how to set profile just for the executable (jar) built by calling bootJar task? Ideally, it would override (or add) a property in application.properties inside the jar or something like that.
I expect he is trying to do exactly what we are. In maven one can do something along these lines:
<profile>
<id>openshift</id>
<properties>
<activatedProperties>openshift</activatedProperties>
</properties>
</profile>
Which can be used to replace #activatedProperties# in application.properties to, e.g., openshift. If this is the case, we solved it by modifying build.gradle as follows:
import org.apache.tools.ant.filters.ReplaceTokens
# whatever you have here
def springProfile = project.hasProperty("profile") ? profile : ""
bootRun {
systemProperties["spring.profiles.active"] = springProfile
}
processResources {
with copySpec {
from "src/main/resources/"
filter(ReplaceTokens, tokens: ["activatedProperties": springProfile])
}
}
and in the application.properties have:
spring.profiles.active=#activatedProperties#
However, I would like to note that we consider this a temporary hack. Build scripts should not be modifying files, even if it's limited to resource files. A more correct approach would be to use templates in a separate location and output those to the resources. Or even better (which is what we will do in due time), fix the deployment setup, so it's part of the deployment configuration, as it should be.
You can provide spring.profiles.active=dev,mysql in your application/bootstrap.properties file of your spring boot application.
Or while running the jar you can provide the active profile as argument,
java -jar fatJar.jar --spring.profiles.active=dev,mysql
Related
I have a SpringBoot application where I have application.properties file outside of project (it's not in usual place src/main/resources).
While building application with gradle clean build, it fails as code is not able to find properties files.
I have tried many command to pass vm args, gradle opts but its not working.
gradle clean build -Djvmargs="-Dspring.config.location=/users/home/dev/application.properties" //not working
It fails on test phase when it creates Spring application context and not able to substitute property placeholders. If I skip test as gradle clean build -x test it works.
Though I can run the app with java -jar api.jar --spring.config.location=file:/users/home/dev/application.properties
Please help how I can pass spring.config.location=/users/home/dev/application.properties in gradle build using command line so that build runs with all Junit tests
If I were you, I would not get involved the actual properties to junit test. So I would create a test properties for the project under src/test/resources/application-test.properties and in junit test I would load the test properties.
Example:
#RunWith(SpringRunner.class)
#SpringBootTest(classes = MyProperties.class)
#TestPropertySource("classpath:application-test.properties")
public class MyTestExample{
#Test
public void myTest() throws Exception {
...
}
}
System properties for running Gradle are not automatically passed on to the testing framework. I presume this is to isolate the tests as much as possible so differences in the environment will not lead to differences in the outcome, unless explicitly configured that way.
If you look at the Gradle API for the Test task, you can see that you can configure system properties through through the systemProperty method on the task (Groovy DSL):
test {
systemProperty "spring.config.location", "/path/to/my/configuration/repository/application.properties"
}
If you also want to read a system property from the Gradle command line and then pass that the test, you have to read it from Gradle first, e.g. as a project property, and then pass that value to the test:
test {
if (project.hasProperty('testconfig')) {
systemProperty 'spring.config.location', project.getProperty('testconfig')
}
}
Run it with gradle -Ptestconfig="/path/to/my/configuration/repository/application.properties" build
However, I would discourage using system properties on the build command line if you can avoid it. At the very least, it will annoy you greatly in the long run. If the configuration file can be in different locations on different machines (depending on where you have checkout out the repository and if it is not in the same relative path to your Spring Boot repository), you may want to specify it in a personal gradle.properties file instead.
I think there is a misunderstanding.
spring.config.location is used at runtime
As you validated:
java -jar api.jar --spring.config.location=file:/users/home/dev/application.properties
spring.config.location is used or required at runtime, not at build time.
When your spring boot app is building, an application.properties is required. An approach could be use an src/main/resources/application.properties with template values, but at runtime you will ignore it spring.config.location=file...
For unit tests
In this case as #nikos-bob said, you must use another properties, commonly inside of your src/test/resources
Environment variables instead external properties
We don't want to have hardcoded values in our main git repository src/main/resources/application.properties so the first idea is use an external properties. But this file must be stored in another git repository (equal to main repository ) or manually created.
Spring and other frameworks give us an alternative: Use environment variables.
So instead of manually external creation of application.properties or store it in our git repository, your spring boot app always must have an application.properties but with environment variables:
spring.datasource.url=jdbc:oracle:thin:#${DATABASE_HOST}:${DATABASE_PORT}:${DATABASE_SID}
spring.datasource.username=${DATABASE_USER}
spring.datasource.password=${DATABASE_PASSWORD}
spring.mail.host = ${MAIL_HOST}
spring.mail.username =${MAIL_USERNAME}
spring.mail.password =${MAIL_PASSWORD}
Advantages:
No manually creation of application.properties allowing us a more easy devops automations
No spring.config.location=file.. is required
I'd like to be able to run bootRun for spring boot, but use a different application.properties file besides the one in src/main/resources/. Is that possible? I'd prefer to not overwrite the file in src/main/resources/, as it would dirty the file.
Is this possible?
You can use profile based configuration selection.
Just set a system environment property:
spring.profiles.active=dev
And now provide application-dev.properties in application resources(src/main/resources/)
By this way you can use different properties for different environment.
