在DispatcherServlet中找不到带有URI [/ SpringRestExample /]的HTTP请求的映射,其名称为' appServlet'

时间:2018-04-21 06:28:42

标签: java spring rest

我试图使用Spring编写基于Rest的服务。在运行我的应用程序时出现以下错误。请在下面找到错误:

警告:org.springframework.web.servlet.PageNotFound - 在DispatcherServlet中找不到带有URI [/ SpringRestExample /]的HTTP请求的映射名称' appServlet'

请找到我的web.xml

的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">

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>

<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<!-- Processes application requests -->
<servlet>
    <servlet-name>appServlet</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

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

请找到我的servlet配置文件

servlet的context.xml中

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


<annotation-driven />


<resources mapping="/resources/**" location="/resources/" />


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

<beans:bean class="org.springframework.web.servlet.mvc.method.annotation.RequestMappingHandlerAdapter">
    <beans:property name="messageConverters">
        <beans:list>
            <beans:ref bean="jsonMessageConverter"/>
        </beans:list>
    </beans:property>
</beans:bean>

<beans:bean id="jsonMessageConverter" class="org.springframework.http.converter.json.MappingJackson2HttpMessageConverter">
</beans:bean>   

<context:component-scan base-package="com.journaldev.spring.controller" />

请找我的控制器类。

EmployeeController.java

@Controller
public class EmployeeController {

private static final Logger logger = LoggerFactory.getLogger(EmployeeController.class);

//Map to store employees, ideally we should use database
Map<Integer, Employee> empData = new HashMap<Integer, Employee>();

@RequestMapping(value = EmpRestURIConstants.DUMMY_EMP, method = RequestMethod.GET)
public @ResponseBody Employee getDummyEmployee() {
    logger.info("Start getDummyEmployee");
    Employee emp = new Employee();
    emp.setId(9999);
    emp.setName("Dummy");
    emp.setCreatedDate(new Date());
    empData.put(9999, emp);
    return emp;
}

@RequestMapping(value = EmpRestURIConstants.GET_EMP, method = RequestMethod.GET)
public @ResponseBody Employee getEmployee(@PathVariable("id") int empId) {
    logger.info("Start getEmployee. ID="+empId);

    return empData.get(empId);
    }

  }

0 个答案:

没有答案