Question

我在Spring 3.2.4上有Spring MVC Web应用程序。我有2个背景。 1. mvc-dispatcher-servlet.xml如下所示：

<context:annotation-config/>
    <context:component-scan base-package="com.ja.dom"/>

    <!-- Tiles 3 config -->

    <bean class="org.springframework.web.servlet.view.UrlBasedViewResolver">
        <!--Don't add suffix or prefix like you do with .jsp files-->
        <property name="viewClass" value="org.springframework.web.servlet.view.tiles3.TilesView"/>
    </bean>

    <bean id="tilesConfigurer" class="org.springframework.web.servlet.view.tiles3.TilesConfigurer" >
        <property name="definitions">
            <value>/WEB-INF/tiles.xml</value>
        </property>
    </bean>

和2. root-context.xml如下：

 <bean class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
     <property name="locations" value="classpath:prop.properties"/>
 </bean>

 <!-- Mail Sender Bean -->

<bean id="mailSender" class="org.springframework.mail.javamail.JavaMailSenderImpl">
     <property name="host" value="smtp.gmail.com" />
     <property name="port" value="587" />
     <property name="username" value="${email.name}" />
     <property name="password" value="${email.password}" />
     <property name="defaultEncoding" value="UTF-8"/>

     <property name="javaMailProperties">
         <props>
             <prop key="mail.smtp.auth">true</prop>
             <prop key="mail.smtp.starttls.enable">true</prop>
             <prop key="mail.debug">true</prop>
         </props>
     </property>
 </bean>

当我运行应用程序时，我看到只有mailSender bean有来自属性文件的参数。我的其他豆类等@Service ..不能注入。怎么了？如何将我的PropertyPlaceholderConfigurer共享到mvc-dispatcher-servlet上下文？

我在服务上注入了这样的属性：

@Value（ “$ {reCaptcha.private.key}”） private String reCaptchaPrivateKey;

和我的web.xml （更新）

<web-app 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_3_0.xsd"
         version="3.0">

    <display-name>Spring MVC Application</display-name>

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

    <servlet>
        <servlet-name>mvc-dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <!--<init-param>-->
            <!--<param-name>contextConfigLocation</param-name>-->
            <!--<param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value>-->
        <!--</init-param>-->
        <load-on-startup>1</load-on-startup>
    </servlet>

    <!-- Creates the Spring Container shared by all Servlets and Filters -->
    <listener>
        <listener-class>
            org.springframework.web.context.ContextLoaderListener
        </listener-class>
    </listener>




    <servlet-mapping>
        <servlet-name>mvc-dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app>

Answer 1

我认为问题在于

<bean class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
    <property name="locations" value="classpath:prop.properties"/>
</bean>

相反，您应该使用PropertySourcesPlaceholderConfigurer

<context:property-placeholder location="classpath:prop.properties" />

哪个

从Spring 3.1开始，XML将不会更长时间注册旧的PropertyPlaceholderConfigurer但新注册介绍了PropertySourcesPlaceholderConfigurer。这个替代品创建的类更灵活，更好地与之交互新引入的环境和物业资源机制;这应该被视为3.1申请的标准。

如何共享2个上下文的propertyplaceholder

1 个答案: