I am creating simple Spring apps using Maven and have 2 config and properties. The hierarchy is:
- package.main
- App.java
- AppConfig.java
- app.properties
- package.main.model
- ModelConfig.java
- model.properties
App.java
public static void main(String[] args) {
ApplicationContext applicationContext = new AnnotationConfigApplicationContext(AppConfig.class);
System.out.println(applicationContext.getEnvironment().getActiveProfiles()[0]);
}
AppConfig.java
#Configuration
#Import(ModelConfig.class)
#PropertySource("classpath:/package/main/app.properties")
public class AppConfig {}
app.properties
spring.profiles.active = prod
ModelConfig.java
#Configuration
#PropertySource("classpath:/package/main/model/model.properties")
#ComponentScan
public class ModelConfig {}
model.properties
spring.profiles.active = dev
Why the model.properties is override the app.properties (the result is dev)?
How to make application.properties like in Spring Boot that cannot overrode by new properties?
it is by the order of beans defined being ModelConfig loaded it will cause the active profile to be set for 'dev'.
BTW, looking at your setup, configuring the active profile of beans via properties file like this not the good idea though, spring.profiles.active should be specified in
application.properties
file, not to loaded via separate property files, so when application starts it loads the bean correctly for the profile
or via command line arguments when you invoke
-Dspring.profiles.active
or programmatically
context.getEnvironment().setActiveProfiles("live");
some good examples here
Related
I am trying to load 2 properties file in spring boot.
One of them contains the metadata (Database connection and other such properties). The other contains business logic (mapping between upstream and downstream Entity. This mapping is different in Dev and Prod, hence can't have a single resource file for these).
I want to use Spring Profiles for different environments (Dev, Stage, Prod).
So, I created 3 different folders in src/main/resources 1 for each environment.
Using spring profies, I am aware how to have env specific application-env.properties file. However, I am unable to move forward on how to use the same for my use case.
PS : Not adding any code snippet, because the question doesn't require one.
Here's an example from the docs:
$ java -jar myproject.jar --spring.config.location=\
optional:classpath:/default.properties,\
optional:classpath:/override.properties
You could also define this in your code before starting Spring Boot:
public static void main(String[] args) {
System.setProperty("spring.config.location", "optional:classpath:/default.properties,optional:classpath:/override.properties");
SpringApplication.run(Application.class, args);
}
To use custom prefixes for your app specific properties you can define #ConfigurationProperties class(es):
#Data
#ConfigurationProperties(prefix = "app.mapper")
public class MapperProperties {
private String foo;
}
and use it in any component:
#Component
#RequiredArgsConstructor
#EnableConfigurationProperties
public class YourComponent {
private final MapperProperties properties;
}
I have this configurations which needs to be used for a spring boot application.
server.port=8085
server.servlet.context-path=/authserver
#data source
spring.jpa.hibernate.ddl-auto=none
spring.datasource.url=<url>
spring.datasource.username=<username>
spring.datasource.password=<password>
spring.datasource.driver-class-name=oracle.jdbc.driver.OracleDriver
spring.jpa.properties.hibernate.dialect=org.hibernate.dialect.Oracle10gDialect
By default spring-boot picks up up the application.properties file located in src/main/resources/
I want to alter this path and direct spring boot to different application.properties file
I can achieve this using
java -jar app.jar --spring.config.location=classpath:/another-location.properties
Is there any any alternative solution I can achieve this without passing args through command line?
I was using this
#PropertySource("file:C:\Users\test\.test\test.properties")
#ConfigurationProperties(prefix = "spring")
public class Configuration {
private String ddlAuto;
private String url;
private String username;
private String password;
private String driverClassName;
}
in my Main class
#SpringBootApplication
#EnableConfigurationProperties(Configuration.class)
public class Application {
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}
There after I tried executing the app commenting out all datasource properties in application.properties under src/main/resources/
But it keeps giving me the error mentioned bellow and application fails to start
I was referring this tutorial : https://www.mkyong.com/spring-boot/spring-boot-configurationproperties-example/
but as it's mentioned I get this error when i start the spring boot application
***************************
APPLICATION FAILED TO START
***************************
Description:
Binding to target org.springframework.boot.context.properties.bind.BindException:
Any help on this would be appreciated
The recommended way to have externalized properties is to use the spring.config.location system property, by starting your application like so:
java -jar -Dspring.config.location=/path/to/my/file.properties app.jar
The reason for this is that you don't add coupling between your code and your filesystem hierarchy.
Before Spring Boot 2.0 this property is additive, meaning that it will complement the default locations. After Spring Boot 2.0, spring.config.location replaces the default locations (e.g. classpath src/main/resources/application.properties). To keep the additive behaviour after 2.0, use spring.config.additional-location instead.
Please see here for official documentation on this matter.
