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.
-
11If someone interested, here is shown how to have multiple ports - https://stackoverflow.com/questions/36357135/configure-spring-boot-with-two-ports – Betlista Feb 08 '18 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 Jun 23 '18 at 07:52
-
your project [application.properties] for add the server.port=8080 – Lahiru Samishka Oct 08 '19 at 11:07
-
set `server.port=8080` in application properties. this configuration is in `ServerProperties.class` class under `org.springframework.boot.autoconfigure.web`. – Atif Mar 11 '20 at 10:12
62 Answers
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

- 10,654
- 8
- 33
- 53

- 60,022
- 51
- 208
- 332
-
75When random port is used, port info can get with `@Value("${local.server.port}")` – azizunsal Jul 23 '15 at 12:46
-
52Actually command line option is --server.port=8090 not -Dserver.port=8090. http://docs.spring.io/spring-boot/docs/current/reference/html/howto-properties-and-configuration.html#howto-use-short-command-line-arguments – alpert Aug 19 '15 at 06:39
-
1As a compliment to this answer: According to the [spring docs](http://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#boot-features-external-config-application-property-files) there are other paths you can put `application.properties` on. In my case that helped a lot. – sargas Oct 02 '15 at 19:37
-
17-Dserver.port=XXXX did not work for me. I used OS environment variable mode: `$ SERVER_PORT=8090 java -jar
` – Soumya Kanti Oct 08 '15 at 07:38 -
It is also worth noting that once you so this it will only matter locally. Once you deploy this application on a server `server.port = 8080 will be ignored. – Mike3355 Apr 26 '16 at 17:28
-
-
16Both (1) `java -Dserver.port=XXXX -jar
` and (2) `java -jar – tan9 Nov 29 '16 at 16:28--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 --server.port=YYYY`, `YYYY` takes precedence over `XXXX`, this is why Spring Boot Externalized Configuration is so charming. -
Many ways to change port on Spring Boot tomcat http://javabycode.com/spring-framework-tutorial/spring-boot-tutorial/spring-boot-change-tomcat-port-command-line.html – David Pham Nov 16 '17 at 09:48
-
-
You can set the server.port property in many different ways. Look at the configuration-documentation for spring boot and find the solution that suits you: https://docs.spring.io/spring-boot/docs/current/reference/html/boot-features-external-config.html – jorgen.ringen Jan 28 '18 at 17:41
-
In case you prefer maven, do this `mvn spring-boot:run -Dserver.port=8888` – old-monk Apr 12 '18 at 15:37
-
-
-Dserver.port did not work but --server.port worked. Also it should be passed to java -jar command and not mvn. So for example after building the projekt using mvn you execute java -jar artifact.jar --server.port=9099 to run the server on another to port – velocity Sep 03 '20 at 12:26
-
-
There is nice explanation of ways to change port of spring boot https://www.javavogue.com/2019/02/how-to-change-the-default-port-in-spring-boot/ – Anuj Dhiman Jan 08 '22 at 09:19
-
Also you can set the management port to expose the actuator on different port – amir bayan Dec 08 '22 at 21:15
-
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

- 10,427
- 6
- 56
- 72
-
In STS 4 it is at run -> run configurations -> main, then scroll down to Table with Parameter Name and Value – serv-inc Apr 28 '19 at 13:00
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:
Pass property through command line argument as application argument
java -jar <path/to/my/jar> --server.port=7788
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}'
Define JVM system property
java -Dserver.port=7788 -jar <path/to/my/jar>
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>
Place property in
./config/application.properties
configuration fileserver.port=7788
and run:
java -jar <path/to/my/jar>
Place property in
./config/application.yaml
server: port: 7788
and run:
java -jar <path/to/my/jar>
Place property in
./application.properties
server.port=7788
and run:
java -jar <path/to/my/jar>
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.

