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很新，所以请耐心等待。

web.xml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页面可以使用以下代码进行登录...

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

这给了404但是在我的Controller中，我已经将控制器映射到/ login并使用下面的代码...

@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@requestmapping应该执行servlet映射而不是web.xml，这是正确的吗？如果需要，还会在下面显示/WEB-INF/spring/appServlet/servlet-context.xml的值。

<?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映射是在web.xml中完成的，还是在控制器类中的@requestmapping中完成的？

第二个问题：如果我要继续添加webxml，那么构建此页面以获得更多页面的最佳方法是什么？我应该为每个网址创建一个控制器吗？我应该为每个网址创建一个servlet上下文吗？

感谢阅读：）

Answer 1

您已将<url-pattern>定义为/，这意味着您的appServlet只会收到对根网址的请求。通过将其更改为/*，appServlet将获得所有传入的请求。这可行，但您也可以考虑创建一个可以映射到网址loginServlet的特定/login/*。

您可以在一个web.xml中定义多个servlet。通过添加更多<servlet-mapping>标记来指定每个servlet的命令。
servlet可能有许多控制器。通常，一个控制器服务于您域的特定部分，例如PersonController，AddressController等
每个控制器通常处理几个逻辑分组在一起的URL，例如/persons/{id}，/persons/search，/persons/add等

SpringMVC servlet映射

1 个答案: