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spring - 如何@Autowire 验证器类中的对象?

转载 作者:IT老高 更新时间:2023-10-28 13:45:42 26 4
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是否可以在 Validation 类中 Autowiring 对象?对于应该是 Autowired 的对象,我一直为 null...

最佳答案

您的验证类是启用的 Spring bean 吗?如果没有,您的对象 Autowiring 总是会为 null。确保您已启用验证类。

并且不要忘记启用注释配置 bean 后处理器(请参阅 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>

如何启用您的验证类作为托管 Spring bean。要么

通过使用xml(如上图)

<beans ...>
<bean class="AccessRequestValidator"/>
<context:annotation-config />
</beans>

改用注解(注意 @Component 就在类的上方)

@Component
public class AccessRequestValidator implements Validator {

}

但是要启用 Spring 注释组件扫描,您必须启用 bean-post 处理器(注意

<beans ...>
<context:annotation-config />
<context:component-scan base-package="<PUT_RIGHT_HERE_WHICH_ROOT_PACKAGE_SHOULD_SPRING_LOOK_FOR_ANY_ANNOTATED_BEAN>"/>
</beans>

在您的 Controller 中,只需执行此操作(不要使用新运算符)

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;
}

}

更新

您的网络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
*/

您的 web.xml 应该看起来像(注意 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>

您的 -servlet.xml 应该看起来像(注意我使用的是 Spring 2.5 - 如果您使用的是 3.0,请替换)

 <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>

关于spring - 如何@Autowire 验证器类中的对象?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/2732993/

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