java - Spring Bean 根据条件创建一个接口(interface)

Question

我的目标是拥有两个 bean 实现，它们的使用取决于类路径中类的属性或存在。我尝试的是 ConditionalOnPropery，见下文，但没有成功。

我有一个像这样的界面

public interface ApplicationConfigurator {

    void configure(WicketApplication wicketApplication);

}

和两个实现类

@Component
@ConditionalOnProperty(
        value = "environment",
        havingValue = "client",
        matchIfMissing = false
)
public class OfflineApplicationConfigurator implements ApplicationConfigurator {

    @Override
    public void configure(WicketApplication wicketApplication) {

    }
}

还有

@Component
    @ConditionalOnProperty(
            value = "environment",
            havingValue = "server",
            matchIfMissing = false
    )
public class OnlineApplicationConfigurator implements ApplicationConfigurator {

    @Autowired
    private LsgConfiguration configuration;

    @Override
    public void configure(WicketApplication wicketApplication) {
        try {
            Class.forName("org.apache.catalina.core.StandardServer");
            configuration.setHttpSchema(new TomcatConnectors().getHttpConnector().getScheme());
        } catch (ClassNotFoundException | ApplicationConfigurationException exception) {
            configuration.setHttpSchema("http");
        }
    }
}

我的配置是基于 xml 的，如下所示:

<bean id="onlineApplicationConfigurator" class="com.kion.lsg.service.core.web.config.OnlineApplicationConfigurator"/>
<bean id="offlineApplicationConfigurator" class="com.kion.lsg.service.core.web.config.OfflineApplicationConfigurator"/>

现在我收到错误消息:

Caused by: org.springframework.beans.factory.NoUniqueBeanDefinitionException: No qualifying bean of type [....service.core.web.config.ApplicationConfigurator] is defined: expected single matching bean but found 2: onlineApplicationConfigurator,offlineApplicationConfigurator

感谢任何帮助

Answer 1

您是否在 application.yml 中指定条件变量“环境”？由于这两个都是组件，因此应该自动扫描，无需在 XML 文件中声明。您可以启用 Debug模式来查看条件评估报告，这将更加深入。