SpringMVC servlet映射

Question

我写了一个非常简单的Spring MVC应用程序。我很抱歉，我对Spring MVC比较陌生，所以请耐心等待。

的web.xml文件如下：

<?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> 

我的第一个问题是，我有一个JSP页面下面的代码登录...

<form action="/login" method="post" > 
Username : <input name="username" type="text" /> 
Password : <input name="password" type="password" /> 
<input type="submit" /> 
</form> 

这给出了一个404，但在我的控制器，我已经将控制器映射到/用以下代码登录...

@Controller 
public class LoginController { 

    private static final Logger logger = LoggerFactory.getLogger(LoginController.class); 

    /** 
    * Simply selects the home view to render by returning its name. 
    */ 
    @RequestMapping(value = "/login", method = RequestMethod.POST) 
    public String home(Locale locale, Model model, String username, String password) { 

     if(username.equalsIgnoreCase("david")) 
     { 
      logger.info("Welcome home! the client locale is "+ locale.toString()); 

      Date date = new Date(); 
      DateFormat dateFormat = DateFormat.getDateTimeInstance(DateFormat.LONG, DateFormat.LONG, locale); 

      String formattedDate = dateFormat.format(date); 

      model.addAttribute("serverTime", formattedDate); 

      return "home"; 
     } 
     else 
     { 
      return "void"; 
     } 

    } 

} 

我的理解是@requestm apping应该执行servlet映射而不是在web.xml中，这是正确的吗？如果需要，/WEB-INF/spring/appServlet/servlet-context.xml的值也显示在下面。

<?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="org.david.myapp" /> 



</beans:beans> 

所以我的第一个问题是：servlet映射在web.xml或在控制器类的@RequestMapping做了什么？

第二个问题：构建这个页面的最佳方法是什么？我应该继续添加到webxml吗？我应该为每个网址创建一个控制器吗？我应该为每个url创建一个servlet-context吗？

感谢您的阅读:)

Answer 1

您已经定义<url-pattern>为/，这意味着你的appServlet只会收到请求到根网址。通过将其更改为/*，appServlet将获得所有传入请求。这将工作，但你也可以考虑创建一个特定的loginServlet，它可以映射到url /login/*。

1. 您可以在单个web.xml中定义多个servlet。通过添加更多<servlet-mapping>标签来指定哪个请求将触及每个servlet。
2. 一个servlet可能有许多控制器。通常，一个控制器为你的域的特定部分提供服务，例如PersonController,AddressController等
3. 每个控制器通常处理逻辑上分组在一起的多个URL，例如， /persons/{id}，/persons/search，/persons/add等