Question

我刚刚迁移到Java 11，Spring 5和Tomcat9。我正在spring-mvc中设置一个简单的HelloWorld程序，其代码如果将编译器的遵从性级别设置为1.8，则可以正常工作。但是使用11我收到状态404错误，显示：

Mar 29, 2019 11:20:52 AM org.springframework.web.servlet.DispatcherServlet noHandlerFound
WARNING: No mapping for GET /spring-mvc-demo/showForm

这是我想要我的应用程序执行的操作。在Tomcat上运行时，将带我到“ main-menu.jsp”页面，浏览器中的URL为：http://localhost:8080/spring-mvc-demo/ 我希望能够将/ showForm添加到浏览器中的该URL并查看“ helloworld-form.jsp”。但是，当我将/ showForm添加到url中时，却收到找不到处理程序的404错误。

这是我从Controller类开始的代码：

@Controller
public class HelloWorldController {

    // need a controller method to show the initial HTML form

    @RequestMapping("/showForm")
    public String showForm() {
        return "helloworld-form";
    }

}

@Controller
public class HomeController {

    @RequestMapping("/")
    public String showPage() {
        return "main-menu";
    }
}

这是我的配置文件： web.xml

<?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_3_1.xsd"
    id="WebApp_ID" version="3.1">

    <display-name>spring-mvc-demo</display-name>
    <absolute-ordering></absolute-ordering>
    <!-- Spring MVC Configs -->

    <!-- Step 1: Configure Spring MVC Dispatcher Servlet -->
    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/spring-mvc-demo-servlet.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <!-- Step 2: Set up URL mapping for Spring MVC Dispatcher Servlet -->
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

</web-app>

和spring-mvc-demo-servlet.xml：

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

    <!-- Step 3: Add support for component scanning -->
    <context:component-scan base-package="com.luv2code.springdemo" />

    <!-- Step 4: Add support for conversion, formatting and validation support -->
    <mvc:annotation-driven/>

    <!-- Step 5: Define Spring MVC view resolver -->
    <bean
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix" value="/WEB-INF/view/" />
        <property name="suffix" value=".jsp" />
    </bean>

</beans>

在此先感谢您的帮助-非常感谢。君士坦丁

找不到处理程序-Java 11问题？

0 个答案: