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
Gradle 5.4.1
apply plugin: application
run {
main = "mypackage.Foo"
}
Run:
gradlew run --args="-Dfoo=bar -Dhello=world"
Trying to pass the system properties using --args when running the application. But they were not set.
gradlew -Dfoo=bar -Dhello=world run --args="arg1"
--args are what's passed to the main method.
You would need explicitly copy system properties from CLI into the corresponding run command/plugin system properties, unfortunatelly:
// The run task added by the application plugin is also of type JavaExec.
tasks.withType(JavaExec) {
// Assign all Java system properties from the command line to the JavaExec task.
systemProperties System.properties
}
The documentation of the gradle application plugin states that I can use applicationDefaultJvmArgs to configure the JVM startup parameters which are written in the start script. When using the Spring Boot Gradle plugin, I get a new distribution called boot, which also contains start scripts. Unfortunately they seem to ignore applicationDefaultJvmArgs. They contain this line:
# Add default JVM options here. You can also use JAVA_OPTS and POCKETKNIFE_OPTS to pass JVM options to this script.
DEFAULT_JVM_OPTS=""
while my original distrubtion contains this line:
# Add default JVM options here. You can also use JAVA_OPTS and POCKETKNIFE_OPTS to pass JVM options to this script.
DEFAULT_JVM_OPTS='"-XX:+UseSerialGC" "-XX:MaxHeapFreeRatio=10" "-XX:MinHeapFreeRatio=10" "-Xms32M" "-Xmx128M" "-Dfile.encoding=UTF-8"'
How can I set the JVM parameters in the Spring Boot distribution?
I think i figured it out: Add this to the build.gradle file:
bootStartScripts {
defaultJvmOpts = project.applicationDefaultJvmArgs
}
It still looks like a bug to me, especially since the bootRun task uses the applicationDefaultJvmArgs.
build.gradle
tasks.withType(Test){
systemProperties=System.properties
println systemProperties['param']
}
Now I can either pass parameters in the command line:
gradle test -Dparam=10
or put them in gradle.properties:
systemProp.param=15
Ideally I would like to put the defaults in the gradle.properties, and be able to overwrite them from the command line. Unfortunately if I do that, the gradle.properties has precedence, and -Dparam=10 is ignored.
Could you offer any solutions on that?
https://issues.gradle.org/browse/GRADLE-2122
It works since 2.12 or 2.13 "the smart way" already!
The example above is working, the command line -D option overdrives the defaults in gradle.properties
I am using gradle 2.12 and sharing how I used it:
test {
// support passing -Dsystem.property=value to bootRun task
systemProperties = System.properties
}
I have JUnit tests that I wanted to skip unless a property was used to include such tests. Using JUnit Assume for including the tests conditionally:
//first line of test
assumeThat(Boolean.parseBoolean(System.getProperty("deep.test.run","false"),true)
Doing this with gradle required that the system property provided at the time of running gradle build, shown here,
gradle build -Ddeep.test.run=true
was indeed passed through to the tests.
Hope this helps others trying out this approach for running tests conditionally.
I'm using Gradle spring-boot plugin and I need to select a spring active profile for the test run.
How do I pass spring.profiles.active system property to the bootRun plugin's task?
What has already failed:
task bootRunLocal {
systemProperty "spring.profiles.active", "local"
System.setProperty("spring.profiles.active", "local")
tasks.bootRun.execute() // I suspect that this task is executed in a separate JVM
}
and some command line magic also fails:
./gradle -Dspring.profiles.active=local bootRun
Could someone kindly help me solve my troubles?
Update from the answers and comments:
I'm able to set the systemProperty and pass it to the spring container by doing :
run {
systemProperty "spring.profiles.active", "local"
}
However, when I do this, the local profile is being set for both bootRun task and bootRunLocal task. I need a way to set this property for bootRunLocal task and call booRun task from bootRunLocal.
That might sound very simple, but I come with peace from the structured world of Maven.
I know I'm late here... but I recently faced this exact issue. I was trying to launch bootRun with spring.profiles.active and spring.config.location set as system properties on the command line.
So, to get your command line "magic" to work, simply add this to your build.gradle
bootRun {
systemProperties System.properties
}
Then running from the command line...
gradle -Dspring.profiles.active=local bootRun
Will set local as the active profile, without needing to define a separate task simply to add the env variable.
task local {
run { systemProperty "spring.profiles.active", "local" }
}
bootRun.mustRunAfter local
Then run gradle command as:
gradle bootRun local
There is no generic way to pass system properties to a task. In a nutshell, it's only supported for tasks that fork a separate JVM.
The bootRunLocal task (as defined above) will not execute in a separate JVM, and calling execute() on a task isn't supported (and would have to happen in the execution phase in any case). Tests, on the other hand, are always executed in a separate JVM (if executed by a Test task). To set system properties for test execution, you need to configure the corresponding Test task(s). For example:
test {
systemProperty "spring.profiles.active", "local"
}
For more information, see Test in the Gradle Build Language Reference.
SPRING_PROFILES_ACTIVE=local gradle clean bootRun
This is according to this and this and it works.
According to the spring-boot-gradle-plugin documentation you should be able to pass arguments like this
./gradlew bootRun --args='--spring.profiles.active=dev'
Seems like this is a new gradle feature since 4.9. I used it in my project and it worked out of the box.
For gradle 2.14 below example works.
I have added as below.
When System.properties['spring.profiles.active'] is null then default profile is set.
bootRun {
systemProperty 'spring.profiles.active', System.properties['spring.profiles.active']
}
command line example
gradle bootRun -Dspring.profiles.active=dev
Just for reference if anyone will have this issue:
Vlad answer didn't quite worked for me but this one works great with 2.4,
task local <<{
bootRun { systemProperty "spring.profiles.active", "local" }
}
local.finalizedBy bootRun
then gradle local
Responding to OP's exact request here ...
How do I pass spring.profiles.active system property to the bootRun plugin's task?
And assuming by "pass" the OP meant "pass from commandline" or "pass from IDE invocation" ... This is how I like to do it.
Add this to build.gradle:
/**
* Task from spring-boot-gradle-plugin, configured for easier development
*/
bootRun {
/* Lets you pick Spring Boot profile by system properties, e.g. gradle bootRun -Dspring.profiles.active=dev */
systemProperties = System.properties
}
Then when you invoke it, use the familiar Java flag for setting a system property
gradle bootRun -Dspring.profiles.active=local
There is one main advantage of sticking to system properties, over the environment variables option (SPRING_PROFILES_ACTIVE=local gradle bootRun) ... and that's easy portability between Linux/OS X (bash, etc.) and Windows (cmd.exe anyway).
I learned this way from this blog post.
(UPDATE: Ah somehow I had missed #Erich's response with same recommendation. Oops! I'm leaving my answer, because of the additional details about portability, etc.)
You can create a new task (in discussed case with name bootRunLocal), that would extend org.springframework.boot.gradle.run.BootRunTask and setup properties before task execution. You can create such a task with following code:
task bootRunLocal(type: org.springframework.boot.gradle.run.BootRunTask) {
doFirst() {
main = project.mainClassName
classpath = sourceSets.main.runtimeClasspath
systemProperty "spring.profiles.active", "local"
}
}
More details can be found here:
https://karolkalinski.github.io/gradle-task-that-runs-spring-boot-aplication-with-profile-activated/
Starting from SpringBoot 2.0.0-M5 setSystemProperties() is no longer a method of the task bootRun.
The build.gradle needs to be updated to
bootRun {
execSpec {
// System.properties["spring.profiles.active"]
systemProperties System.properties
}
}
This is as springBoot's run task uses org.gradle.process.JavaExecSpec
This works for me using Gradle 4.2
This works:
SPRING_PROFILES_ACTIVE=production ./gradlew app-service:bootRun
with run command you can add to build file run { systemProperties = System.properties } and start with gradle run -Dspring.profiles.active=local
Another way which doesn't require any support from the gradle task: Set the JAVA_TOOL_OPTIONS environment variable:
JAVA_TOOL_OPTIONS='-Dfoo=bar' gradle ...
Or if the variable might already contain anything useful:
JAVA_TOOL_OPTIONS="$JAVA_TOOL_OPTIONS -Dfoo=bar" gradle ...
// defualt value
def profiles = 'dev'
bootRun {
args = ["--spring.profiles.active=" + profiles]
}
Then you can simply pick a specific version when starting a gradle task, like
./gradlew bootRun -P dev
"dev" is gonna to take place "prod"