Spring: overriding one application.property from command line

Question

I have an application.properties file with default variable values. I want to be able to change ONE of them upon running with mvn spring-boot:run. I found how to change the whole file, but I only want to change one or two of these properties.

Andy Wilkinson · Accepted Answer · 2020-06-16T11:55:14.807

199

You can pass in individual properties as command-line arguments. For example, if you wanted to set server.port, you could do the following when launching an executable jar:

java -jar your-app.jar --server.port=8081

Alternatively, if you're using mvn spring-boot:run with Spring boot 2.x:

mvn spring-boot:run -Dspring-boot.run.arguments="--server.port=8081"

Or, if you're using Spring Boot 1.x:

mvn spring-boot:run -Drun.arguments="--server.port=8081"

You can also configure the arguments for spring-boot:run in your application's pom.xml so they don't have to be specified on the command line every time:

<plugin>
    <groupId>org.springframework.boot</groupId>
    <artifactId>spring-boot-maven-plugin</artifactId>
    <configuration>
        <arguments>
            <argument>--server.port=8085</argument>
        </arguments>
    </configuration>
</plugin>

edited Jun 16 '20 at 11:55

answered May 05 '16 at 14:17

Andy Wilkinson

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@yuli I've edited your question and updated my answer with some options for when you're using `mvn spring-boot:run` – Andy Wilkinson May 05 '16 at 14:33
I get `org.kohsuke.args4j.CmdLineException: "--server.port=8181" is not a valid option` – clueless user May 05 '16 at 15:37
Spring Boot doesn't use args4j, so that failure must be coming from your code – Andy Wilkinson May 05 '16 at 16:48
Note that: 1. Arguments should be comma separated 2. Each argument should be prefixed with -- eg. -Dspring-boot.run.arguments="--server.port=8081,--customArgument=custom" – George Kargakis Aug 16 '20 at 15:35
if the argument is a custom argument inside application.properties, say myValue, will it still work? – HKIT Aug 23 '20 at 16:51
This worked for me java -jar -D"server.port=8081" your-app.jar – Amandeep Apr 20 '23 at 08:59

score 23 · Answer 2 · answered Aug 22 '19 at 17:31

23

To update a little things, the Spring boot 1.X Maven plugin relies on the --Drun.arguments Maven user property but the Spring Boot 2.X Maven plugin relies on the -Dspring-boot.run.arguments Maven user property.

So for Spring 2, you need to do :

mvn spring-boot:run -Dspring-boot.run.arguments="--server.port=8081"

And if you need to pass multiple arguments, you have to use , as separator and never use whitespace between arguments :

mvn spring-boot:run -Dspring-boot.run.arguments="--server.port=8081,--foo=bar"

About the the maven plugin configuration and the way of passing the argument from a fat jar, it didn't change.
So the very good Andy Wilkinson answer is still right.

answered Aug 22 '19 at 17:31

davidxxx

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Cool!. System properties like -Dproperty1=value1 -Dproperty2=value2 can also be passed on the command line via mvn spring-boot:run -Dspring-boot.run.arguments="--property1=value1,--property2=value2" – Karthick Meenakshi Sundaram Nov 01 '19 at 12:50
Not working for me. If I put a whitespace the application will not detect the arguments, and if I put a whitespace the SDK will not read it. Still looking for how to add application.properties args. – May 31 '20 at 15:32

score 23 · Answer 3 · answered Jun 16 '20 at 11:57

23

Quick update:

if you are using the latest versions of spring-boot 2.X and maven 3.X, the below command line will override your server port:

java -jar -Dserver.port=9999   your_jar_file.jar

answered Jun 16 '20 at 11:57

Ikbel

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Why in the world did they change it to this syntax of all things? -D? – Spencer Sutton Jan 28 '21 at 23:08
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They're not defining that syntax. That's the `java` executable's syntax for defining *system properties* (the kind you can get from System.getProperty()). Spring just grabs those in addition to its other sources. – Joshua Taylor Aug 06 '21 at 20:27

score 10 · Answer 4 · answered Oct 17 '20 at 03:08

You can set an environment variable to orverride the properties. For example, you have an property name test.props=1 . If you have an environment variable TEST_PROPS spring boot will automatically override it.

export TEST_PROPS=2
mvn spring-boot:run

You can also create a json string with all the properties you need to override and pass it with -Dspring.application.json or export the json with SPRING_APPLICATION_JSON.

mvn spring-boot:run -Dspring.application.json='{"test.props":"2"}'

Or just pass the properties with -Dtest.props=2

mvn spring-boot:run -Dtest.props=2

Tested on spring boot 2.1.17 and maven 3.6.3

score 6 · Answer 5 · answered Jun 14 '20 at 05:12

If not working with comma, to override some custom properties or spring boot properties in multiple mode, use whitespace instead of comma, like this code bellow:

mvn spring-boot:run -Dspring-boot.run.arguments="--server.port=8899 --your.custom.property=custom"

score 3 · Answer 6 · answered Mar 22 '22 at 06:52

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

score 2 · Answer 7 · edited Aug 31 '22 at 00:54

2

If you have the jar file after doing mvn clean install, you can override any property that you have in your application.yml using the double -, like this:

java -jar name_of_your_jar_file.jar --parameter=value

For example, if you need to change your server port when starting you server, you can write the following:

java -jar name_of_your_jar_file.jar --server.port=8888

edited Aug 31 '22 at 00:54

Jakob Sachs

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answered Aug 25 '22 at 18:40

Carlos Ruiz

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score 1 · Answer 8 · edited Feb 12 '20 at 07:00

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In Spring Boot we have provision to override properties as below

mvn spring-boot:run -Dspring-boot.run.arguments=--server.port=8082

edited Feb 12 '20 at 07:00

Jaydeep

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answered Feb 12 '20 at 05:39

Durgesh Kumar

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userdr725 · Answer 9 · 2022-11-30T17:35:18.393

I Was making a mistake with the syntax of commandline command , while passing command-line arguments, I was wrapping multiple arguments between " " and this was the issue. I simply ran the same command having multiple arguments separated by a space without wraaping them between "" and it worked just fine.

Please note this answer is for cases where we are trying to run this scenario from a jar file(not using mvn).

Correct Command: java -jar myJar.jar --com.arg1=10 --com.arg2=1

Incorrect Command: java -jar myJar.jar "--com.arg1=10 --com.arg2=1"

Spring: overriding one application.property from command line

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