Question

我在一个控制器中有2个请求映射，而第二个请求映射无法找到（在DispatcherServlet中找不到带有URI [/LeeSong_login_v2/verify.htm]的HTTP请求的映射，名称为“dispatcher”）

我的调度员 - servlet：

<mvc:annotation-driven/> 
<context:component-scan base-package="controller" />

<bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/>
<!-- <bean class="com.LeeSong.controller.loginController" id="loginController" /> -->

<!--
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>
            <prop key="login.htm">loginController</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" />

<bean name="loginController"
      class="org.springframework.web.servlet.mvc.ParameterizableViewController"
      p:viewName="login" />

控制器：

@Controller
public class loginController{  

@RequestMapping(value = "/login", method = RequestMethod.GET)
public ModelAndView login(){
    return new ModelAndView("login");
}

@RequestMapping(value = "/verify", method = RequestMethod.POST)
public ModelAndView verify(
        @RequestParam("uname") String userid,
        @RequestParam("pass") String pwd) {
    ModelAndView mv = new ModelAndView("home");
    System.out.println("begin logging...");

    try {
        List<Users> lst = LeeSongDAO.layDS();
        System.out.println("INPUT DETAILS: user: " + userid + " pass: " + pwd);

        for (Users lst1 : lst) {
            System.out.println("user: " + lst1.getUsername() + " pass: " + lst1.getPassword());
            if (userid.equals(lst1.getUsername())) {
                if (pwd.equals(lst1.getPassword())) {
                    System.out.println("credentials correct!");
                    return mv;
                }
            }
        }
        System.out.println("wrong credentials!");
    } catch (Exception e) {
    }

    return new ModelAndView("wronglogin");
}

}

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>

我仍然是春天新手，我可能会在这里做一些新手错误。任何帮助将不胜感激。谢谢！在发布后约2小时内，我试着回答一些评论，需要休息一下。

Answer 1

好吧，菜鸟错了。调度员中的这一行：

<context:component-scan base-package="controller" />

＆＃34;控制器＆＃34;必须特定于您的控制器的地址。就我而言：＆＃34; com.springtest1.controller＆＃34;

在DispatcherServlet中找不到带有URI []的HTTP请求的映射，名称为“dispatcher”

1 个答案: