I have a Springboot project and I know I can send a application.properties file as a argument, but is this possible using a jar file?
I built my jar file using maven and in my application I have this piece of code that runs the programm if the user sent the argument run.
Is there any method that allows me to set the application properties if I receive it through argument? Or does the override of the file happens automatically as it does when I use the command
mvn spring-boot:run -Dspring.config.location=your.properties
if (args[0].equals("run")) {
ConfigurableApplicationContext ctx = SpringApplication.run(MigrationsApplication.class, args);
int exitCode = SpringApplication.exit(ctx, () -> 0);
System.exit(exitCode);
}
For jar you can either pass one of the properties or the complete or its location as beow.
we can configure the location directly in the command line:
java -jar app.jar --spring.config.location=file:///Users/home/config/jdbc.properties
We can also pass a folder location where the application will search for the file:
java -jar app.jar --spring.config.name=application,jdbc --spring.config.location=file:///Users/home/config
And, an alternative approach is running the Spring Boot application through the Maven plugin. There, we can use a -D parameter:
mvn spring-boot:run -Dspring.config.location="file:///Users/home/jdbc.properties"
Related
In my spring boot batch (2.7.3) application.properties file I have:
rs.input.path=/opt/ingestiondata/rs
Afterthat, when I do a mvn clean install on my windows machine, I get the jar file in my target folder. When I try to do java -jar myjar.jar on my local windows command prompt, it give (as expected, as there is no such path) exception - java.nio.file.NoSuchFileException: \opt\ingestiondata\rs
Then I move the same jar file to the linux box there when I do java -jar myjar.jar my key rs.input.path get evaluated to a windows path - c:\users\ajay\some\dir.
What can be wrong here? As I am using the same jar. Its odd but its what happening since last couple of hours. Tried and verified killall java etc etc and now running out of options. Any pointers/help will be greatly appreciated. Must be something trivial and something horrible I am expecting.
Update:
As asked by Abhijit, this is how this is being used.
#Value("${rs.input.path}")
private String inputPath;
#Bean
ItemReader<File> myReader() throws IOException {
List<File> files = Files.walk(Paths.get(inputPath))
.filter(Files::isRegularFile)
.map(Path::toFile)
.collect(Collectors.toList());
return new IteratorItemReader<>(files);
}
I don't think in a spring application,
for any property,
there can't be multiple values;
unless you are using multiple profile specific 'application.properties' files which are specific to each environment.
Option 1:
"""
Check where your property is residing:
'rs.annual.path=/your/linux/path'
Suppose, let's assume it's inside
application-dev.properties
So, you need to select "dev" as Spring profile.
Try running the jar again, by pass passing profile.
(Spring_Property)
java -jar myjar.jar --spring.profiles.active=dev
(Vm_Argument)
2.java -jar -Dspring.profiles.active=dev myjar.jar
"""
Option 2:
"""
Since you are running your jar file inside linux.
Try passing path directly as argument.
(Spring_Property)
java -jar myjar.jar --rs.annual.path=/your/linux/path
You can pass the property as argument. So the Argument-property will taken precedence over Application-property.
"""
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 want to run Micronaut server from Gradle command line with "local" environment variables.
The regular command
.\gradlew.bat run
will use default variables defined in application.yml file.
I want to override some of them with values for my local environment and therefore need to specify system property micronaut.environments=local to use overriding values from application-local.yml file.
.\gradlew.bat run -Dmicronaut.environments=local
The command above won't work as Gradle will take only -Dmicronaut for the system property and the rest ".environments=local" will be considered as another task name:
Task '.environments=local' not found in root project 'abc'
What would be the correct way to pass such system property to the java process?
Command below works for unix, probably it should work also for windows:
MICRONAUT_ENVIRONMENTS=local gradle run
or use gradle wrapper
MICRONAUT_ENVIRONMENTS=local .\gradlew.bat run
P.S. also, you can find the same approach for Spring Boot
My approach is to add a gradle task.
task runLocal(type: JavaExec) {
classpath = sourceSets.main.runtimeClasspath
main = "dontdrive.Application"
jvmArgs '-Dmicronaut.environments=local'
}
then start with:
./gradlew runLocal
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
}
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"