I have created a symlink for a spring boot executable jar and i am able to start the application. I want to customize the logs path, pid folder etc.
Went through the customizing the startup of this script.
But I could not find where to store the custom script and how it can be linked to executable application jar. Could you please assist?
Create a script with name your-app.service, place this script in /etc/systemd/system directory.
Installation as a systemd Service, using Java System Properties (VM Arguments):
[Unit]
Description= Spring Boot App
After=syslog.target
[Service]
User=myapp
ExecStart=java -Dspring.application.name=example -Dlogging.file=/opt/spring-boot-app/log/app.log -Dspring.pid.file=/opt/spring-boot-app/app.pid -jar /opt/spring-boot-app/app.jar
SuccessExitStatus=200
[Install]
WantedBy=multi-user.target
Or pass throught via program arguments:
java -jar /opt/spring-boot-app/app.jar --spring.application.name=example --logging.file=/opt/spring-boot-app/log/app.log --spring.pid.file=/opt/spring-boot-app/app.pid
References in here and here.
logging.file= # Log file name (for instance, myapp.log). Names can
be an exact location or relative to the current directory.
spring.pid.file= # Location of the PID file to write (if
ApplicationPidFileWriter is used).
Related
I am planning to use a native image with Spring Boot 3. My environment-specific properties are stored in the application.properties file. Sample file
spring.config.activate.on-profile=dev
server.port=9092
#---
spring.config.activate.on-profile=local
server.port=9093
I build the native executable using the following command
./mvnw -Pnative native:compile -Dspring.profiles.active=local
and run the executable using the below command
./target/<app-executable> -Dspring.profiles.active=local
In the logs I see
No active profile set, falling back to 1 default profile: "default"
Am I doing something wrong or Are profiles not supported with Spring Boot 3 native image?
Able to make it work by passing the arguments as -- instead of using -D.
For building the native image, no need to pass -D.
./mvnw -Pnative native:compile
And to run the executable with system properties, use the below command
./target/<app-executable> --spring.profiles.active=local
This picked up the server.port=9093.
I run my web application with Payara with java -jar webapp.war
How do I set where the this Payara instance would log, for example I want to set it to /var/log/mywebapp.log or just relative to the Uber Jar file, what would be the command to pass?
My goal is to tail the logs from ssh so the log needs to have a path.
You can do it using from command line java -jar payara-micro.jar --logToFile /path/file.log or you can do it programmatically using PayaraMicro.getInstance().setUserLogFile("/path/file.log").bootStrap();, but I don't know if you need to use embbeded payara for do this on the main method. You can even create a logging.properties file and configure it, so you use --logProperties logging.properties argument when run your jar file.
I have spring boot application with executable jar-file
So, I tried to start to from cmd in windows by command
java -jar -Dspring.config.location="application.yaml" MyService.jar
and it sucessfully started
(I try to use external application.yaml to start my application)
In my application all properties.yaml like datebase mocked like this
url: ${DATA_BASE_URL}
and external yaml contains all values.
But, when I try to start application using this command on centOS (or other linux), there is fail when try to start, spring boot used ${DATA_BASE_URL} instead external application.yaml values.
How to fix it?
On linux I used
nohup java -jar -Dspring.config.location="application.yaml" MyService.jar
According to the official documentation https://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#boot-features-external-config-application-property-files
you should use double dashes
nohup java -jar --spring.config.location="application.yaml" MyService.jar
I have a Spring Boot application and I'm runnig it java -jar command on CMD. But I need to use external config file that is application.properties in C:\ folder. To do this I'm running the application with this command:
java -jar app.jar --spring.config.name=application --spring.config.location=file:///C:/
But command doesn't cares the spring.config.location property and application.properties file can't be found. Because of this the application not working.
How can I run the app with spring.config.name and spring.config.location parameters in Windows CMD?
On Linux you should use file://sample-app/application.yml
On Windows you should use file:///C:/sample-app/application.yml
You try placing the jar and application.properties files in same folder and run java -jar yourapplication.jar
See if it works.
I have a Java Spring Boot application which is deployed on an Azure Webservice, it's all working. What I really want to do at this moment is the following:
"java -jar -Dspring.profiles.active=test app.jar", I want to run the application based on that command via the imput field (startup command) on Azure Cloud, see next image:
The problem is that it gives me the following error: Unable to access jarfile app.jar, I find this information in the logs.
I had to set the Startup Command to
java -jar /home/site/wwwroot/app.jar --server.port=80
for my Linux based WebApp service to pick up the jar properly and start my Spring Boot 2.2 application. Note the path is absolute and starts from /home, I found that info here: https://learn.microsoft.com/en-us/azure/app-service/faq-app-service-linux#built-in-images.