Question

我认为我误解了Spring启动配置文件的功能。我的spring启动应用程序中有两个独立的启动类（ApplicationLocal和ApplicationProduction），这些类注释如下

@SpringBootApplication
@EnableSwagger2
@Import(value={Config.class})
@ComponentScan(basePackages={"com.abc.*"})
@Profile("local")
public class ApplicationLocal extends SpringBootServletInitializer {
//

@SpringBootApplication
@EnableSwagger2
@Import(value={Config.class})
@ComponentScan(basePackages={"com.abc.*"})
@Profile("production")
public class ApplicationProduction extends SpringBootServletInitializer {
//

并且我希望在运行时通过Spring启动来获取相应的启动类，具体取决于-Dspring.profiles.active jvm arguement的值。但除非我在pom.xml中提到start-class，否则它不适用于我喜欢这个..

<properties>
    <start-class>com.abc.web.service.ApplicationLocal</start-class>
</properties>

如果我没有在pom.xml中提及start-class，那么我会收到以下错误：（我按照以下方式运行应用程序

mvn spring-boot:run  -Drun.jvmArguments=" -Dspring.profiles.active=local"

我收到以下错误

[ERROR] Failed to execute goal org.springframework.boot:spring-boot-maven-plugin:1.3.3.RELEASE:run (default-cli) on project payee-list-ws: Execution default-cli of goal org.springframework.boot:spring-boot-maven-plugin:1.3.3.RELEASE:run failed: Unable to find a single main class from the following candidates [com.abc.web.service.ApplicationProduction, com.abc.web.service.ApplicationLocal] -> [Help 1]

Answer 1

您可以为每个env创建一个主类和两个配置：

@SpringBootApplication
@EnableSwagger2
@Import(value={Config.class})
@ComponentScan(basePackages={"com.abc.*"})
public class Application extends SpringBootServletInitializer {
//

@Configuration
@Profile("production")
public class ConfigProduction {
    //...
}

@Configuration
@Profile("Local")
public class ConfigLocal {
    //...
}

当主类将扫描bean时，Spring将根据您的实际配置文件选择正确的配置。

如何根据活动弹簧配置文件切换弹簧启动启动类

1 个答案: