CODEWITHSUNDEEP
javaspringconfigurationspring-boot
salvador
salvador

Reputation: 1089

Override yml configuration in spring-boot with command line arguments

I have a spring-boot application that is configured with a yml file. Is it possible to override these properties when executing the jar? For example let say I have the input variable in yml file set to user1 and I want to execute the jar with user2. Is it possible to do something like this? 

java -jar --input=user2

Upvotes: 10

Views: 12698

Answers (2)

Joy
Joy

Reputation: 21

this worked for me

java -Dspring.profiles.active=test-environment -jar config-server-0.0.1-SNAPSHOT.jar

If we see below syntax, it tells options should be included between java and -jar. Also, options are listed below as well.

Reason why this command works. Also, it takes the parameters instead of the application.yml configuration.

C:\Users\joy> java -?

Usage: java [-options] -jar jarfile [args...] (to execute a jar file) where options include:

-d32          use a 32-bit data model if available
-d64          use a 64-bit data model if available
-server       to select the "server" VM
              The default VM is server. 
 -cp <class search path of directories and zip/jar files>
 -classpath <class search path of directories and zip/jar files>
               A ; separated list of directories, JAR archives,
               and ZIP archives to search for class files.
 **-D<name>=<value>
               set a system property**

Note: Operating system is windows

Upvotes: 0

eis
eis

Reputation: 53462

To elaborate answer by cLyric, you can do this:

java -jar yourapp.jar --input=user2

Or if you want to provide using json, you can do

java -jar yourapp.jar --spring.application.json='{"input":"user2"}'

Or if you're in unix/linux,

SPRING_APPLICATION_JSON='{"input":"user2"}' java -jar yourapp.jar

Upvotes: 4

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