Question

您好，我是Spring Mvc。

我无法解决这个问题

Dispatcher servlet（servlet-context.xml）

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

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“＆gt;

<!-- 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>

控制器类

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";
  }



}

查看页面（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>

请帮我解决这个问题。

按下保存按钮后，我得到以下url被称为

 http://localhost:8080/model/customer

并抛出上述错误。

Answer 1

您的两种方法已映射到/。但您的网址为/customer。

此外，这没有意义：

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

表单只能有一个操作。如果要调用方法addCustomer，它应该是

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

Answer 2

你也可以使用spring taglib来帮助你生成你的URL。

示例：

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

Answer 3

你的组件扫描标签扫描“com.projects.model”包但你的控制器是在“com.myprojects.controller”包中更改了组件扫描来自

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

到

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

然后试试。

org.springframework.web.servlet.PageNotFound - 找不到带URI的HTTP请求的映射

3 个答案: