沒有爲我的案例的URI找到URI的映射

Question

我的案例是： 我有一個主頁，在我的主頁上有一個名爲「登錄」的鏈接，當我點擊「登錄」鏈接時，瀏覽器顯示「HTTP狀態[404] - [未找到]」。 我想要的是：當我點擊「登錄」鏈接時，我想讓網站直接進入一個新的頁面，這是登錄頁面。謝謝！

homepage.jsp：

<li><a class="drop" href="#">My Account</a> 
 
\t  \t <ul> 
 
\t  \t \t <li><a href="login.html">Sign In</a></li> 
 
\t  \t \t <li><a href="#">Create Account</a>></li> 
 
\t  \t </ul> 
 
\t  </li>

web.xml中：

<?xml version="1.0" encoding="UTF-8"?> 
 
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
 
\t \t xmlns="http://java.sun.com/xml/ns/javaee" 
 
\t \t xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" 
 
\t \t version="3.0"> 
 
    <display-name>TooO</display-name> 
 
    <welcome-file-list> 
 
    <welcome-file>homepage.jsp</welcome-file> 
 
    </welcome-file-list> 
 

 
\t <context-param> 
 
     <param-name>contextConfigLocation</param-name> 
 
     <param-value>/WEB-INF/spring/root-context.xml</param-value> 
 
    </context-param> 
 
    <listener> 
 
     <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> 
 
    </listener> 
 

 
    <servlet> 
 
     <servlet-name>crunchify</servlet-name> 
 
     <servlet-class> 
 
      org.springframework.web.servlet.DispatcherServlet 
 
     </servlet-class> 
 
     <load-on-startup>1</load-on-startup> 
 
    </servlet> 
 
    
 
    <servlet> 
 
     <servlet-name>login</servlet-name> 
 
     <servlet-class> 
 
      org.springframework.web.servlet.DispatcherServlet 
 
     </servlet-class> 
 
     <load-on-startup>1</load-on-startup> 
 
    </servlet> 
 
    <servlet-mapping> 
 
     <servlet-name>login</servlet-name> 
 
     <url-pattern>/login.jsp</url-pattern> 
 
     <url-pattern>/login.html</url-pattern> 
 
     <url-pattern>*.html</url-pattern> 
 
    </servlet-mapping> 
 
    
 
</web-app>

登錄-servlet.xml中：

<?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.xsd 
 
     http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd 
 
     http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.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/jsp/" /> 
 
     <beans:property name="suffix" value=".jsp" /> 
 
    </beans:bean> 
 
    <context:component-scan base-package="com.crunchify.controller" /> 
 
</beans:beans>

LoginController.java：

package com.crunchify.controller; 
 

 
import org.springframework.stereotype.Controller; 
 
import org.springframework.ui.Model; 
 
import org.springframework.web.bind.annotation.ModelAttribute; 
 
import org.springframework.web.bind.annotation.RequestMapping; 
 
import org.springframework.web.bind.annotation.RequestMethod; 
 
    
 
@Controller 
 
public class LoginController { 
 
    @RequestMapping(value = "/login", method = RequestMethod.GET) 
 
    public String init(Model model) { 
 
     model.addAttribute("msg", "Please Enter Your Login Details"); 
 
     return "login"; 
 
    } 
 
    
 
    @RequestMapping(method = RequestMethod.POST) 
 
    public String submit(Model model, @ModelAttribute("loginBean") LoginBean loginBean) { 
 
     if (loginBean != null && loginBean.getUserName() != null & loginBean.getPassword() != null) { 
 
      if (loginBean.getUserName().equals("chandra") && loginBean.getPassword().equals("chandra123")) { 
 
       model.addAttribute("msg", "welcome" + loginBean.getUserName()); 
 
       return "success"; 
 
      } else { 
 
       model.addAttribute("error", "Invalid Details"); 
 
       return "login"; 
 
      } 
 
     } else { 
 
      model.addAttribute("error", "Please enter Details"); 
 
      return "login"; 
 
     } 
 
    } 
 
}

Here is the screenshot of the 404 error web page

Answer 1

您初始化 Spring Servlets - 這不是它的工作方式。

讓它只有一個SpringDispatcher並分配很簡單

<servlet> 
    <servlet-name>crunchify</servlet-name> 
    <servlet-class> 
     org.springframework.web.servlet.DispatcherServlet 
    </servlet-class> 
    <load-on-startup>1</load-on-startup> 
</servlet> 


<servlet-mapping> 
    <servlet-name>crunchify</servlet-name> 
    <url-pattern>/crunch/*</url-pattern> 
</servlet-mapping> 

現在所有來電事端像緊縮URL模式/ *將被引導到春天。

鑑於你@RequestMapping在你的LoginController

@RequestMapping(value = "/login", method = RequestMethod.GET) 
public String init(Model model) { 

現在，這將導致進入以下網址以實際到達登錄控制器：

/crunch/login 
^ ^
    |  will be routed into LoginController by Spring 
    will be routed to Spring by Servlet Container 

現在修改相應的登錄鏈接：

<ul> 
     <li><a href="crunch/login">Sign In</a></li> 
     <li><a href="#">Create Account</a>></li> 
    </ul> 

再試一次。