为什么春天没有检测到我的@components?

时间:2013-07-10 23:54:28

标签: spring spring-mvc

这个很棘手 - 至少对我来说。组件扫描似乎不起作用

这是web.xml

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >
<web-app>
    <display-name>Geomajas GWT face example application</display-name>
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            classpath:org/geomajas/spring/geomajasContext.xml
            classpath:org/geomajas/plugin/rasterizing/DefaultRasterizedPipelines.xml
            WEB-INF/applicationContext.xml
<!--            WEB-INF/applicationContext2.xml -->
<!--            WEB-INF/layer*.xml -->
<!--            WEB-INF/map*.xml -->
            WEB-INF/layerOsm.xml
            WEB-INF/mapOsm.xml
<!--            WEB-INF/applicationContext2.xml -->

       </param-value>
    </context-param>



    <filter>
        <filter-name>CacheFilter</filter-name>
        <filter-class>org.geomajas.servlet.CacheFilter</filter-class>
    </filter>

    <filter-mapping>
        <filter-name>CacheFilter</filter-name>
        <url-pattern>*</url-pattern>
    </filter-mapping>

    <listener>
        <listener-class>org.geomajas.servlet.PrepareScanningContextListener</listener-class>
    </listener>

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

    <servlet>
        <servlet-name>GeomajasServiceServlet</servlet-name>
        <servlet-class>org.geomajas.gwt.server.GeomajasServiceImpl</servlet-class>
    </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>classpath*:META-INF/geomajasWebContext.xml</param-value>
            <description>Spring Web-MVC specific (additional) context files.</description>
        </init-param>
        <load-on-startup>3</load-on-startup>
    </servlet>

<!-- SpringGwt remote service servlet --> 
    <servlet>
        <servlet-name>springGwtRemoteServiceServlet</servlet-name>
        <servlet-class>org.spring4gwt.server.SpringGwtRemoteServiceServlet</servlet-class>
        <init-param>
        <param-name>contexConfigLocation</param-name>
        <param-value>WEB-INF/applicationContext2.xml</param-value>
        <description>j</description>
        </init-param>
    </servlet>

    <servlet-mapping>
        <servlet-name>GeomajasServiceServlet</servlet-name>
        <url-pattern>/showcase/geomajasService</url-pattern>
    </servlet-mapping>

    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/d/*</url-pattern>
    </servlet-mapping>



    <servlet-mapping>
        <servlet-name>springGwtRemoteServiceServlet</servlet-name>
        <url-pattern>/showcase/springGwtServices/test</url-pattern>

    </servlet-mapping>


    <welcome-file-list>
        <welcome-file>index.html</welcome-file>
    </welcome-file-list>

</web-app>

我的服务:

@Service("test")
public class ProjServiceImpl extends RemoteServiceServlet implements ProjService {
    private static final Log LOG = LogFactory.getLog(ProjServiceImpl.class);



    @Autowired
    PoiCategDAO poiCategDAO;

    public String greetServer(String input) throws IllegalArgumentException {
        // Verify that the input is valid. 
        if (!FieldVerifier.isValidName(input)) {
            // If the input is not valid, throw an IllegalArgumentException back to
            // the client.
            throw new IllegalArgumentException(
                    "Name must be at least 4 characters long");
        }
        //RequestContextUtils.getWebApplicationContext(request);
//      String serverInfo = getServletContext().getServerInfo();
//      String userAgent = getThreadLocalRequest().getHeader("User-Agent");



        return "Hello, " + input + "!<br><br>I am running .<br><br>It looks like you are using:<br>";
        }

    @Override
    //@Transactional(propagation = Propagation.REQUIRED, rollbackFor = Exception.class)
    public void testdao(Integer id) throws IllegalArgumentException {
        // TODO Auto-generated method stub
        PoiCateg X=poiCategDAO.findById(id);    
    PoiCateg z=poiCategDAO.findById(id);
    }

}

和applicaationContext2.xml

<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p"
    xmlns:aop="http://www.springframework.org/schema/aop" xmlns:context="http://www.springframework.org/schema/context"
     xmlns:tx="http://www.springframework.org/schema/tx"
    xsi:schemaLocation="
                http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.0.xsd
                http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
                http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd
                http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.0.xsd">

    <context:annotation-config />

<tx:annotation-driven transaction-manager="transactionManager" /> 
<context:component-scan base-package="ne.projl.*" />
<!--    <context:component-scan base-package="FULLY QUALIFIED" /> -->

<bean class="org.springframework.orm.jpa.LocalEntityManagerFactoryBean" id="entityManagerFactory">
        <property name="persistenceUnitName" value="MyPUnit" />
    </bean>

    <bean class="org.springframework.orm.jpa.JpaTransactionManager" id="transactionManager">
        <property name="entityManagerFactory" ref="entityManagerFactory" />
    </bean>


</beans>

请注意,如果我将context:component-scan放在applicationContext.xml中(而不是applicatinContext2.xml,则会检测到我的服务bean)。如果有其他信息我应该提供告诉。

4 个答案:

答案 0 :(得分:2)

如果我没记错的话,组件扫描中的基础包只需要一个包名。然后将包括所有底层子包。

因此,正确的设置应为

<context:component-scan base-package="ne.projl" />

答案 1 :(得分:1)

可能有很多原因,一个常见的原因是您在错误的包装下创建了@Component类。

作为您的上下文扫描配置集,Class必须在此包

<context:component-scan base-package="ne.projl.*" />

答案 2 :(得分:0)

发布例外(如果有的话)可以为这里的每个人提供更多的问题洞察力。在我看来就像一个能见度问题。 如果您引用在applicationContext2中扫描的组件,则在通过contextConfigLocation加载的某个组件中,显然它不可见。 。check this link for more info on visibility of application context 。我无法在这个问题上添加评论。所以我发表了我的观察结果作为答案..对不起,

答案 3 :(得分:0)

我不确定,但是你的web.xml没有说明这个

<!--            WEB-INF/applicationContext2.xml -->

是不是已经注释掉了?

你在applicationContext.xml文件中做了applicationContext2的导入资源吗?

相关问题