2013-05-11 81 views
1

嗨,我是新的Spring Mvc。org.springframework.web.servlet.PageNotFound - 沒有找到HTTP請求與URI的映射

,我不不能解決這個問題

調度的servlet(servlet的context.xml中)

<?xml version="1.0" encoding="UTF-8"?> 

http://www.springframework.org/schema/mvc/spring-mvc的.xsd http://www.springframework.org/schema/beanshttp://www.springframework.org/schema/beans/spring-beans.xsd http://www.springframework.org/schema/contexthttp://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/views/" /> 
    <beans:property name="suffix" value=".jsp" /> 
</beans:bean> 

<context:component-scan base-package="com.projects.model" /> 

<beans:bean id="dataSource" 
    class="org.springframework.jdbc.datasource.DriverManagerDataSource"> 

    <beans:property name="driverClassName" value="com.mysql.jdbc.Driver" /> 
    <beans:property name="url" value="jdbc:mysql://localhost:3306/customerdb" /> 
    <beans:property name="username" value="root" /> 
    <beans:property name="password" value="root" /> 
</beans:bean> 

<beans:bean id="sessionFactory" class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean"> 
    <beans:property name="dataSource" ref="dataSource" /> 
    <beans:property name="packagesToScan" value="com.projects.model" /> 

    <beans:property name="hibernateProperties"> 
     <beans:props> 
      <beans:prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</beans:prop> 
      <beans:prop key="hibernate.show_sql">true</beans:prop> 
     </beans:props> 
    </beans:property> 

    </beans:bean> 


<beans:bean id="transactionManager" class="org.springframework.orm.hibernate3.HibernateTransactionManager"> 
    <beans:property name="sessionFactory" ref="sessionFactory" /> 
</beans:bean> 

的web.xml

<?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> 

控制器類

package com.myprojects.controller; 


import java.util.Locale; 

import org.slf4j.Logger; 
import org.slf4j.LoggerFactory; 
import org.springframework.stereotype.Controller; 
import org.springframework.ui.Model; 
import org.springframework.validation.BindingResult; 
import org.springframework.web.bind.annotation.ModelAttribute; 
import org.springframework.web.bind.annotation.RequestMapping; 
import org.springframework.web.bind.annotation.RequestMethod; 
import org.springframework.web.servlet.ModelAndView; 

import com.projects.model.CustomerDao; 
import com.projects.model.Customer; 


@Controller 
public class HomeController { 

private static final Logger logger = LoggerFactory.getLogger(HomeController.class); 

/** 
* Simply selects the home view to render by returning its name. 
*/ 

private CustomerDao dao; 



@RequestMapping(value = "/", method = RequestMethod.GET)   
public String customer(Locale locale, Model model) { 
    logger.info("Welcome home! The client locale is {}.", locale); 

    return "home"; 
} 

@RequestMapping(value = "/", method = RequestMethod.POST) 
    public String addCustomer(@ModelAttribute("customer") Customer customer, BindingResult result) { 

    dao.save(customer); 

    return "home"; 
    } 



} 

查看頁面(針對home.jsp)

<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %> 
<%@ page session="false" %> 
<html> 
<head> 
<title>Home</title> 
</head> 
<body> 


<P> The time on the server is ${serverTime}. </P> --%> 

<form action="addCustomer"> 
Cust_Id : 
<input type="text"> 
<br> 
Name : 
<input type="text"> 
<br> 
Age : 
<input type="text"> 

<input type="submit" value="Save"> 
</form> 
</body> 
</html> 

請幫我解決這個問題。

後,我按保存按鈕,我得到以下網址被稱爲

http://localhost:8080/model/customer 

,並拋出了上述的錯誤。

回答

0

您的兩種方法映射到/。但是你的網址是/customer

而且,這沒有任何意義:

<form action="customer" action="addCustomer"> 

表單只能有一個動作。如果你想要調用addCustomer方法,它應該是

<form action="<c:url value='/'/>" method="post"> 
0

你也可以使用spring taglib來幫助你生成你的URL。

實施例:

<spring:url value="/model/customer" var="url" /> 
<a href="${url}">Link</a> 
0

你的部件掃描標籤掃描 「com.projects.model」 包但控制器處於 「com.myprojects.controller」 包從

更改組件掃描
<context:component-scan base-package="com.projects.model" /> 

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

然後嘗試。

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