SpringMVC servlet映射

Question

我寫了一個非常簡單的Spring MVC應用程序。我很抱歉，我對Spring MVC比較陌生，所以請耐心等待。

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> 

我的第一個問題是，我有一個JSP頁面下面的代碼登錄...

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

這給出了一個404，但在我的控制器，我已經將控制器映射到/用以下代碼登錄...

@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應該執行servlet映射而不是在web.xml中，這是正確的嗎？如果需要，/WEB-INF/spring/appServlet/servlet-context.xml的值也顯示在下面。

<?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> 

所以我的第一個問題是：servlet映射在web.xml或在控制器類的@RequestMapping做了什麼？

第二個問題：構建這個頁面的最佳方法是什麼？我應該繼續添加到webxml嗎？我應該爲每個網址創建一個控制器嗎？我應該爲每個url創建一個servlet-context嗎？

感謝您的閱讀:)

Answer 1

您已經定義<url-pattern>爲/，這意味着你的appServlet只會收到請求到根網址。通過將其更改爲/*，appServlet將獲得所有傳入請求。這將工作，但你也可以考慮創建一個特定的loginServlet，它可以映射到url /login/*。

1. 您可以在單個web.xml中定義多個servlet。通過添加更多<servlet-mapping>標籤來指定哪個請求將觸及每個servlet。
2. 一個servlet可能有許多控制器。通常，一個控制器爲你的域的特定部分提供服務，例如PersonController,AddressController等
3. 每個控制器通常處理邏輯上分組在一起的多個URL，例如， /persons/{id}，/persons/search，/persons/add等