java - Spring Web 应用程序在 tomcat-7.0.37 上出现 404 错误

Question

我制作了 Spring Web 应用程序示例项目。我使用基于 java 的配置和 servlet 3.0 web.xml 完成了所有基本配置。但在完成所有这些配置后，当我点击 url 时，它给了我 404 错误。可能是什么问题呢。有人可以帮我吗？

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/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;
    }
}

Controller :

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;
    }
}

当我尝试访问此网址时

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

它给了我 404 错误。但是当我直接访问jsp页面时，它显示该页面。

Answer 1

当 Spring 引用手册中的所有示例都使用以 / 开头的绝对 URL 时，首先使用相对 URL。我认为它可以工作，但帮自己一个忙并遵循 Spring 示例......

但实际问题是您将 Spring DispatcherServlet 映射到 /practise/*。这是完全正确的，但是您还必须声明您的 Controller 来响应 /practise/... :

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

编辑:

作为替代方案，您可以将 Controller 映射到 /homeController 并将调度程序 servlet 映射到 /