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Any way to create a permanent JUnit config. on IntelliJ?

I try to test my spring boot application via JUnit 5.
The test methods need some environment variables. Even I put them on the list, each test method creates a new JUnit config so I have to put them again.
I saw the document on JetBrains.
I can not save the JUnit config. because each config. depends on the method.
Is there any way to create a permanent JUnit config. on IntelliJ?

You may edit the Junit run configuration in the "Run configuration templates for new projects":

This is a bad idea. Setting up your environment variables in something like IntelliJ will make this unportable or unrunnable in a Jenkins environment.
Because you're using Spring Boot, you have more flexibility than you realize. Rework your code so that the environment variables can be injected in through application properties. Then, you can add the application properties to your test through the #SpringBootTest annotation.

Spring Boot not working with dependent application.properties

I'm working with the Spring Boot 2.2.9.RELEASE. I have the main app and some plain starter (which just uses spring-actuator functionality) with some properties in its some-starter/src/main/resources/application.properties file:
management.server.port=9000
management.server.ssl.enabled=false
management.endpoints.enabled-by-default=false
management.endpoint.health.enabled=true
management.endpoints.web.base-path=/
management.endpoints.web.path-mapping.health=health
I've imported the starter into my main app and I believe that the healthcheck endpoint should work with the port 9000 and with the path /health (smth like that localhost:9000/health).
But it doesn't. However, it works in case of the same properties in my main app main-app/src/main/resources/application.properties.
Is it problem with the property overriding in Spring Boot? Should i configure my resources via something like maven-resources-plugin?

When application.properties is loaded from the classpath, the first one on the classpath is loaded and any others on the classpath are ignored. In your case, the file in main-app/src/main/resources/application.properties will appear on the classpath before the application.properties in the jar of some-starter.
As its name suggests, application.properties is intended for configuring your application and it shouldn't be used in a starter. You should either configure all of the properties in your application, or you could update your starter to include an EnvironmentPostProcessor that is registered via spring.factories and adds some default properties to the Environment.

adding new property based on spring profile

If spring boot is run in override profile , can we have application-override.properties having properties like foo.baz that is not defined in application.properties ?
application.properties
foo.bar=1
application-override.properties
spring.profiles.include=default
foo.baz=1

That is correct. When you have new properties in application-override.properties and and the override profile is the active profile, then yes in your program the properties from application.properties as wel as application-override.properties is loaded.
Using spring.profiles.include=default in your override profile is not needed.
In the case of loading multiple specific profiles with same properties:
Also, in the context of property overriding with profiles, something to keep in mind when you have multiple active profiles and they contain the same property. The last profile in the list will be used.
Let's say you start up your program with mvn spring-boot:run -Drun.profiles=profile1,profile2
Both application-profile1.properties and application-profile2.properties
contains the property my.custom-property=x (for profile1) and my.custom-property=y (for profile2). The value of my.custom-property will be y, as that was the last profile in the provided profiles.

You can create configuration class for your custom profile and load the appropriate properties file in it like this:
#Configuration
#Profile("override")
#PropertySource("classpath:application-override.properties")
public class OverrideConfig {
}
This way, all the configuration you do in OverrideConfig (including taking properties from application-override.properties), will only load if override profile is enabled in application.properties like this:
spring.profiles.active=override

Long story short : Spring boot overrides values of properties with same name according to their evaluation order. But here you don't override any property, you add a new.
That is still simpler : Spring boot just adds it into the Spring Environment.
Just run the app by specifying this profile and makes sure that the properties are located in the locations expected by Spring Boot.
Example from a fat jar (Java system property) :
java -Dspring.profiles.active=override -jar foo.jar
Example from the source code (Maven property) :
mvn spring-boot:run -Dspring-boot.run.profiles=override

Yes, you can do this by simply add the profile name to the application.properties:
application-override.properties
Then you can load profile from the command line:
java -jar foo.jar --spring.profiles=override
source: https://docs.spring.io/spring-boot/docs/current/reference/html/howto-properties-and-configuration.html#howto-change-configuration-depending-on-the-environment
Spring will load the application.properties first followed by any application-{profile}.properties.
Another option is to use yaml, and load everything into one file:
foo:
bar: 1
---
spring:
profiles: override
foo:
baz: 1
---
spring:
profiles: otherOverride
foo:
bar: 2
baz: 2

Spring Boot refuses to pick up application test properties [duplicate]