- 3,490
- 2
- 18
- 21
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);
});
}
}

- 2,481
- 2
- 27
- 41

- 2,179
- 2
- 18
- 27
-
2This is working and very useful, when you have port in your own config file and want to set it during runtime. – Xdg Nov 08 '15 at 09:48
-
6This was helpful when i needed to deploy an application to a AWS Elastic Beanstalk service, to get the port from an environment variable. – Martin Hansen Nov 19 '15 at 12:51
-
This is super useful when all you want is a self-contained unit or integration test, +1. – Priidu Neemre Apr 28 '16 at 10:34
-
Very useful when the env variable for port is already defined under a different name. – higuaro Aug 18 '16 at 08:27
-
2Is'nt it the `@Configuration` instead of `@Controller`? Please update if so. – Lucky Sep 13 '16 at 11:52
-
It should be `@Configuration`, but if you import it using `@Import` it works, not sure if it is working with auto configuration search. – kap Mar 23 '17 at 13:00
-
For a beginner: I tried creating a file `PortConfiguration.java` in the package folder and pasted this code here. However, I'm geting an error from VS code: Configuration cannot be resolved to a type. Any suggestions? – PeMa Mar 02 '18 at 07:12
-
interesting, I didn't know that it is possible to do it programmatically. Thanks for info – Heybat Dec 17 '20 at 23:24
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:

- 15,253
- 21
- 95
- 158

- 1,314
- 1
- 10
- 17
-
5Starting from Spring Boot 2, you should use `spring-boot.run.jvmArguments`. – mapm Apr 12 '18 at 02:57
-
1
-
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 – Manuel Jordan Jul 16 '23 at 16:41
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

- 1,879
- 1
- 17
- 13
-
Using the HashMap will work only if no port is set in applications.properties or .yml. – Milgo Apr 25 '19 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! – Thomas Schütt Oct 08 '21 at 12:43
In case you are using application.yml
add the Following lines to it
server:
port: 9000
and of course 0 for random port.

- 60,022
- 51
- 208
- 332

- 1,193
- 7
- 12
-
1this didn't seem to work. I used server.port in the application.yml and it worked – yathirigan Mar 05 '15 at 13:11
-
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

- 2,451
- 1
- 19
- 31
Include below property in application.properties
server.port=8080

- 3,669
- 5
- 35
- 50

- 552
- 4
- 6
-
This doesn't work. What application.properties? Which properties can overwrite this one? How can I be sure Spring is picking it up? – Philip Rego Dec 10 '21 at 16:24
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

- 397
- 3
- 6
-
2`System.setProperty("server.port", 80);` is another way to achieve the same. – hd1 Feb 11 '17 at 20:34
-
@hd1, I added our answers to the main answer, check it out and modify as you se fit please – Luis Mauricio Dec 30 '19 at 14:40
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

- 4,410
- 6
- 26
- 47

- 471
- 4
- 8
-
-
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). – SeverityOne Jun 03 '21 at 09:01
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() {
}
}
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.
-
1Direct link: http://docs.spring.io/spring-boot/docs/current-SNAPSHOT/reference/htmlsingle/#howto-configure-tomcat – Erik Martino Nov 13 '14 at 10:24
-
Here's a current link to the programmatic override section: https://docs.spring.io/spring-boot/docs/current-SNAPSHOT/reference/htmlsingle/#boot-features-customizing-embedded-containers – jocull Jun 14 '18 at 15:33
In application.properties
file present in resources:
server.port=8082

- 6,052
- 10
- 43
- 117

- 1,090
- 1
- 12
- 25
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");
Add this in your application.properties
file
server.port= 8080

- 3,669
- 5
- 35
- 50

- 159
- 1
- 3
-
2
-
Why to add same asnwer one year later?!? and `server.port 8080` is wrong syntax for Java property file... – Betlista Feb 08 '18 at 13:19
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

- 884
- 2
- 11
- 29
As everyone said, you can specify in application.properties
server.port = 9000 (could be any other value)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)

- 1,374
- 1
- 16
- 24
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:

- 4,491
- 1
- 27
- 39
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

- 901
- 12
- 20
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.

