Spring MVC 404未找到错误

时间：2016-01-12 04:05:39

标签： java spring jsp spring-mvc

我正在尝试在Spring MVC中运行一个项目。这是代码

的index.jsp 

IplImage* image = cvCreateImage(cvSize(mat.cols, mat.rows), 8, 3);
IplImage ipltemp = mat;
cvCopy(&ipltemp, image);

RegistrationController.java 

<%@page contentType="text/html" pageEncoding="UTF-8"%>
<%@taglib prefix="form" uri="http://www.springframework.org/tags/form" %>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
    "http://www.w3.org/TR/html4/loose.dtd">

<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>Welcome to Spring Web MVC project</title>
    </head>

    <body>
        <h1>Spring 3 Register!</h1>
        <a href="register.htm">click</a>
        <form:form action="${pageContext.request.contextPath}/register.htm" method="POST" modelAttribute="userForm">
            <table>
                <tr>
                    <td colspan="2" align="center">Spring MVC Form Demo - Registration</td>
                </tr>
                <tr>
                    <td>User Name</td>
                    <td><form:input path="username" /></td>
                </tr>
                <tr>
                    <td>Password</td>
                    <td><form:password path="password" /></td>
                </tr>
                <tr>
                    <td>Email</td>
                    <td><form:input path="email" /></td>
                </tr>
                <tr>
                    <td>BirthDate (mm/dd/yyyy)</td>
                    <td><form:input path="birthDate" /></td>
                </tr>
                <tr>
                    <td>Profession</td>
                    <td><form:select path="profession" items="${professionList}" /></td>
                </tr>
                <tr>
                    <td colspan="2" align="center"><input type="submit" value="Register" /></td>
                </tr>

            </table>
        </form:form>
    </body>
</html>

User.java 

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

package RegisterInfo;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;

/**
 *
 * @author Harshit Shrivastava
 */
import RegisterInfo.model.User;

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.ModelAttribute;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.ui.Model;

@Controller
@RequestMapping(value = "/register")
public class RegistrationController {

    @RequestMapping(method = RequestMethod.GET)
    public String viewRegistration(Model model)
    {
        User userForm = new User();
                model.addAttribute("userForm", new User());

                /*List<String> professionList = new ArrayList();
                professionList.add("Developer");
                professionList.add("Designer");
                professionList.add("IT Manager");
                model.put("professionList", professionList);*/
        return "index";
    }

    @RequestMapping(method = RequestMethod.POST)
    public String processRegistration(@ModelAttribute("userForm") User user, Map<String, Object> model)
    {       
        System.out.println("Username : " + user.getUserName());
        model.put("userForm", new User());
        return "index";
    }
}

的web.xml 

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

package RegisterInfo.model;

/**
 *
 * @author Harshit Shrivastava
 */
import java.util.Date;

public class User {
    private String username;
    private String password;
    private String email;
    private Date birthDate;
    private String profession;

    public String getUserName()
    {
        return username;
    }
    public void setUserName(String username)
    {
        this.username = username;
    }
        public String getPassword()
    {
        return password;
    }
    public void setPassword(String password)
    {
        this.password = password;
    }
        public String getEmail()
    {
        return email;
    }
    public void setEmail(String email)
    {
        this.email = email;
    }
        public Date getBirthDate()
    {
        return birthDate;
    }
    public void setBirthDate(Date birthDate)
    {
        this.birthDate = birthDate;
    }
        public String getProfession()
    {
        return profession;
    }
    public void setProfession(String profession)
    {
        this.profession = profession;
    }
}

调度-servlet.xml中

<?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/dispatcher-servlet.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>index.htm</welcome-file>
    </welcome-file-list>
</web-app>

项目结构：

enter image description here

在上面的程序中，我总是得到<?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p" xmlns:aop="http://www.springframework.org/schema/aop" xmlns:tx="http://www.springframework.org/schema/tx" xmlns:context="http://www.springframework.org/schema/context" xmlns:mvc="http://www.springframework.org/schema/mvc" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.1.xsd http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.1.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.1.xsd http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd "> <bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/> <context:component-scan base-package="SpringRegister" /> <mvc:annotation-driven /> <!-- 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> </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" /> </beans> 。我试过了

http://localhost:8080/RegisterInfoMaven/index

http://localhost:8080/RegisterInfoMaven/index.htm

都不起作用。

当我删除web.xml＆amp;来自项目的dispatcher-servlet.xml，然后404 not found status加载正常，但是当我放http://localhost:8080/RegisterInfoMaven/index.jsp时，它又开始给出web.xml & dispatcher-servlet.xml错误。

404 not found个文件有什么问题？

2 个答案:

答案 0 :(得分：5)

首先。 web.xml文件中的映射：

<servlet-mapping>
   <servlet-name>dispatcher</servlet-name>
   <url-pattern>*.html</url-pattern>
</servlet-mapping>

表示，该调度程序servlet将仅处理以.html结尾的请求。将其更改为下一个：

<servlet-mapping>
   <servlet-name>dispatcher</servlet-name>
   <url-pattern>/</url-pattern>
</servlet-mapping>

之后，调度程序servlet将处理所有传入的请求。

二。解析器来自dispatcher-servlet.xml的声明：

<bean id="viewResolver"
      class="org.springframework.web.servlet.view.InternalResourceViewResolver"
      p:prefix="/WEB-INF/jsp/"
      p:suffix=".jsp" />

表示当Controller的方法返回String值时，解析器会在任意值之前添加/WEB-INF/jsp并在.jsp之后添加jsp。然后将查找具有此名称的文件。因此，您需要在WEB-INF目录中创建@RequestMapping(value = "/index")目录并将jsp文件放在那里。

第三。当您输入类似 http://localhost:8080/index 的内容时，spring会尝试查找标有@RequestMapping(value = "/index")的Controller 类或方法，但不是文件 index.jsp。

因此，您必须声明Controller的方法，标有NtQueryInformationProcess。您的请求字符串看起来会像 http://localhost:8080/index 一样，没有任何扩展名。

答案 1 :(得分：0)

尝试在web.xml中编写您的网址：

  <servlet-mapping>
    <servlet-name>dispatcher</servlet-name>
    <url-pattern>/*.htm</url-pattern>

和

    <welcome-file-list>
    <welcome-file>/index.htm</welcome-file>
   </welcome-file-list>

并尝试替换它：

  <form:form action="${pageContext.request.contextPath}/register.htm" method="POST" modelAttribute="userForm"

通过此帖子发布正确的路径${pageContext.request.contextPath}/register：

  <form:form action="${pageContext.request.contextPath}/register" method="POST" modelAttribute="userForm">

并添加值User userForm代替User user，如下所示：

   public String processRegistration(@ModelAttribute("userForm") User userForm, Map<String, Object> model)

并更改base-package之类的内容：

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