是否可以在Validation类中自动装配对象?对于应该是Autowired的对象,我一直都为null ...
Art*_*ald 22
你的Validation类是一个启用的Spring bean ??? 如果没有,您将始终为自动装配的对象获取null.确保已启用验证类.
并且不要忘记启用Annotation配置bean后处理器(参见<context:annotation-config />元素)
<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-2.5.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-2.5.xsd">
<context:annotation-config />
</beans>
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如何将Validation类作为托管Spring bean启用.或
1°使用xml(如上图所示)
<beans ...>
<bean class="AccessRequestValidator"/>
<context:annotation-config />
</beans>
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2°通过使用注释(注意@Component就在类之上)
@Component
public class AccessRequestValidator implements Validator {
}
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但是要启用Spring注释组件扫描,必须启用bean-post处理器(注意<context:component-scan element)
<beans ...>
<context:annotation-config />
<context:component-scan base-package="<PUT_RIGHT_HERE_WHICH_ROOT_PACKAGE_SHOULD_SPRING_LOOK_FOR_ANY_ANNOTATED_BEAN>"/>
</beans>
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在您的控制器内部,只需这样做(不要使用新的运算符)
选择以下策略之一
public class MyController implements Controller {
/**
* You can use FIELD @Autowired
*/
@Autowired
private AccessRequestValidator accessRequestValidator;
/**
* You can use PROPERTY @Autowired
*/
private AccessRequestValidator accessRequestValidator;
private @Autowired void setAccessRequestValidator(AccessRequestValidator accessRequestValidator) {
this.accessRequestValidator = accessRequestValidator;
}
/**
* You can use CONSTRUCTOR @Autowired
*/
private AccessRequestValidator accessRequestValidator;
@Autowired
public MyController(AccessRequestValidator accessRequestValidator) {
this.accessRequestValidator = accessRequestValidator;
}
}
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UPDATE
您的Web应用程序结构应如下所示
<CONTEXT-NAME>/
WEB-INF/
web.xml
<SPRING-SERVLET-NAME>-servlet.xml
business-context.xml
classes/
/com
/wuntee
/taac
/validator
AccessRequestValidator.class
lib/
/**
* libraries needed by your project goes here
*/
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您的web.xml应该是(NOTICE contextConfigLocation context-param和ContextLoaderListener)
<web-app version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<context-param>
<param-name>contextConfigLocation</param-name>
<!--If your business-context.xml lives in the root of classpath-->
<!--replace by classpath:business-context.xml-->
<param-value>
/WEB-INF/business-context.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name><SPRING-SERVLET-NAME></servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name><SPRING-SERVLET-NAME></servlet-name>
<url-pattern>*.htm</url-pattern>
</servlet-mapping>
</web-app>
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你的<SPRING-SERVLET-NAME> -servlet.xml应该是这样的(注意我使用的是Spring 2.5 - 如果你使用3.0则替换)
<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-2.5.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-2.5.xsd">
<!--ANY HANDLER MAPPING-->
<!--ANY VIEW RESOLVER-->
<context:component-scan base-package="com.wuntee.taac"/>
<context:annotation-config/>
</beans>
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