- 467
- 8
- 10
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.

- 1,558
- 2
- 27
- 41
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
)
Hope this one help
application.properties=> server.port=8090 application.yml=> server port:8090

- 315
- 3
- 8
You can add the port in below methods.
Run -> Configurations section
In
application.xml
addserver.port=XXXX

- 1,707
- 2
- 20
- 31

- 471
- 6
- 17
-
1Do you mean `application.yml` and what IDE are you using? Please be specific. – Lucky Sep 13 '16 at 11:56
Just have a application.properties
in src/main/resources
of the project and give there
server.port=****
where ****
refers to the port number.

- 2,843
- 4
- 29
- 44

- 131
- 1
- 4
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);
}
}

- 15,341
- 5
- 46
- 64

- 182
- 2
- 13
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"

- 771
- 10
- 5
By Default Spring-web module provides an embedded tomcat server runs on port number 8080.
You can change it as follows -
A) If you are using gradle then use can set the property in your application.yml :
server:
port: 8042
B) If you are using maven then you can set the property in your application.properties :
server.port: 8042
C) When you have port in your own config file and want to set it during runtime.
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(8042);
}
}
By Implementing EmbeddedServletContainerCustomizer interface - Spring 1.x
@Component
public class CustomizationBean implements EmbeddedServletContainerCustomizer {
@Override
public void customize(ConfigurableEmbeddedServletContainer container) {
// customize here
container.setPort(8042);
}
}

- 571
- 5
- 13
-
I agree A) and B) should be your options if you are not doing anything weird – Javi Vazquez Jul 27 '21 at 20:23
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.

- 46,221
- 15
- 164
- 151
You can set that in application.properties under /src/main/resources/
server.port = 8090

- 5,596
- 5
- 42
- 50
Mostly springboot runs on port:8080
because of embedded Tomcat. In some it may throw an error port 8080 already in use
. To avoid this kind of issues we can config the server port.
Using application.properties
add server.port=9898
On runtime config
run your application with below arguments.
spring-boot:run -Drun.jvmArguments='-Dserver.port=8081'

- 396
- 3
- 8
- 21
-
1This answer is outdated. You need to use the following arguments `mvn spring-boot:run -Dspring-boot.run.jvmArguments='-Dserver.port=8081'` , see https://stackoverflow.com/a/66370832/179014 – asmaier Nov 25 '22 at 13:43
-
I've confirmed it works in SB 3.1.1 - BTW is not necessary use `''`, therefore is `mvn spring-boot:run -Dspring-boot.run.jvmArguments=-Dserver.port=8081` works – Manuel Jordan Jul 16 '23 at 16:24
Using mvn shell command line, spring-boot 2:
mvn spring-boot:run -Dspring-boot.run.jvmArguments='-Dserver.port=8085'

- 1,040
- 2
- 10
- 18
You can configure your port in application.properties file in the resources folder of your spring boot project.
server.port="port which you need"

- 3,669
- 5
- 35
- 50

- 116
- 1
- 3
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 namedPORT
, 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 namedPORT
, 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.
- For Gradle:
- 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

- 898
- 8
- 20

- 1,367
- 10
- 11
Open the application.properties file in your spring boot application. And add the below property in the properties file.
server.port = 1443
This will be working fine and you can set any port number as per your wish.

- 83
- 6

- 124
- 1
- 4
"server.port=8080" will only works if your running application as a jar through main method,
This configuration will not work if your running this application as a war file through tomcat container.

