java—如何从xml文件迁移到编程配置

1yjd4xko  于 2021-07-13  发布在  Java
关注(0)|答案(0)|浏览(128)

我从应用程序开始的地方开始上课:

@SpringBootApplication
public class MainCore extends SpringBootServletInitializer{

    public static void main(String[] args){
        SpringApplication.run(MainCore.class, args);
    }

}

以及这两个xml文件:
web.xml文件:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd"
         id="WebApp_ID"
         version="4.0">

    <display-name>Fooname</display-name>

    <servlet>
        <servlet-name>fooname</servlet-name>
        <servlet-class>
            org.springframework.web.servlet.DispatcherServlet
        </servlet-class>
        <load-on-startup>0</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>fooname</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

    <session-config>
        <session-timeout>1</session-timeout>
    </session-config>

</web-app>

foonName-servlet.xml:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:mvc="http://www.springframework.org/schema/mvc"
       xmlns:context="http://www.springframework.org/schema/context"
       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.xsd
        http://www.springframework.org/schema/mvc
        http://www.springframework.org/schema/mvc/spring-mvc.xsd
        http://www.springframework.org/schema/context
        http://www.springframework.org/schema/context/spring-context.xsd">

    <mvc:annotation-driven />
    <context:component-scan
            base-package="com.web.fooname.controller" />
    <mvc:default-servlet-handler />

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

    <!-- Handles HTTP GET requests for /resources/**by efficiently serving up static resources in the ${webappRoot}/resources directory -->
    <mvc:resources mapping="/resources/**" location="/resources/" />

</beans>

我想删除这两个文件并将它们转换为代码。
有人能帮我吗?我知道这不是堆栈溢出的目的,但我读了很多教程,但不理解它们。

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