I am using spring boot and I have added another spring boot app as Maven dependency in my project. The problem I am facing is that when I run the application, it picks the property file of the dependency instead of my current application. For example if I run the app using dev profile, application-dev.property file is picked from dependency instead of the running application.
I tried to debug the EnableEncryptablePropertySourcesPostProcessor file and below is the screenshot of the list of property file picked.
Check this out but you have a few options:
Simply specify the config file name:
java -jar myproject.jar --spring.config.name=myproject
And basically you can have myproject-dev.properties
Or directly specify the config files you wanna import:
java -jar myproject.jar --spring.config.location=classpath:/default.properties,classpath:/override.properties
Use PropertySource annotation to refer the properties file in your main application file as shown below
#PropertySource(value = { "file:/path/to/folder/file.properties" })
If you have same property in the multiple properties file then one in the classpath will get more preference
**In Application.java file it should be something like this
#PropertySource("classpath:application.properties")
Related
I have a external configuration file(out side jar). I try to run and expected
that value in external file will override value in internal file(application.properties in \resource\ - in jar file).
I read Documentation and try this:
java -jar ccgame-1.0.jar --spring.config.location=classpath:/application.properties,file:/production.properties
This not working.
My jar file at \target\ directory and my production.properties too(at \target\)
How can I resolve my problem?
Where should I put external config file ?
And what I have to do ?
Starting from Spring Boot 2.0 it's possible to use property spring.config.additional-location. With this property, you can set external config file, but properties from that config will only override the corresponding ones from internal config, leaving other properties unchanged.
More about it in docs.
If you need to completely override the whole config, then continue to use spring.config.location property instead.
By convention, Spring Boot looks for an externalized configuration file – application.properties or application.yml – in 4 predetermined locations in the following order of precedence:
/config subdirectory of the current directory
The current directory
Classpath /config package
The classpath root
You can place your application.properties in any of the 4 locations without needing to give the location of application.properties while executing the jar. If you want to given any other custom location , then you will have to provide the path of the config location while executing the jar:
java -jar -Dspring.config.location=<path-to-file> myProject.jar
Source: https://www.baeldung.com/spring-properties-file-outside-jar
I want to read Spring's boot active application.properties file a jar file that I add as Maven dependency.
We mange environment params via 3-4 files under publish folder, for example:
-publish
--some.project-application.properties.test
--some.project-application.properties.dev
in the project root so I cannot use PropertySources(different file name per env).
So is there is a way to read the active(in use) application.properties file?
Is using #Value annotation will work without any configuration?
I think you should use spring naming conventions(application.properties, application-dev.properties), so that spring will automatically take the correct property file from the class path.
Also, refer : How to resolve placeholder in properties file with values from another properties file in spring boot application
I have the problem with configuration file that is located in other directory than my jar file.
I use #PropertySource for loading properties.
#PropertySource(ignoreResourceNotFound = true, value = "${ext.properties.dir:classpath:}/properties.properties")
I try to run jar using following command:
java -jar import-0.0.1-SNAPSHOT.jar -Dext.properties.dir=file:/C:\Users\Admin\Desktop\
The following error pring in logs: Properties location
Properties location
[${ext.properties.dir:classpath:}/properties.properties] not resolvable:
class path resource [properties.properties] cannot be opened because it
does not exist
How can I fix this error?
I saw that you are using spring boot application, according to the spring documentation you can try to use this environment property:
--spring.config.location=file:/path/location/file-name.properties
The final instruction would be:
java -jar import-0.0.1-SNAPSHOT.jar --spring.config.location=file:C:\Users\Admin\Desktop\import.properties
Spring Boot looks for application.properties on classpath. You don't need an explicit #PropertySource annotation. And #PropertySource doesn't have the capability to resolve SPEL in the path, which should be evident from the error message you're getting.
Use src/main/resources/application.properties. When you want to use the external file, use spring.config.location like #juan-calvopina-m suggested in his answer.
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;
I would like to add a configuration directory to the classpath for a spring boot application at start up, so it can load xml files from the configuration directory.
ie /var/application/config contains
test.xml, dev.xml
The xml will contain mapping information that is required by the application; this is different from application.properties.
I would like to load them at startup.
I am using ClassPathResource to load the files.
Please advise.
You can define your own classpath by the command-line. Lets suppose your jar is myapp.jar and you wand add one extra directory /var/application/config/, so you can execute with the following command line:
java -cp myapp.jar:/var/application/config/ -Dloader.main=myapp.Application org.springframework.boot.loader.PropertiesLauncher
ps: if you are using Windows use ; instead of : to separate your classpath items.
From the Spring Boot Reference Guide, add your config location:
java -jar myproject.jar --spring.config.location=classpath:/var/application/config/