- 99
- 6
If you are going to run apps as jar file in command environment, just type "SERVER_PORT=*** " as prefix. The full command to execute will look like below:
SERVER_PORT=8080 java -jar ***.jar
If you wanna run app in background in Linux, command with 'nohup' will look like below:
SERVER_PORT=8080 nohup java -jar ***.jar &

- 338
- 2
- 13
There are three ways to do it
1 Set server.port property in application.properties
file
server.port = 8090
2 Set server port property in application.yml
file
server:
port: 8090
3 Set the property as system property in main
method
System.setProperty("server.port","8090");

- 3,669
- 5
- 35
- 50

- 79
- 5
Just set the environment variable SERVER_PORT
.
(The examples works on Linux
)
Start via java -jar:
SERVER_PORT=9093 java -jar target/eric-sc-dummy.jar
Start via maven spring-boot plugin:
SERVER_PORT=9093 mvn spring-boot:run
Tips:
- If you add other sub commands before the
java -jar
ormvn
command, then you need to addexport
to set env in a separate command, and split them via;
, to make sure it's available to sub process.
e.g:
export SERVER_PORT=9093; export MAVEN_OPTS="-Xmx256m -Xms64m"; mvn spring-boot:run

- 22,183
- 20
- 145
- 196
You can configure the port in the application.property file or application.yaml file which is in src/main/resources .
server.port=8080

- 290
- 3
- 5
Apart from all the answers, I would like to point out that most IDE (IntelliJ with Spring plugin, STS) have a feature where it suggests all the configuration keys supported by SpringBoot. (i.e. all the opinionated configuration keywords)

- 2,165
- 1
- 27
- 26
Server port declare in two types
1.static type
server.port=8080. // your port number
Dynamic type
server.port=0. // randomly generate port number. server.port=${PORT:0}

- 7,808
- 5
- 32
- 49

- 41
- 1
- 7
Via
application.properties
server.port = 8082
(or any new port number)
via
application.yml
server
port: 8082

- 638
- 1
- 11
- 15
The default port is : 8080 but we can customize the port number in application.properties as shown below
spring.mvc.view.prefix=/WEB-INF/jsp/
spring.mvc.view.suffix=.jsp
server.port = 5050 -- #here we can give any port number.

- 2,994
- 1
- 30
- 33
This question is the first result if you google for Gradle Spring Port.
If you use gradle you can do something like this if you have the Spring Boot Gradle Plugin already applied:
bootRun {
args += ["--server.port=[PORT]"]
}
For a more sophisticated answer please see my answer here.

- 1,312
- 3
- 16
- 32
If you are working over boot projects and you wanna configure the port you can give the input in the application.properties file like NOTE:properties file should be under src/main/resource
Spring properties
server.port=9999 If you using the CMD then follow this command -Dserver.port=9999 For default port its server.port=0 Make sure no port is using this port number
Similar to https://stackoverflow.com/a/36865796/1587329 and https://stackoverflow.com/a/40799750/1587329, a gradle one-liner would be
SERVER_PORT=9090 gradle bootRun

- 35,772
- 9
- 166
- 188
Programmatically, with spring boot 2.1.5:
import org.springframework.boot.web.server.WebServerFactoryCustomizer;
import org.springframework.boot.web.servlet.server.ConfigurableServletWebServerFactory;
import org.springframework.stereotype.Component;
@Component
public class CustomizationBean implements WebServerFactoryCustomizer<ConfigurableServletWebServerFactory> {
@Override
public void customize(ConfigurableServletWebServerFactory server) {
server.setPort(9000);
}
}

- 7,806
- 9
- 63
- 100
This worked for me :
Added a custom container class :
@Component
public class CustomContainer implements EmbeddedServletContainerCustomizer {
@Override
public void customize(ConfigurableEmbeddedServletContainer configurableEmbeddedServletContainer) {
configurableEmbeddedServletContainer.setPort(8888);
}
}
But this was still not using port 8888.
Then I set "scanBasePackages" property like this on "@SpringBootApplication" class on my main method: (scanBasePackages = {"custom container package"})
@SpringBootApplication(scanBasePackages = {"com.javabrains.util"})
public class CourseApiApp {
public static void main (String args []) {
SpringApplication.run(CourseApiApp.class, args);
}
}
And it started picking up port set in Custom Container.