I have a spring boot application that I can package in a war that I want to deploy to different environments. To automate this deployment it'd be easier to have the configuration file externalized.
Currently everything works fine with a application.properties file in src/main/resources. Then I use ´mvn install´ to build a war deployable to tomcat.
But I would like to use a .yml file that does not need to be present on mvn install but that would be read from during deployment of the war and is in the same or a directory relative to my war.
24. externalized configuration shows where spring boot will look for files and 72.3 Change the location of external properties of an application gives more detail on how to configure this but I just do not understand how to translate this to my code.
My application class looks like this:
package be.ugent.lca;
Updated below
Do I need to add a #PropertySource to this file? How would I refer to a certain relative path?
I feel like it's probably documented in there as most spring boot documentation but I just don't understand how they mean me to do this.
EDIT
Not sure if this should be a separate issue but I think it's still related.
Upon setting the os variable the error of yaml file not found went away. Yet I still get the same error again as when I had no application .properties or .yml file.
Application now looks like this:
#Configuration
**#PropertySource("file:${application_home}/application.yml")**
#ComponentScan({"be.ugent.lca","be.ugent.sherpa.configuration"})
#EnableAutoConfiguration
#EnableSpringDataWebSupport
public class Application extends SpringBootServletInitializer{
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
The application_home OS variable
$ echo $application_home
C:\Masterproef\clones\la15-lca-web\rest-service\target
My application.yml file(part it complains about):
sherpa:
package:
base: be.ugent.lca
Error upon java -jar *.war
All variations upon:
Caused by: java.lang.IllegalArgumentException: Could not resolve placeholder 'sherpa.package.base' in string value "${sherpa.package.base}"
at org.springframework.util.PropertyPlaceholderHelper.parseStringValue(PropertyPlaceholderHelper.java:174)
at org.springframework.util.PropertyPlaceholderHelper.replacePlaceholders(PropertyPlaceholderHelper.java:126)
at org.springframework.core.env.AbstractPropertyResolver.doResolvePlaceholders(AbstractPropertyResolver.java:204)
at org.springframework.core.env.AbstractPropertyResolver.resolveRequiredPlaceholders(AbstractPropertyResolver.java:178)
at org.springframework.context.support.PropertySourcesPlaceholderConfigurer$2.resolveStringValue(PropertySourcesPlaceholderConfigurer.java:172)
at org.springframework.beans.factory.support.AbstractBeanFactory.resolveEmbeddedValue(AbstractBeanFactory.java:808)
at org.springframework.beans.factory.support.DefaultListableBeanFactory.doResolveDependency(DefaultListableBeanFactory.java:1027)
at org.springframework.beans.factory.support.DefaultListableBeanFactory.resolveDependency(DefaultListableBeanFactory.java:1014)
at org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor$AutowiredFieldElement.inject(AutowiredAnnotationBeanPostProcessor.java:545)
... 142 more

Using external properties files
The answer lies in the Spring Boot Docs, I'll try to break it down for you.
First of all, no you should not use #PropertySource when working with Yaml configuration, as mentioned here under the Yaml shortcomings :
YAML files can’t be loaded via the #PropertySource annotation. So in the case that you need to load values that way, you need to use a properties file.
So, how to load propery files? That is explained here Application Property Files
One is loaded for you: application.yml , place it in one of the directories as mentioned in the link above. This is great for your general configuration.
Now for your environment specific configuration (and stuff like passwords) you want to use external property files, how to do that is also explained in that section :
If you don’t like application.properties as the configuration file name you can switch to another by specifying a spring.config.name environment property. You can also refer to an explicit location using the spring.config.location environment property (comma-separated list of directory locations, or file paths).
So you use the spring.config.location environment property.
Imagine you have an external config file: application-external.yml in the conf/ dir under your home directory, just add it like this:
-Dspring.config.location=file:${home}/conf/application-external.yml as a startup parameter of your JVM.
If you have multiple files, just seperate them with a comma. Note that you can easily use external properties like this to overwrite properties, not just add them.
I would advice to test this by getting your application to work with just your internal application.yml file , and then overwrite a (test) property in your external properties file and log the value of it somewhere.
Bind Yaml properties to objects
When working with Yaml properties I usually load them with #ConfigurationProperties, which is great when working with for example lists or a more complex property structure. (Which is why you should use Yaml properties, for straightforward properties you are maybe better of using regular property files). Read this for more information: Type-Safe Configuration properties
Extra: loading these properties in IntelliJ, Maven and JUnit tests
Sometimes you want to load these properties in your maven builds or when performing tests. Or just for local development with your IDE
If you use IntelliJ for development you can easily add this by adding it to your Tomcat Run Configuration : "Run" -> "Edit Configurations" , select your run configuration under "Tomcat Server" , check the Server tab and add it under "VM Options".
To use external configuration files in your Maven build : configure the maven surefire plugin like this in your pom.xml:
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-surefire-plugin</artifactId>
<configuration>
<argLine>-Dspring.config.location=file:${home}/conf/application-external.yml</argLine>
</configuration>
</plugin>
When running JUnit tests in IntelliJ:
Run → Edit Configurations
Defaults → JUnit
add VM Options -> -ea -Dspring.config.location=file:${home}/conf/application-external.yml

Yes, you need to use #PropertySource as shown below.
The important point here is that you need to provide the application_home property (or choose any other name) as OS environment variable or System property or you can pass as a command line argument while launching Spring boot. This property tells where the configuration file (.properties or .yaml) is exactly located (example: /usr/local/my_project/ etc..)
#Configuration
#PropertySource("file:${application_home}config.properties")//or specify yaml file
#ComponentScan({"be.ugent.lca","be.ugent.sherpa.configuration"})
#EnableAutoConfiguration
#EnableSpringDataWebSupport
public class Application extends SpringBootServletInitializer{
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}