If you want to provide files at a different location the use this environment property
spring.config.location=<path>
If you wish to use different name then application in property file name the use this environment property:
spring.config.name=<new_name>
For more info check this link:
https://docs.spring.io/spring-boot/docs/current/reference/html/boot-features-profiles.html
Here is what my bootRun task looks like that does this
bootRun {
systemProperty "spring.config.location", "file:$projectDir/spring-config/"
main = springBootAppClass
}
Which specifies to use the spring-config directory in the project's root folder.
If using in conjunction with #PropertySource it looks like this
#PropertySource("${spring.config.location}/persistence.properties")
Try this command,
./gradlew bootRun --args='--spring.profiles.active=dev'
I too was looking for the solution, found it in official documentation
https://docs.spring.io/spring-boot/docs/current/gradle-plugin/reference/html/#running-your-application
I got it to work like this:
tasks.named("bootRun") {
systemProperty "spring.config.location", "file:$projectDir/myConfigFolder"
mainClass = "my.project.MyMainClass"
}
I want to pass variable from maven command line to spring boot main class,but i have no idea,here is my maven pom file:
<properties>
<maven.tomcat.home>Here should add argument from command line</maven.tomcat.home>
</properties>
Here is my spring boot main class:
#PropertySource(value ={"file:#maven.tomcat.home#/em/easymobile-application.properties"})
public class Application {
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}
So when i run maven by command line as below ,the application will deploy as war file to tomcat
mvn clean install -Dmaven.tomcat.home=/usr/local/apache-tomcat-8.0.33/webapps
Then i start up tomcat,the error says easymobile-application.properties is not found,so do you know how to load the easymobile-application.properties from maven command line using #PropertySource or something else.Thanks!!
I don't think that's the right approach, assuming at some point you get that to work, what happens if you need / have to deploy the webapp to /usr/local/apache-tomcat-8.0.xx? Would you need to compile the artifact again? That's not good. A better approach is to only produce 1 binary / artifact that could be deployed / installed in every environment: staging / prod / etc..
Like mentioned before, if the app needs files path configuration, I would suggest to use a config properties file instead of a Maven property.
Thanks! I have solved the problem using a simple way ,i just add the path of the folder which contains application.properties in tomcat\bin\catalina.sh
CLASSPATH=:/usr/local/apache-tomcat-8.0.33
the #PropertySource points to "classpath:em/easymobile-application.properties"
I’m using Gradle 2.7 on Mac Yosemite. I have the following files:
src/main/environment/dev/context.xml
src/main/environment/qa/context.xml
src/main/environment/prod/context.xml
What I would like is if I run a build gradle -Pqa build, the appropriate context.xml file above is copied into my WAR (into the WEB-INF/classes directory is fine). How do I set this up with gradle?
There're many ways of solving the problem. You can configure sourceSets, or include or exclude particular resources when building war file. You can also have single context.xml and perform resource filtering with ReplaceTokens filter.
I've chosen sourceSets:
apply plugin: 'war'
ext.env = project.hasProperty('env') ? project.env : 'dev'
sourceSets {
main {
resources {
srcDir "src/main/environment/$env"
}
}
}
The trick is to include/process only the env being passed. If no env is passed dev will be picked for further processing. Have a look a the demo.
You would have to do that using the environment variable. Have the system properties in a file. Read them in the build.gradle and based on it include the context.xml into the war.
I want to use the applicationContext.xml in my src/main/resources directory from within my test harness in src/test/java. How do I load it? I have tried:
#RunWith(SpringJUnit4ClassRunner.class)
#ContextConfiguration(locations="classpath:applicationContext.xml")
public class TestService {
...
}
but get a file not found error. I'm using Maven and Spring. Thanks.
Try this (note the asterisk):
#ContextConfiguration("classpath*:applicationContext.xml")
The Maven test classpath uses the files in target/test-classes. That folder contains Java classes from src/test/java and resources from src/test/resources.
The way to go is to create a test specific app context and store it under src/main/resources.
You may try to reference the file directly using file: i.e. something like file:src/main/resources/applicationContext.xml but to me this is an ugly hack.
Also, you can of course use the Maven resources plugin to copy applicationContext.xml prior to test execution.
Here's how I do it, it may or may not be the best way for you. The main thing is it works in both Eclipse and Maven:
Keep exactly one copy of each applicationContext-xxx.xml file per project. NEVER copy-and-paste them or their contents, it'll create a maintenance nightmare.
Externalize all environmental settings to properties files (e.g. database-prod.properties and database-test.properties) and place them in src/main/resources and src/test/resources respectively. Add this line to your app contexts:
<context:property-placeholder location="classpath:**/*.properties"/>
Create a superclass for all test classes and annotate it with a context configuration:
#RunWith(SpringJUnit4ClassRunner.class)
#ContextConfiguration(locations = {"classpath:applicationContext.xml"})
#Ignore
public class SpringEnabledTest {
// Inheritable logger
protected Logger logger = LoggerFactory.getLogger(this.getClass());
}
Add <additionalClasspathElements> to your maven-surefire-plugin configuration to make sure surefire picks up appContext and the right properties files. I've created an example here.
Add the location(s) of the app context files and src/test/resources to your Eclipse classpath so you can execute unit tests in Eclipse as well.
NEVER add src/main/resources to your Eclipse classpath, it's only a convenient place for Maven to package additional source files, it should have no bearing on Eclipse. I often leave this directory blank and create additional folders (e.g. env/DEV, env/UAT and env/PROD) outside of the src/ folder and pass a parameter to the build server and let it know from which folder it needs to copy files to src/main/resources.
Add the src folder to the classpath of your testing tool. If it's in Eclipse, I think you can do it from the project properties. You may have to change it to classpath:**/applicationContext.xml as well.