我正在嘗試從頭開(kāi)始創(chuàng)建一個(gè)非?；镜膽?yīng)用程序的 Spring RESTful API，但我無(wú)法訪問(wèn)控制器。我可以訪問(wèn) JSP 文件但不能訪問(wèn)控制器。我也嘗試過(guò)注釋，@RestController但它沒(méi)有用。我在Tomcat 8上運(yùn)行。錯(cuò)誤是：源服務(wù)器沒(méi)有找到目標(biāo)資源的當(dāng)前表示或不愿意透露存在的表示。WEB.xml    <?xml version="1.0" encoding="UTF-8"?><web-app version="2.5" 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_2_5.xsd">  <listener>    <listener-class>        org.springframework.web.context.ContextLoaderListener    </listener-class>  </listener>      <servlet>        <servlet-name>dispatcher</servlet-name>        <servlet-class>            org.springframework.web.servlet.DispatcherServlet        </servlet-class>        <load-on-startup>1</load-on-startup>    </servlet>    <servlet-mapping>        <servlet-name>dispatcher</servlet-name>        <url-pattern>/</url-pattern>    </servlet-mapping></web-app>調(diào)度程序-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.controller />    <mvc:default-servlet-handler /></beans>