Three Ways to Start a Spring Boot Application

Spring Boot provides three main startup methods: executing a class that contains a main method, running the packaged JAR with java -jar, and using the spring-boot-maven-plugin.

1. Execute a class with a main method

This is the simplest approach, often done directly from an IDE like IntelliJ IDEA, which automatically loads configuration files from the classpath.

@RestController
@EnableAutoConfiguration
public class Example {
    @RequestMapping("/")
    public String home() {
        return "Hello World";
    }
    public static void main(String[] args) {
        /**
         * SpringApplication will automatically load <code>application.properties</code> from several locations:
         * 1. <b>/config</b> subdirectory of the current directory;
         * 2. The current directory;
         * 3. A classpath <code>/config</code> package;
         * 4. The classpath root.
         */
        SpringApplication.run(Example.class, args);
    }
}

In IDEA you can also set application parameters via the run configuration.

2. Use java -jar command

java -jar jar_path --param

jar_path

refers to the location of the packaged JAR file, and --param represents command‑line arguments, e.g.: java -jar example.jar --server.port=8081 This overrides the port defined in application.properties.

3. Start via spring-boot-maven-plugin

Add the following plugin configuration to your Maven project (the parent spring-boot-starter-parent already provides the appropriate version):

<plugin>
  <groupId>org.springframework.boot</groupId>
  <artifactId>spring-boot-maven-plugin</artifactId>
  <!-- version is inherited from the parent -->
</plugin>

After setting up, run the application from the project root: mvn spring-boot:run To pass arguments, consult the plugin’s help: mvn spring-boot:help -Ddetail Then specify arguments using -Drun.arguments, for example:

mvn spring-boot:run -Drun.arguments="--server.port=8888"

This command passes the specified parameters to the application at runtime.