I am building my Spring Boot application using Maven, so I can start it with:
java -jar myjar-1.0-SNAPSHOT.jar --spring.profiles.active=prod
I want to have a directory first on the classpath that would allow me to place some files on the filesystem without having to unzip the jar to change them.
I have tried using loader.path, but it does not seem to work.
java -Dloader.path="config/*" -jar myjar-1.0-SNAPSHOT.jar --spring.profiles.active=prod
The config dir is a subdirectory of where the jar is located. I am trying to load a keystore file which is injected as a Resource in my application. There is such a file in the src/main/resources, but that only works in my IDE, not when packaged as a jar. So I want to put a file first on the classpath so that that one is found first on the classpath.
You can use loader.path but only if the Main-Class is PropertiesLauncher (so it depends how you built the JAR file). Maybe you need to re-build the JAR with packaging=ZIP in the Boot plugin (e.g. docs here)? Can you not set the path to the keystore as a "file:" URL?
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I have a spring-boot executable WAR file which I launch using java -jar . Now I want to add the flexibility of having it reference JAR files outside of WAR file so as to enable customization of code.
So e.g. I want to have to flexibility of having a jar being present in d:\mylib\my.jar and reference the classes in my.jar from code within the WAR file.
Is this possible?
If yes, how?
I tried adding Class-Path and even editing the spring-boot-lib path in manifest.mf but no use.
I have a Spring Boot application using Google Pub Sub API. I need to inject Google credentials and other properties using file credentials.json. I put the file in my src/main/resources (otherwise, it will not put the file in the built jar) like this:
spring.cloud.gcp.credentials.location=file:src/main/resources/credentials.json
However, when I build the jar, this file is placed in the root directory and this path is no longer valid. So I am able to run my application from Eclipse, since by the that time, the file is still in my resources directory but I can't run it as a standalone jar after built because the path is suddently just file:credentials.json.
Is there some easy way how to specify the path as relative so it works both in IDE and when running my jar? I can inject the path through env. variables but I would do so only if absolutely necessary.
If you use the classpath prefix then Spring will look for the file on your classapth.
If you put the file in src/main/resources then Maven will, by default, copy it to the root of your classpath and it will then be addressable as follows:
spring.cloud.gcp.credentials.location=classpath:credentials.json
This should hold true whether ...
You are running in your IDE; your IDE's Maven integration will copy the file from src/main/resources to the root of your classpath - typically target/classes
You are running a built JAR; Maven will copy the file from src/main/resources to the root of your JAR
I'm using spring boot, the config file bootstrap.yml located at src/main/resources. When I build a jar file, the file bootstrap.yml was also included in the jar. That caused I can not modify the file once the jar was build.
So how should I move the bootstrap.yml file out of the jar, I tried remove it out and using -classpath to indicate the file, but it doesn't work. I tried command as following:
java -classpath /user/local/bootstrap.yml -jar spring_test.jar
You can place bootstrap.yml outside the jar simply, and startup the application with regular command such as
nohup java -jar xx.jar >nohup.log &
It will work as you wish, and config in this file is loaded with higher precedence than that one (also name as bootstrap.yml) inside the jar.
I have a spring boot jar(name : myjar) located in the directory structure
D:/hello/myjar
The fat jar contains profile specific application.properties in src/main/resources for e.g. application-local.properties
I want to override the attributes defined in the application-local.properties inside the jar with application-local.properties outside the jar
Hence I created application-local.properties and kept it in the same folder as jar i.e. D:/hello
However when I run my jar using the command :
java -jar -Dspring.profiles.active=local D:/hello/myjar.jar
it still picks the properties which is inside the jar. Am I missing something ?
Try this use -D before -jar
java -Dspring.profiles.active=local -jar D:/hello/myjar.jar
let's see https://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#boot-features-external-config
I think it was more of point that from where I was running my jar from the command prompt.
If I run the jar from the folder where my jar is present, it does picks the profile specific files present outside the jar.
From the documentation of Spring Boot, I can see the directory of the executable jar will be the working directory when you start the Spring Boot application. I would like to understand, whether there are configurations/flags to explode/unpack the jar on deployment, so that I can find access the contents of the executable jar in the file system ?
There are not flags to explode the jar automatically but you can extract it yourself using the jar command. This reference may be helpful http://docs.spring.io/spring-boot/docs/current/reference/html/executable-jar.html in understanding the structure and what you can do.