Spring Web应用程序在tomcat-7.0.37上给出404错误

Question

我制作了Spring web应用程序示例项目。我做了所有的基本配置，基于java的配置和servlet 3.0web.xml.但在所有这些配置后，当我打网址它给我404错误。会有什么问题。有人能帮帮我吗？

网状物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_3_0.xsd"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee;
    http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
    id="WebApp_ID" version="3.0">
    <display-name>SpringPracs</display-name>
    <!-- <welcome-file-list> -->
    <!-- <welcome-file>/jspfiles/index.jsp</welcome-file> -->
    <!-- </welcome-file-list> -->

    <servlet>
        <servlet-name>home</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextClass</param-name>
            <param-value>org.springframework.web.context.support.AnnotationConfigWebApplicationContext
            </param-value>
        </init-param>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>home.practise.configuration.SpringConfigs</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
        <servlet-mapping>
            <servlet-name>home</servlet-name>
            <url-pattern>/practise/*</url-pattern>
        </servlet-mapping>

    </servlet>
</web-app>

配置java类：

package home.practise.configuration;

import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.ComponentScan;
import org.springframework.context.annotation.Configuration;
import org.springframework.context.annotation.PropertySource;
import org.springframework.web.servlet.ViewResolver;
import org.springframework.web.servlet.config.annotation.EnableWebMvc;
import org.springframework.web.servlet.view.InternalResourceViewResolver;

@Configuration
@ComponentScan(basePackages = "home.practise.application")
@EnableWebMvc
@PropertySource("classpath:practise.properties")
public class SpringConfigs {
    @Bean
    public ViewResolver viewResolver() {
        InternalResourceViewResolver internalResourceViewResolver = new InternalResourceViewResolver();
        internalResourceViewResolver.setPrefix("/jspfiles/");
        internalResourceViewResolver.setSuffix(".jsp");
        return internalResourceViewResolver;
    }
}

控制器：

package home.practise.application;

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.servlet.ModelAndView;

@Controller
@RequestMapping("homeController")
public class HomeController {
    @RequestMapping("homePage.ht")
    public ModelAndView takeMeToHomePage(){
        ModelAndView modelAndView = new ModelAndView("index");
        return modelAndView;
    }
}

当我试图访问这个网址

http://localhost:8080/SpringPracs/practise/homeController/homePage.ht

它给我404错误。虽然当我直接访问jsp页面时，它显示该页面。

Answer 1

首先，当Spring参考手册中的所有示例使用以/开头的绝对URL时，使用相对URL。我认为它可以工作，但帮你自己一个忙，并遵循Spring的例子...

但实际问题是您将SpringDispatcherServlet映射到了/practice/*。这是完全正确的，但是您还必须声明您的控制器以响应/practice/ ：

@Controller
@RequestMapping("/practise/homeController")
public class HomeController {
    @RequestMapping("/homePage.ht")
    public ModelAndView takeMeToHomePage(){
    ...
    }
}

编辑：

或者，您可以将控制器映射到/homeController，并将dispatcher servlet映射到/