Question

当我尝试运行弹簧网络应用程序时，我收到以下错误：

org.springframework.web.servlet.DispatcherServlet noHandlerFound
WARNING: No mapping found for HTTP request with URI [/HibwenateWeb/index] in DispatcherServlet with name 'customerdispatcher

我无法加载主页。我经历了很多文章，但是在添加我没有得到nohandler错误但仍然没有加载主页之后，大多数人都建议在调度程序中添加<mvc:default-servlet-handler />。我想知道这个问题。任何人都可以帮我解决这个问题吗？

我的控制器

package com.springforbeginners.controller;

import java.util.Map;

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

@Controller
public class CustomerController {
    @RequestMapping(value="/index")
    public String listCustomers(ModelAndView mav) {
        System.out.println("Hi i am in controller");
        //map.put("customer", new Customer());
        //map.put("customerList", customerService.listCustomer());
        mav.setViewName("customer");
        return "customer";
    }

}

customerdispatcher-servlet.xml中

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:aop="http://www.springframework.org/schema/aop" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:tx="http://www.springframework.org/schema/tx" xmlns:mvc="http://www.springframework.org/schema/mvc"
    xsi:schemaLocation="http://www.springframework.org/schema/beans  

        http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
        http://www.springframework.org/schema/aop 
        http://www.springframework.org/schema/aop/spring-aop-3.1.xsd
        http://www.springframework.org/schema/context 
        http://www.springframework.org/schema/context/spring-context-3.1.xsd 
        http://www.springframework.org/schema/tx 
        http://www.springframework.org/schema/tx/spring-tx-3.1.xsd
         http://www.springframework.org/schema/mvc
        http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd">
    <mvc:annotation-driven />
    <context:annotation-config />
    <context:component-scan base-package="com.springforbeginners" />

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

的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
    <servlet>
        <servlet-name>customerdispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>customerdispatcher</servlet-name>
        <url-pattern>/index</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>30</session-timeout>
    </session-config>

</web-app>

Answer 1

我在你的结构中写了一个非常简单的prj，就在这里。 pom依赖

<dependencies>
        <dependency>
            <groupId>javax.servlet</groupId>
            <artifactId>servlet-api</artifactId>
            <version>2.5</version>
        </dependency>
        <dependency>
            <groupId>org.springframework</groupId>
            <artifactId>spring-webmvc</artifactId>
            <version>4.2.5.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>javax.servlet</groupId>
            <artifactId>jstl</artifactId>
            <version>1.2</version>
        </dependency>
    </dependencies>

WEB-INF下的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
    <display-name>MyFirstSpringMVC</display-name>
    <servlet>
        <servlet-name>customerdispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>customerdispatcher</servlet-name>
        <url-pattern>/index/*</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>30</session-timeout>
    </session-config>
</web-app>

WEB-INF

下的customerdispatcher-servlet.xml

<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:mvc="http://www.springframework.org/schema/mvc"
    xsi:schemaLocation="
        http://www.springframework.org/schema/beans     
        http://www.springframework.org/schema/beans/spring-beans-3.2.xsd
        http://www.springframework.org/schema/mvc 
        http://www.springframework.org/schema/mvc/spring-mvc-3.2.xsd
        http://www.springframework.org/schema/context 
        http://www.springframework.org/schema/context/spring-context-3.2.xsd">
    <context:component-scan base-package="com.*"/>
    <bean
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix">
            <value>/WEB-INF/jsp/</value>
        </property>
        <property name="suffix">
            <value>.jsp</value>
        </property>
    </bean>
    <!-- <mvc:annotation-driven /> --> 
</beans>

/ WEB-INF / jsp /

下的customer.jsp

<%@ page language="java" contentType="text/html; charset=UTF-8"
    pageEncoding="UTF-8"%>
<!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>Insert title here</title>
</head>
<body>
this is customer.jsp
</body>
</html>

CustomerController

package com.springforbeginners.controller;

import java.util.Map;

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

@Controller
public class CustomerController {

    @RequestMapping(value="/index")
    public String listCustomers(ModelAndView mav) {
        System.out.println("Hi i am in controller");
        return "customer";
    }
}

这是结果。 http://imgur.com/P51TSbk

由于您以默认方式声明customerdispatcher-servlet.xml（“dispatcher name”+“ - servlet.xml”），因此您无需添加spring contextloader或设置init-param以使调度程序servlet工作。

更多信息http://docs.spring.io/spring/docs/current/spring-framework-reference/html/mvc.html#mvc-servlet

希望这可以帮到你。

顺便说一下，在将url-pattern声明为根路径时

    <servlet-mapping>
        <servlet-name>customerdispatcher</servlet-name>
        <url-pattern>/</url-pattern> <= it's slash only, not /*
    </servlet-mapping>

这种微妙的差异使我花了一个小时才找到它......

Answer 2

您必须将以下元素添加到您的web.xml。

<!-- The definition of the Root Spring Container -->
<context-param>  
    <param-name>contextConfigLocation</param-name> 
    <!-- provide the path for your customerdispatcher-servlet.xml -->
    <param-value>/WEB-INF/customerdispatcher-servlet.xml</param-value>  
</context-param>  

<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>  
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>  
</listener>

此外，您必须将url-pattern元素更改为：

<url-pattern>/</url-pattern>

警告：找不到包含URI

2 个答案: