1078

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.

4
  • 12
    If someone interested, here is shown how to have multiple ports - stackoverflow.com/questions/36357135/…
    – Betlista
    Commented Feb 8, 2018 at 13:21
  • if you use "yml" file for configuration then you can use this server: port: 8081 Also annotate you main class as "@SpringBootApplication" and remove @ EnableAutoConfiguration
    – Keaz
    Commented Jun 23, 2018 at 7:52
  • your project [application.properties] for add the server.port=8080 Commented Oct 8, 2019 at 11:07
  • set server.port=8080 in application properties. this configuration is in ServerProperties.class class under org.springframework.boot.autoconfigure.web.
    – Atif
    Commented Mar 11, 2020 at 10:12

65 Answers 65

1584

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 a random port use:

server.port=0

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

server:
  port: 8090
17
  • 77
    When random port is used, port info can get with @Value("${local.server.port}")
    – azizunsal
    Commented Jul 23, 2015 at 12:46
  • 52
    Actually command line option is --server.port=8090 not -Dserver.port=8090. docs.spring.io/spring-boot/docs/current/reference/html/…
    – alpert
    Commented Aug 19, 2015 at 6:39
  • 1
    As a compliment to this answer: According to the spring docs there are other paths you can put application.properties on. In my case that helped a lot.
    – sargas
    Commented Oct 2, 2015 at 19:37
  • 17
    -Dserver.port=XXXX did not work for me. I used OS environment variable mode: $ SERVER_PORT=8090 java -jar <path/to/my/jar> Commented Oct 8, 2015 at 7:38
  • 17
    Both (1) java -Dserver.port=XXXX -jar <path/to/my/jar> and (2) java -jar <path/to/my/jar> --server.port=YYYY works. The first command defines server.port system property, and the second command pass the property through the command line arguments (String... args in the main method). Moreover, if you run with java -Dserver.port=XXXX -jar <path/to/my/jar> --server.port=YYYY, YYYY takes precedence over XXXX, this is why Spring Boot Externalized Configuration is so charming.
    – tan9
    Commented Nov 29, 2016 at 16:28
275

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

application.properties file

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

Change port with a vm arg

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

1
  • In STS 4 it is at run -> run configurations -> main, then scroll down to Table with Parameter Name and Value
    – serv-inc
    Commented Apr 28, 2019 at 13:00
215

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.

0
141

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);
        });
    }
}
8
  • 2
    This is working and very useful, when you have port in your own config file and want to set it during runtime.
    – Xdg
    Commented Nov 8, 2015 at 9:48
  • 6
    This was helpful when i needed to deploy an application to a AWS Elastic Beanstalk service, to get the port from an environment variable. Commented Nov 19, 2015 at 12:51
  • This is super useful when all you want is a self-contained unit or integration test, +1. Commented Apr 28, 2016 at 10:34
  • Very useful when the env variable for port is already defined under a different name.
    – higuaro
    Commented Aug 18, 2016 at 8:27
  • 2
    Is'nt it the @Configuration instead of @Controller? Please update if so.
    – Lucky
    Commented Sep 13, 2016 at 11:52
114

If you would like to run it locally, use this -

mvn spring-boot:run -Drun.jvmArguments='-Dserver.port=8085'

As of Spring Boot 2.0, here's the command that works:

mvn spring-boot:run -Dspring-boot.run.arguments=--server.port=8085

clues were at:

3
  • 5
    Starting from Spring Boot 2, you should use spring-boot.run.jvmArguments.
    – mapm
    Commented Apr 12, 2018 at 2:57
  • 1
    -Dspring-boot.run.arguments works with Spring Boot 3 Commented Apr 3, 2023 at 19:13
  • For the first command is valid wihout '' too - therefore mvn spring-boot:run -Drun.jvmArguments=-Dserver.port=8085 works. Tested on SB 3.1.1 Commented Jul 16, 2023 at 16:41
99

You can set port in java code:

HashMap<String, Object> props = new HashMap<>();
props.put("server.port", 9999);

new SpringApplicationBuilder()
    .sources(SampleController.class)                
    .properties(props)
    .run(args);

Or in application.yml:

server:
    port: 9999

Or in application.properties:

server.port=9999

Or as a command line parameter:

-Dserver.port=9999
2
  • Using the HashMap will work only if no port is set in applications.properties or .yml.
    – Milgo
    Commented Apr 25, 2019 at 11:28
  • I have a 2nd "application" class as a utility, and I wanted to only set a different port in that one. Therefore all the config file based approaches were not helpful for me. Close to giving up I found your programatic approach. Thank you! Commented Oct 8, 2021 at 12:43
63

In case you are using application.yml add the Following lines to it

server:
     port: 9000

and of course 0 for random port.

2
  • 1
    this didn't seem to work. I used server.port in the application.yml and it worked
    – yathirigan
    Commented Mar 5, 2015 at 13:11
  • Also, make sure to keep a space between the : and the value. Commented Jul 20, 2022 at 16:28
54

As explained in Spring documentation, there are several ways to do that:

Either you set the port in the command line (for example 8888)

-Dserver.port=8888 or --server.port=8888

Example : java -jar -Dserver.port=8888 test.jar

Or you set the port in the application.properties

server.port=${port:4588}

or (in application.yml with yaml syntax)

server:
   port: ${port:4588}

If the port passed by -Dport (or -Dserver.port) is set in command line then this port will be taken into account. If not, then the port will be 4588 by default.

If you want to enforce the port in properties file whatever the environment variable, you just have to write:

server.port=8888
43

Include below property in application.properties

server.port=8080
1
  • This doesn't work. What application.properties? Which properties can overwrite this one? How can I be sure Spring is picking it up? Commented Dec 10, 2021 at 16:24
31

When you need a programatically way of doing it, you can set it during startup:

System.getProperties().put( "server.port", 80 );
SpringApplication.run(App.class, args);

This might help for things like environment dependent port. Have a nice day

2
  • 2
    System.setProperty("server.port", 80); is another way to achieve the same.
    – hd1
    Commented Feb 11, 2017 at 20:34
  • @hd1, I added our answers to the main answer, check it out and modify as you se fit please Commented Dec 30, 2019 at 14:40
21

if you are using gradle as the build tool, you can set the server port in your application.yml file as:

server:
  port: 8291

If you are using maven then the port can be set in your application.properties file as:

server.port: 8291
2
  • for application.properties its server.port = 8291 Commented Dec 5, 2019 at 9:52
  • What do Maven and Gradle have to do with whether you use a properties or YAML file? The build process (Maven) is completely disparate from the application framework (Spring Boot). Commented Jun 3, 2021 at 9:01
20

To extend other answers:

There is a section in the docs for testing which explains how to configure the port on integration tests:


At integration tests, the port configuration is made using the annotation @SpringBootTest and the webEnvironment values.


Random port:

@SpringBootTest(webEnvironment=WebEnvironment.RANDOM_PORT)

You can inject the value using @LocalServerPort which is the same as @Value("${local.server.port}").

  • Example:

Random port test configuration:

@RunWith(SpringRunner.class
@SpringBootTest(webEnvironment = WebEnvironment.RANDOM_PORT)
public class ExampleTest {
   ...
   @LocalServerPort //to inject port value
   int port;
}

Defined port:

@SpringBootTest(webEnvironment=WebEnvironment.DEFINED_PORT)

It takes the value from server.port if is defined.

  • If is defined using @TestPropertySource(properties = "server.port=9192"), it overrides other defined values.
  • If not, it takes the value from src/test/resources/application.properties (if exists).
  • And finally, if it is not defined it starts with the default 8080.

Example:

Defined port test configuration:

@RunWith(SpringRunner.class)
@SpringBootTest(webEnvironment = SpringBootTest.WebEnvironment.DEFINED_PORT)
@TestPropertySource(properties = "server.port=9192")
public class DemoApplicationTests {

    @Test
    public void contextLoads() {
    }

}
19

You can specify port by overriding EmbeddedServletContainerFactory bean within your configuration (java based or xml). There you can specify port for used embedded servlet container. Please, see Spring Boot - Core "Embedded Servlet Container Support" paragraph and example there. Hope this helps.

2
19

In application.properties file present in resources:

server.port=8082
19

There are three ways to do it depending on the application configuration file you are using

a) If you are using application.properties file set

server.port = 8090

b) If you are using application.yml file set server port property in YAML format as given below

server:
     port: 8090

c) You can also Set the property as the System property in the main method

System.setProperty("server.port","8090");
15

There are many other stuffs you can alter in server configuration by changing application.properties. Like session time out, address and port etc. Refer below post

ref: http://docs.spring.io/spring-boot/docs/1.4.x/reference/html/common-application-properties.html

I used few of them as below.

server.session.timeout=1
server.port = 3029
server.address= deepesh
15

Add this in your application.properties file

server.port= 8080
2
  • 2
    Welcome to SO :-) please look at How to Answer
    – JimHawkins
    Commented Jun 17, 2016 at 10:03
  • Why to add same asnwer one year later?!? and server.port 8080 is wrong syntax for Java property file...
    – Betlista
    Commented Feb 8, 2018 at 13:19
14
  1. As everyone said, you can specify in application.properties
    server.port = 9000 (could be any other value)

  2. If you are using spring actuator in your project, by default it points to
    8080, and if you want to change it, then in application.properties mention
    management.port = 9001 (could be any other value)

14

In the application.properties file, add this line:

server.port = 65535

where to place that fie:

24.3 Application Property Files

SpringApplication loads properties from application.properties files in the following locations and adds them to the Spring Environment:

A /config subdirectory of the current directory
The current directory
A classpath /config package
The classpath root

The list is ordered by precedence (properties defined in locations higher in the list override those defined in lower locations).

In my case I put it in the directory where the jar file stands.

From:

https://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#boot-features-external-config-application-property-files

14

By default spring boot app start with embedded tomcat server start at default port 8080. spring provides you with following different customization you can choose one of them.

NOTE – you can use server.port=0 spring boot will find any unassigned http random port for us.

1) application.properties

server.port=2020

2) application.yml

server:  
     port : 2020

3) Change the server port programatically

3.1) By implementing WebServerFactoryCustomizer interface - Spring 2.x

@Component
public class MyTomcatWebServerCustomizer implements WebServerFactoryCustomizer<TomcatServletWebServerFactory> {

    @Override
    public void customize(TomcatServletWebServerFactory factory) {
        // customize the factory here
        factory.setPort(2020);
    }
}

3.2) By Implementing EmbeddedServletContainerCustomizer interface - Spring 1.x

@Component
public class CustomizationBean implements EmbeddedServletContainerCustomizer {

    @Override
    public void customize(ConfigurableEmbeddedServletContainer container) {
        // customize here
        container.setPort(2020);
    }
}

4) By using command line option

 java -jar spring-boot-app.jar -Dserver.port=2020
12

Indeed, the easiest way is to set the server.port property.

If you are using STS as IDE, from version 3.6.7 you actually have Spring Properties Editor for opening the properties file.

This editor provides autocomplete for all Spring Boot properties. If you write port and hit CTRL + SPACE, server.port will be the first option.

1
  • 1
    Or Also if you are using Intellij IDEA the autocomplete works too. ;)
    – Lucky
    Commented Sep 13, 2016 at 11:53
12

By default, spring-web module provides an embedded tomcat server that is running under the port number 8080. If you need to change the port number of the application then go to application.properties file and configure the port number by using server.port property.

  server.port= 9876

then your application is running under the port 9876.

11

Hope this one help

application.properties=> 

server.port=8090

application.yml=> 

server
  port:8090
1
  • application.yml => server: port: 8090 Commented May 23, 2018 at 17:36
11

Using property server.port=8080 for instance like mentioned in other answers is definitely a way to go. Just wanted to mention that you could also expose an environment property:

SERVER_PORT=8080

Since spring boot is able to replace "." for "_" and lower to UPPER case for environment variables in recent versions. This is specially useful in containers where all you gotta do is define that environment variable without adding/editing application.properties or passing system properties through command line (i.e -Dserver.port=$PORT)

1
  • SERVER_PORT=8081 mvn spring-boot:run Commented Sep 17, 2018 at 4:11
9

You can add the port in below methods.

  1. Run -> Configurations section

  2. In application.xml add server.port=XXXX

1
  • 1
    Do you mean application.yml and what IDE are you using? Please be specific.
    – Lucky
    Commented Sep 13, 2016 at 11:56
9

Just have a application.properties in src/main/resources of the project and give there

server.port=****

where **** refers to the port number.

9

1.1 Update via a properties file.

/src/main/resources/application.properties

server.port=8888

Update via a yaml file.

   server:

     port: 8888

EmbeddedServletContainerCustomizer

@Component
public class CustomContainer implements EmbeddedServletContainerCustomizer {

    @Override
    public void customize(ConfigurableEmbeddedServletContainer container) {

        container.setPort(8888);

    }

}
9

Providing the port number in application.properties file will resolve the issue

 server.port = 8080

"port depends on your choice, where you want to host the application"

9

Running by Gradle:

  • Run in default port(8080): ./gradlew bootRun
  • Run in provided port(8888): ./gradlew bootRun --args='--server.port=8888'
  • If we have any variable in the application.properties file named PORT, run this: PORT=8888 ./gradlew bootRun

Running by Maven:

  • Run in default port(8080): mvnw spring-boot:run
  • Run in provided port(8888): mvnw spring-boot:run -Dspring-boot.run.jvmArguments='-Dserver.port=8085'
  • Run in provided port(8888): mvn spring-boot:run -Dspring-boot.run.arguments='--server.port=8085'
  • Run in provided port(8888) with other custom property: mvn spring-boot:run -Dspring-boot.run.arguments="--server.port=8899 --your.custom.property=custom"
  • If we have any variable in the application.properties file named PORT, run this: SERVER_PORT=9093 mvn spring-boot:run

Using java -jar:

  • Create the .jar file:
    • For Gradle: ./gradlew clean build. We will find the jar file inside: build/libs/ folder.
    • For Maven: mvn clean install. We will find the jar file inside:target folder.
  • Run in default port(8080): java -jar myApplication. jar
  • Run in provided port(8888): java -jar myApplication.jar --port=8888
  • Run in provided port(8888): java -jar -Dserver.port=8888 myApplication.jar
  • Run in provided port(8888) having variable SERVER_PORT in application.properties file: SERVER_PORT=8888 java -jar target/myApplication.jar
1
  • this is the best answer, plesae upvote to make it to the top Commented May 28 at 9:58
8

You can also use SERVER_PORT environment variable to configure Spring Boot port. Just set the environment variable and restart the app:

set SERVER_PORT=9999 // on windows machine
export SERVER_PORT=9999 // on linux

Please note that if you do not set those environment variables system wide, you should run the boot app on the same session.

Not the answer you're looking for? Browse other questions tagged or ask your own question.