- 1,203
- 18
- 16
In spring boot you can easily configure the service exposing port in application.properties
file.
server.port=portnumber
if you don't set a port specifically, then it will try to open the service with port 80
. in case if it is already used, service will not be started on the embedded container.

- 3,669
- 5
- 35
- 50

- 977
- 2
- 13
- 33
if your port number can be random you can use random function in your application.properties
server.port=${random.int(4)}

- 328
- 1
- 15
-
I can't think of a scenario where you would want to have random port number – Sudip Bhandari Jun 28 '19 at 08:48
in the file application.properties add the following: server.port=8888 THE PROT NEEDED ALWAYS MENTIONED HERE

- 709
- 8
- 3
Aside from adding the port on application.properties, you can also easily achieve multiple ports for different environments, by combining the property files approach with Spring profiles. Specifically, we can create a property file for each environment.
For example, we'll have an application-dev.properties file with this content:
server.port=8081
Then you can add another application-qa.properties file with a different port:
server.port=8082

- 814
- 6
- 10
Configure the port details in Config file or application properties.
e.g.
port =8876

- 7,252
- 10
- 25
- 39

- 35
- 1
- 5
If you are use the spring command line interface (CLI) use the --
to separate commands from the spring
command arguments, to change the port:
spring run hello.groovy -- --server.port=9000

- 824
- 2
- 17
- 35
Many parameters including server port can be changed in multiple ways. However there is an order of precedence as outlined below:
- First precedence is assigned to custom code like below:
@Component
public class CustomConfiguration implements WebServerFactoryCustomizer {
@Override
public void customize(ConfigurableServletWebServerFactory factory) {
factory.setPort(9090);
}
}
Here we are setting server port as 9090 which is hard coded in the code. To avoid hardcoding, we can assign a value from environment using @Value annotation in a bean class and the use it here.
Second precedence is assigned to command line arguments like below:
java -jar target/spring-boot-0.0.1-SNAPSHOT.jar --server.port=8092
Here we are telling the server to start listening at 8092. Note if we use both the above approaches together, it will ignore command line argument as custom code is given the first precedence.
Third precedence is assigned to OS environment variable. If none of the above two approaches is taken up, Spring will take server port from environment property. In case of deployment on Kubernetes, a property set under env section in Deployment yaml will be used.
Fourth precedence is assigned to profile specific application.properties file.
Fifth precedence is assigned to value assigned in application.properties file (which by default Spring Boot tries to find src/main/resources/config, if not found, then tries to find under src/main/resources).
Most manageable and useful approach can be combination of first and third approach. You can use an environment property and use that custom code.
Sample code:
import org.springframework.beans.factory.annotation.Value;
import org.springframework.stereotype.Component;
@Component
public class EnvironmentCustomizer {
@Value("${server.port}")
private int serverPort;
public void setServerPort(int serverPort) {
this.serverPort = serverPort;
}
public int getServerPort() {
return serverPort;
}
}
@Configuration
public class CustomConfiguration
{
@Autowired
EnvironmentCustomizer envCustomizer;
@Bean
WebServerFactoryCustomizer<ConfigurableServletWebServerFactory> webServerPortCustomizer() {
return factory -> factory.setPort(envCustomizer.getServerPort());
}
}

- 168
- 2
- 11
Here's a Kotlin approach for setting the port in code:
fun main(args: Array<String>) {
runApplication<TestApplication>(*args) {
setDefaultProperties(mapOf(
"server.port" to 8081
))
}
}

- 4,656
- 1
- 29
- 36
In my case adding statement
server.port=${port:8081}
override the default tomcat server port.

- 25
- 1
- 1
- 5