如何连接jsp和控制器
我想用Java和Spring编写一个简单的web应用程序。因此,我已经设置了应用程序,但我有一个问题。这就是 web.xml如何连接jsp和控制器,jsp,spring-mvc,jakarta-ee,Jsp,Spring Mvc,Jakarta Ee,我想用Java和Spring编写一个简单的web应用程序。因此,我已经设置了应用程序,但我有一个问题。这就是 web.xml <?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
<?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>SpringMVCApp</display-name>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>springmvcapp</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>springmvcapp</servlet-name>
<url-pattern>*.html</url-pattern>
</servlet-mapping>
</web-app>
现在,当我尝试单击并打开registorZone.jsp时,出现了一个问题
除此之外,我想知道我使用的连接jsp和控制器的方法是否正确
谢谢这是您告诉调度程序您的spring上下文在哪里的方式:
<servlet>
<servlet-name>springmvcapp</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
classpath:springmvcapp-servlet.xml
</param-value>
</init-param>
</servlet>
springmvcapp
org.springframework.web.servlet.DispatcherServlet
1.
上下文配置位置
类路径:springmvcapp-servlet.xml
像这样在控制器中添加@RequestMapping
@RequestMapping(value = "/registrazione")
public ModelAndView registrazione(HttpServletRequest request, HttpServletResponse response) throws Exception {
ModelAndView model = new ModelAndView("registrazione");
model.addObject("msg", "hello world");
return model;
}
我在Controlare的名称上插入了@RequestMapping(“/registrazione”)代码。它是有效的。但我的解决方案是否正确?这取决于url的配置方式。如果它是基本url,则在控制器级别使用它,否则在方法级别使用它。如果有效,请接受它作为答案。
package com.springmvcapp.controller;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.multiaction.MultiActionController;
@Controller
//@RequestMapping("/RegistrazioneController")
public class RegistrazioneController extends MultiActionController{
public ModelAndView registrazione(HttpServletRequest request, HttpServletResponse response) throws Exception {
ModelAndView model = new ModelAndView("registrazione");
model.addObject("msg", "hello world");
return model;
}
@RequestMapping(value = "/sendDati")
public String getDataFromForum(@RequestParam String nome){
System.out.println("pippo");
return nome;
}
public String getViewName() throws Exception {
return "registrazione";
}
public String getPageName() throws Exception {
return "registrazione";
}
}
<servlet>
<servlet-name>springmvcapp</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
classpath:springmvcapp-servlet.xml
</param-value>
</init-param>
</servlet>
@RequestMapping(value = "/registrazione")
public ModelAndView registrazione(HttpServletRequest request, HttpServletResponse response) throws Exception {
ModelAndView model = new ModelAndView("registrazione");
model.addObject("msg", "hello world");
return model;
}