[英]Get values from JSP with Spring MVC
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中 ,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);
}
}
还有servlet-context.xml代码:
<?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"/>
尝试访问此JSP时,出现了消息表格tomcat服务器:
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'
有人可以告诉我,如何通过控制器访问JSP中填写到表单中的值,然后将它们打印出来-例如通过记录器吗?
这可能在您的需求中很晚,但是您应该实现一个Controller并使用在web.xml中定义的Spring DispatcherServlet。 如果继续使用xml应用程序上下文配置,则可以在applicationContext文件中定义控制器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还允许针对应用程序上下文使用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)简化Java代码。 我建议您阅读ViralPatel提供的教程或Craig Walls的《 Spring in Action》一书中的示例。 这些帮助我们的一些较新的开发人员更好地了解了Spring。
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