@Autowired不适用于Spring 3

Question

我环顾四周寻找解决方案,找不到.

我正在使用Tomcat,Spring 3和罐子:

org.springframework.aop-3.0.5.RELEASE.jar
org.springframework.asm-3.0.5.RELEASE.jar
org.springframework.beans-3.0.5.RELEASE.jar
org.springframework.context-3.0.5.RELEASE.jar
org.springframework.core-3.0.5.RELEASE.jar
org.springframework.expression-3.0.5.RELEASE.jar
org.springframework.jdbc-3.0.5.RELEASE.jar
org.springframework.orm-3.0.5.RELEASE.jar
org.springframework.test-3.0.5.RELEASE.jar
org.springframework.transaction-3.0.5.RELEASE.jar
org.springframework.web-3.0.5.RELEASE.jar

我的代码是这样的:

public class EmailResource {
@Autowired
EmailManager emailManager;
}

在applicationContext我有:

<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"
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">

<bean
    class="org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor" />
<bean id="emailManager" class="com.mycompany.manager.impl.EmailManagerImpl" />
<context:component-scan base-package="com.mycompany.component" />

和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>classpath:/applicationContext.xml</param-value>
    </context-param>


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

但是,emailManager始终为null!我错过了什么？

EDITED

EmailResource是用于休息调用的jersey servlet,定义如下:

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>com.sun.jersey.config.property.packages</param-name>
        <param-value>com.mycompany.resource</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>Jersey Web Application</servlet-name>
    <url-pattern>/api/*</url-pattern>
</servlet-mapping>

Answer 1

您需要使用Jersey/Spring连接器让Jersey在启动时识别您的Spring上下文.

更换:

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>com.sun.jersey.config.property.packages</param-name>
        <param-value>com.mycompany.resource</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

附:

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
    <init-param>
   ....
    </init-param>
</servlet>

你还需要jersey-spring依赖:

<dependency>
    <groupId>com.sun.jersey.contribs</groupId>
    <artifactId>jersey-spring</artifactId>
    <version>${jersey.version}</version>
</dependency>