How to configure port for a Spring Boot application

How do I configure the TCP/IP port listened on by a Spring Boot application, so it does not use the default port of 8080.

Solution 1:

As said in docs either set server.port as system property using command line option to jvm -Dserver.port=8090 or add application.properties in /src/main/resources/ with

server.port=8090

For random port use

server.port=0

Similarly add application.yml in /src/main/resources/ with

server:
  port : 8090

Solution 2:

There are two main ways to change the port in the Embedded Tomcat in a Spring Boot Application.

Modify application.properties

First you can try the application.properties file in the /resources folder:

server.port = 8090

Modify a VM option

The second way, if you want to avoid modifying any files and checking in something that you only need on your local, you can use a vm arg:

Go to Run -> Edit Configurations -> VM options

-Dserver.port=8090

Additionally, if you need more information you can view the following blog post here: Changing the port on a Spring Boot Application

Solution 3:

Since Spring Boot provides various configuration externalization mechanism (through various PropertySource implementations and/or processors wired into Environment object in order), you can set any property outside of your jar archive through following methods:

1. Pass property through command line argument as application argument

   java -jar <path/to/my/jar> --server.port=7788
   

2. From property in SPRING_APPLICATION_JSON (Spring Boot 1.3.0+)

   • Define environment variable in U*IX shell:

     SPRING_APPLICATION_JSON='{"server.port":7788}' java -jar <path/to/my/jar>
     

   • By using Java system property:

     java -Dspring.application.json='{"server.port":7788}' -jar <path/to/my/jar>
     

   • Pass through command line argument:

     java -jar <path/to/my/jar> --spring.application.json='{"server.port":7788}'
     

3. Define JVM system property

   java -Dserver.port=7788 -jar <path/to/my/jar>
   

4. Define OS environment variable

   • U*IX Shell

     SERVER_PORT=7788 java -jar <path/to/my/jar>
     

   • Windows

     SET SERVER_PORT=7788
     java -jar <path/to/my/jar>
     

5. Place property in ./config/application.properties configuration file

   server.port=7788
   

   and run:

    java -jar <path/to/my/jar>
   

6. Place property in ./config/application.yaml

   server:
       port: 7788
   

   and run:

    java -jar <path/to/my/jar>
   

7. Place property in ./application.properties

   server.port=7788
   

   and run:

    java -jar <path/to/my/jar>
   

8. Place property in ./application.yaml

   server:
       port: 7788
   

   and run:

    java -jar <path/to/my/jar>
   

You can combine above methods all together, and the former configuration in the list take precedence over the latter one.

For example:

SERVER_PORT=2266 java -Dserver.port=5566 -jar <path/to/my/jar> --server.port=7788

The server will start and listen on port 7788.

This is very useful providing default properties in PropertySources with lower precedence (and usually packaged in the archive or coded in the source), and then override it in the runtime environment. And it is the design philosophy of Spring Boot:

Be opinionated out of the box, but get out of the way quickly as requirements start to diverge from the defaults.

SERVER_NAME to server.name conversion was done by Relaxed Binding.

Solution 4:

Also, you can configure the port programmatically.

For Spring Boot 2.x.x:

@Configuration
public class CustomContainer implements WebServerFactoryCustomizer<ConfigurableServletWebServerFactory> {
  public void customize(ConfigurableServletWebServerFactory factory){
    factory.setPort(8042);
  }
}

For older versions:

@Configuration
public class ServletConfig {
    @Bean
    public EmbeddedServletContainerCustomizer containerCustomizer() {
        return (container -> {
            container.setPort(8012);
        });
    }
}