Статус: 404-Не найдено

Question

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
    http://www.springframework.org/schema/beans/spring-beans-2.5.xsd">

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

Диспетчер-сервлет-контекст. xml при запуске проекта выдается ошибка 404

Web . xml Это мой веб XML файл, помогающий мне настроить диспетчер-сервлет в качестве фронт-контроллера, оба эти файла находятся в WebContent \ WEB-INF \ web. xml папка

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
        xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" version="3.1">
            <listener>
                <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
            </listener>

            <context-param>
                <param-name>contextConfigLocation</param-name>
                <param-value>/WEB-INF/dispatcher-servlet-context.xml</param-value>
            </context-param>

            <servlet>
                <servlet-name>dispatcher-servlet</servlet-name>
                <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
                <init-param>
                    <param-name>contextConfigLocation</param-name>
                    <param-value></param-value>
                </init-param>
                <load-on-startup>1</load-on-startup>
            </servlet>

            <servlet-mapping>
                <servlet-name>dispatcher-servlet</servlet-name>
                <url-pattern>/*</url-pattern>
            </servlet-mapping>
            <welcome-file-list>
                <welcome-file>/index.htm</welcome-file>
            </welcome-file-list>

        </web-app>

Answer 1

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
        xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" version="3.1">
            <listener>
                <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
            </listener>
            <servlet>
                <servlet-name>dispatcher-servlet</servlet-name>
                <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
                <init-param>
                    <param-name>contextConfigLocation</param-name>
                    <param-value>/WEB-INF/dispatcher-servlet.xml</param-value>
                </init-param>
                <load-on-startup>1</load-on-startup>
            </servlet>
            <servlet-mapping>
                <servlet-name>dispatcher</servlet-name>
                <url-pattern>/*</url-pattern>
            </servlet-mapping>          
           <welcome-file-list>
                <welcome-file>/index.htm</welcome-file>
            </welcome-file-list>
        </web-app>

попробуйте это и измените файл сервлета на dispatcher-servlet. xml

<`context`-param> is at different level from dispatcher-servlet. what you do is you put your servlet inside it.