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我在/target/m2e-wtp/web-resources中有 maven 和我的 JSP insert.jsp 生成的 Spring 项目，它看起来像这样：

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!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>Library</title>
</head>
<body>
    <h1>Insert record</h1>
    <form action="InsertBook" method="POST">
        Title: <input type="text" name="title"/><br>
        Author: <input type="text" name="author"/><br>
        Category: <select name="category">
            <c:forEach var="cat" items="${categories}">
                <option>${cat}</option>
            </c:forEach>
        </select>
        <input type="submit" value="insert"/>
    </form>
</body>

控制器位于/src/main/java/com/mypackage/web/InsertBook.java和代码在这里：

package com.mypackage.web;

import java.io.IOException;

import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;

/**
 * Servlet implementation class InsertBook
 */
public class InsertBook extends HttpServlet {
@Override
   protected void doGet(HttpServletRequest request, HttpServletResponse response){
     logger.info("GOT IT.");
   }
private static final Logger logger = LoggerFactory.getLogger(HomeController.class);

@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    if (request.getCharacterEncoding() == null) {
        request.setCharacterEncoding("UTF-8");
    }

    String title = request.getParameter("title");
    String author = request.getParameter("author");
    logger.info("GOT IT.");


    RequestDispatcher rd = request.getRequestDispatcher("register");
    rd.forward(request, response);
}
}

servlet-context.xml 代码在这里：

<?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/views/" />
    <beans:property name="suffix" value=".jsp" />
</beans:bean>

<context:component-scan base-package="com.mypackage.web" />
<resources location="/resources/**" mapping="/src/webapp/resources"/>

在尝试访问此 JSP 时，我收到了来自 tomcat 服务器的消息：

INFO : com.mypackage.web.HomeController - Welcome home! The client locale is cs.
WARN : org.springframework.web.servlet.PageNotFound - No mapping found for HTTP request         with URI [/web/<c:url value=] in DispatcherServlet with name 'appServlet'

谁能告诉我，如何通过控制器访问在 JSP 中填写到表单中的值，然后将它们打印出来 - 例如通过记录器？

4

1 回答 1

1

这可能是您需要的后期，但您应该实现一个控制器并使用您在 web.xml 中定义的 Spring DispatcherServlet。如果您继续使用 xml 应用程序上下文配置，那么您将在 applicationContext 文件中定义您的控制器 bean 和其他 bean（例如 DAO、服务或其他组件）以及 URLViewResolver 之类的项目。（Spring 3 还允许应用程序上下文的 Java 配置方法）。 

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns="http://java.sun.com/xml/ns/javaee"
    xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    id="WebApp_ID" version="2.5">
    <display-name>SpringMVC</display-name>
    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>

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

...

<bean id="viewResolver"
    class="org.springframework.web.servlet.view.UrlBasedViewResolver">
    <property name="viewClass"
        value="org.springframework.web.servlet.view.JstlView" />
    <property name="prefix" value="/WEB-INF/jsp/" />
    <property name="suffix" value=".jsp" />
</bean>

<bean id="defaultMyDatasource" class="org.apache.commons.dbcp.BasicDataSource"
    destroy-method="close">
    <property name="driverClassName" value="${myDriverClass}" />
    <property name="url" value="${myURL}" />
    <property name="username" value="${myLogin}" />
    <property name="password" value="${myPassword}" />
</bean>
 ...

事情是让 Spring（以及您决定使用的任何 ORM）来简化您的 Java 代码。我建议查看ViralPatel提供的教程或 Craig Walls 的 Spring in Action 书中的示例。这些帮助我们的一些新开发人员更好地了解了 Spring。

于 2013-04-18T22:28:25.890 回答