问题

I have 2 Main entry points in a single application.

The first main starts the server, maps the controllers and starts some worker threads. These workers receive messages from cloud queues.

In case of increased load, I want to be able to add additional workers to do my job. So I have a second Main entry point in my application which I want to be able to start without starting the default server in spring-boot (as a client application) so as to avoid port conflict (and obviously which will lead to failure).

How do I achieve this?

回答1:

Launching from the command line with `server` and `client` profiles

To use the same jar and same entry point with 2 different profiles, you should simply provide the Spring profile at runtime, to have distinct application-${profile}.properties loaded (and potentially Conditional Java config triggered).

Define 2 spring profiles (client and server):

Each will have its own application-${profile}.properties
The client's properties will disable the web container

Have a single SpringBootApp and entry point:

@SpringBootApplication
public class SpringBootApp {
    public static void main(String[] args) {
        new SpringApplicationBuilder()
            .sources(SpringBootApp.class)
            .run(args);
    }
}

Make this class your main-class.

src/main/resources/application-server.properties:

spring.application.name=server
server.port=8080

src/main/resources/application-client.properties:

spring.application.name=client
spring.main.web-environment=false

Launch both profiles from the command line:

$ java -jar -Dspring.profiles.active=server YourApp.jar
$ java -jar -Dspring.profiles.active=client YourApp.jar

You may have @Configuration classes triggered conditionally based on the active profile too:

@Configuration
@Profile("client")
public class ClientConfig {
    //...
}

Launching from the IDE with `server` and `client` profiles

Launchers:

@SpringBootApplication
public class SpringBootApp {
}

public class LauncherServer {
    public static void main(String[] args) {
        new SpringApplicationBuilder()
            .sources(SpringBootApp.class)
            .profiles("server")
            .run(args);
    }
}

public class ClientLauncher {
    public static void main(String[] args) {
        new SpringApplicationBuilder()
            .sources(SpringBootApp.class)
            .profiles("client")
            .web(false)
            .run(args);
    }
}

You may specify additional configuration classes (specific to client or server):

new SpringApplicationBuilder()
    .sources(SpringBootApp.class, ClientSpecificConfiguration.class)
    .profiles("client")
    .web(false)
    .run(args);

src/main/resources/application-server.properties:

spring.application.name=server
server.port=8080

src/main/resources/application-client.properties:

spring.application.name=client
#server.port= in my example, the client is not a webapp

Note, you may also have 2 SpringBootApp (ClientSpringBootApp, ServerSpringBootApp), each with its own main, it's a similar setup which allow you to configure different AutoConfiguration or ComponentScan:

@SpringBootApplication
@ComponentScan("...")
public class ServerSpringBootApp {
    public static void main(String[] args) {
        new SpringApplicationBuilder()
            .sources(ServerSpringBootApp.class)
            .profiles("server")
            .run(args);
    }
}

//Example of a difference between client and server
@SpringBootApplication(exclude = SecurityAutoConfiguration.class) 
@ComponentScan("...")
public class ClientSpringBootApp {
    public static void main(String[] args) {
        new SpringApplicationBuilder()
            .sources(ClientSpringBootApp.class)
            .profiles("client")
            .web(false)
            .run(args);
    }
}

来源：https://stackoverflow.com/questions/39205430/how-to-run-spring-boot-as-a-client-application

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