNameHandlerMapping above, but
for the index controller we are using ParameterizableViewController, so we must
define an explicit mapping for it.
-->
<bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="index.htm">indexController</prop>
</props>
</property>
</bean>
<bean id="viewResolver"
class="org.springframework.web.servlet.view.InternalResourceViewResolver"
p:prefix="/WEB-INF/jsp/"
p:suffix=".jsp" />
<!--
The index controller.
-->
<bean name="indexController"
class="org.springframework.web.servlet.mvc.ParameterizableViewController"
p:viewName="index" />
</beans>
But where do I need to insert some entries to let my RegisterController do work?
回答1:
Since you gave *.htm in url pattern, the dispatcher-servlet will only recognize *.htm requests. Change your web.xml
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
And above your controller, there's no need to give RequestMapping. Your methods will do the work for you
@Controller
public class RegisterController {
And above your method, you should give value so that your method can find it
@RequestMapping(value="/register.htm" method=RequestMethod.GET)
public String viewRegistration(Model model)
回答2:
According to Dispatcher Servlet mapping, your RequestMapping will go to the controller with HttpRequest with .htm extension.. In your case, you have mentioned *.htm
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.htm</url-pattern>
</servlet-mapping>
So in the controller, for the appropriate request, add like this..
@RequestMapping(value="/register.htm" method=RequestMethod.GET)
public String viewRegistration(Model model){
and In the top of the controller, add like this,
@Controller
@RequestMapping("/")
public class RegisterController {
add this in spring.xml,
<context:component-scan base-package="RegisterController" />
including package name of Controller should be there in component-scan base-package...
回答3:
Edit servlet mapping in you web.xml as:
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
And also add below piece of code in your spring configuration file:
<mvc:annotation-driven></mvc:annotation-driven>
<context:component-scan base-package="your.packagename.RegisterController"></context:component-scan>
来源：https://stackoverflow.com/questions/30604072/spring-mapping-url-to-controller-and-view