There is a very simple way to achieve this.
Inside your original application.properties file you can just specify the following line:
spring.config.import=file:Directory_To_The_File/Property_Name.properties
It will automatically sync all the properties from the external property file.
Now lets say that you have a situation where you need to get properties from multiple property files. In that case, you can mention the same line in the external property file which in turn will take the remaining properties from the second property file and so on.
Consider the following example.
application.properties:
spring.config.import=file:Resources/Custom1.properties
Custom1.properties:
server.port=8090
.
.
.
spring.config.import=file:Resources/Custom2.properties

One of the easiest way to use externalized property file using system environment variable is, in application.properties file you can use following syntax:
spring.datasource.url = ${OPENSHIFT_MYSQL_DB_HOST}:${OPENSHIFT_MYSQL_DB_PORT}/"nameofDB"
spring.datasource.username = ${OPENSHIFT_MYSQL_DB_USERNAME}
spring.datasource.password = ${OPENSHIFT_MYSQL_DB_PORT}
Now, declare above used environment variables,
export OPENSHIFT_MYSQL_DB_HOST="jdbc:mysql://localhost"
export OPENSHIFT_MYSQL_DB_PORT="3306"
export OPENSHIFT_MYSQL_DB_USERNAME="root"
export OPENSHIFT_MYSQL_DB_PASSWORD="123asd"
This way you can use different value for same variable in different environments.

Use below code in your boot class:
#PropertySource({"classpath:omnicell-health.properties"})
use below code in your controller:
#Autowired
private Environment env;

Using Spring boot/cloud with Amazon AWS lambda does not inject values

I have an AWS lambda RequestHandler class which is invoked directly by AWS. Eventually I need to get it working with Spring Boot because I need it to be able to retrieve data from Spring Cloud configuration server.
The problem is that the code works if I run it locally from my own dev environment but fails to inject config values when deployed on AWS.
#Configuration
#EnableAutoConfiguration
#ComponentScan("my.package")
public class MyClass implements com.amazonaws.services.lambda.runtime.RequestHandler<I, O> {
public O handleRequest(I input, Context context) {
ApplicationContext applicationContext = new SpringApplicationBuilder()
.main(getClass())
.showBanner(false)
.web(false)
.sources(getClass())
.addCommandLineProperties(false)
.build()
.run();
log.info(applicationContext.getBean(SomeConfigClass.class).foo);
// prints cloud-injected value when running from local dev env
//
// prints "${path.to.value}" literal when running from AWS
// even though Spring Boot starts successfully without errors
}
}
#Configuration
public class SomeConfigClass {
#Value("${path.to.value}")
public String foo;
}
src/main/resources/bootstrap.yml:
spring:
application:
name: my_service
cloud:
config:
uri: http://my.server
failFast: true
profile: localdev
What have I tried:
using regular Spring MVC, but this doesn't have integration with #Value injection/Spring cloud.
using #PropertySource - but found out it doesn't support .yml files
verified to ensure the config server is serving requests to any IP address (there's no IP address filtering)
running curl to ensure the value is brought back
verified to ensure that .jar actually contains bootstrap.yml at jar root
verified to ensure that .jar actually contains Spring Boot classes. FWIW I'm using Maven shade plugin which packages the project into a fat .jar with all dependencies.
Note: AWS Lambda does not support environment variables and therefore I can not set anything like spring.application.name (neither as environment variable nor as -D parameter). Nor I can control the underlying classes which actually launch MyClass - this is completely transparent to the end user. I just package the jar and provide the entry point (class name), rest is taken care of.
Is there anything I could have missed? Any way I could debug this better?

After a bit of debugging I have determined that the issue is with using the Maven Shade plugin. Spring Boot looks in its autoconfigure jar for a META-INF/spring.factories jar see here for some information on this. In order to package a Spring Boot jar correctly you need to use the Spring Boot Maven Plugin and set it up to run during the maven repackage phase. The reason it works in your local IDE is because you are not running the Shade packaged jar. They do some special magic in their plugin to get things in the right spot that the Shade plugin is unaware of.
I was able to create some sample code that initially was not injecting values but works now that I used the correct plugin. See this GitHub repo to check out what I have done.
I did not connect it with Spring Cloud but now that the rest of the Spring Boot injection is working I think it should be straightforward.
As I mentioned in the comments you may want to consider just a simple REST call to get the cloud configuration and inject it yourself to save on the overhead of loading a Spring application with every request.
UPDATE: For Spring Boot 1.4.x you must provide this configuration in the Spring Boot plugin:
<configuration>
<layout>MODULE</layout>
</configuration>
If you do not then by default the new behavior of the plugin is to put all of your jars under BOOT-INF as the intent is for the jar to be executable and have the bootstrap process load it. I found this out while addressing adding a warning for the situation that was encountered here. See https://github.com/spring-projects/spring-boot/issues/5465 for reference.