[英]Get Spring view without mapping a Controller to a URL pattern
[英]Spring Mapping Url to Controller and View
我在Netbeans 8.0.2中有标准的Spring 4.x文件。 我在controllers
-package中添加了一个控制器RegisterController.java
。 我还添加了一个模型User
,其中包含有关该用户的一些基本信息。 接下来,我制作了2个.jsp
文件Registration.jsp
和RegistrationSuccess.jsp
。
这是我的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";
}
}
现在,转到myProject/register
结果为404。但是我很困惑Spring如何管理路由。 有一个如下所示的web.xml :
<?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>
我认为这意味着带*.htm
每个URL都将进入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>
但是我需要在哪里插入一些条目来让RegisterController正常工作?
由于您以网址格式提供了* .htm,因此分派器servlet将仅识别* .htm请求。 更改您的web.xml
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
在您的控制器上方,无需提供RequestMapping
。 您的方法将为您效劳
@Controller
public class RegisterController {
在您的方法之上,您应该给予价值,以便您的方法可以找到它
@RequestMapping(value="/register.htm" method=RequestMethod.GET)
public String viewRegistration(Model model)
根据Dispatcher Servlet映射,您的RequestMapping将以扩展名为.htm的HttpRequest进入控制器。在您的情况下,您提到的* .htm
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.htm</url-pattern>
</servlet-mapping>
因此,在控制器中,对于适当的请求,请像这样添加。
@RequestMapping(value="/register.htm" method=RequestMethod.GET)
public String viewRegistration(Model model){
然后在控制器顶部添加这样的内容,
@Controller
@RequestMapping("/")
public class RegisterController {
在spring.xml中添加它,
<context:component-scan base-package="RegisterController" />
包括Controller的软件包名称应在component-scan基本软件包中存在...
将web.xml
servlet映射编辑为:
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
并且在您的spring配置文件中添加以下代码:
<mvc:annotation-driven></mvc:annotation-driven>
<context:component-scan base-package="your.packagename.RegisterController"></context:component-scan>
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