I am able to make it work properly on Spring Boot 2.1.2.RELEASE. This is what I have done:
I have a test.properties in my /tmp folder with the following content:
test.myprop=hello
I also have the usual property file in the resources folder:
myprop=world
I have created a class for the custom property file:
#Configuration
#PropertySource("file:/tmp/test.properties")
#ConfigurationProperties(prefix = "test")
public class TestConfig {
private String myprop;
public String getMyprop() {
return myprop;
}
public void setMyprop(String myprop) {
this.myprop = myprop;
}
}
And then in my main class I have enabled to configuration properties:
#EnableConfigurationProperties(TestConfig.class)
#SpringBootApplication
public class MyApp {
public static void main(String[] args) {
SpringApplication.run(MyApp.class, args);
}
}
Now I have this test controller:
#RestController
public class TestController {
#Value("${test.myprop}")
private String externalisedProp;
#Value("${myprop}")
private String prop;
#GetMapping("test")
public void test() {
System.out.println("externalised: " + externalisedProp);
System.out.println("not externalised" + prop);
}
}
Which, once called, is properly printing:
externalised: hello
not externalised: world
My TestConfig class is in the same package as the MyApp main class.
What I have done is very similar, almost identical, to your solution, are you sure your path is correct? Also, I can see that the content of your property file is not matching what you have in your config class, the prefix is different. Maybe that is the problem?
Edit:
I have tried to remove the #Configuration annotation from my property class (which you do not have as well) and it is not able to pick up the externalised properties anymore. The error is different though but you should try to add it.
I have to load file from outside of my classpath.
The location depends of env properties:
in dev properties I want to load file from resources folder
in prod properties I want to load file from path (/location/file)
What is the best way to do it?
A possible solution is to use configuration properties and the use of Resource. For example, define your properties like this:
#ConfigurationProperties(prefix = "app")
public class SomeProperties {
private Resource file;
// Getters + Setters
}
Then enable your configuration properties by using the #EnableConfigurationProperties annotation on any class, for example your main class:
#SpringBootApplication
#EnableConfigurationProperties(SomeProperties.class)
public class Application {
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}
To configure the file location, you can use the following in development:
app.file=classpath:test.txt
And in the production environment you could use:
app.file=file:/usr/local/test.txt
And now you can just autowire the SomeProperties class within any other service. The Resource class has a getFile() method that allows you to retrieve the file, but in addition it contains several other useful methods as well.
To my surprise I have had a difficult time finding an answer to this question. I have Seen many examples where you can use #PropertySource to load a specific properties file for a class. I have also seen examples where you can easily add different property files in spring boot projects. But what I want to do is to do this for a spring project that is NOT spring boot and load a properties file so that the values of this file can be injected in classes annotated with #Component which is dependent on the server environment. So for example if I am on development server I want a particular properties file loaded and on production a different properties file. The reason that I am doing it like this is because my data and service layers are their own modules. These modules contain their own unit tests and can be imported as their own modules in other spring boot projects. I need properties files to be loaded to serve these modules which use spring but not spring boot. I have tried the following, but this does not work.
#Configuration
#Profile("test")
#EnableJpaRepositories("com.hi.repository")
#EnableTransactionManagement
#EnableScheduling
public class InfrastructureConfig {
...
#Bean
public static PropertySourcesPlaceholderConfigurer placeHolderConfigurer() {
PropertySourcesPlaceholderConfigurer configurer = new PropertySourcesPlaceholderConfigurer();
Map<String, String> env = System.getenv();
String propertiesFile=null;
String e = env.get("SERVER_ENV");
if (e.equals("dev")) {
propertiesFile = "environment/development.properties";
} else if (e.equals("prod")) {
propertiesFile = "environment/production.properties";
}
configurer.setLocation(new ClassPathResource(propertiesFile));
return configurer;
}
Then I have a test which looks like this
#RunWith(SpringJUnit4ClassRunner.class)
#ContextConfiguration(locations = {"classpath:/spring/DealServiceTest-context.xml"})
#ActiveProfiles("test")
public class LogTest {
private static final Logger log = LogManager.getLogger(LogTest.class);
#Autowired
PathsService pathsService;
#Autowired
Environment environment;
#Test
public void testBeans(){
System.out.println("********** WASSUP from LogTest");
System.out.println(environment.getProperty("imageBucket"));
}
Although the test prints out null which indicates to me the properties file has not been loaded and prepared for its values to be injected. How can I achieve this?
You don't really need to set properties yourself, but you can do this using spring configuration. Check the documentation: https://docs.spring.io/spring-boot/docs/current/reference/html/boot-features-external-config.html#boot-features-external-config-profile-specific-properties
If you're using spring boot - all you need to do is create multiple properties file for your environments. And only for properties you need to override.
So your main properties file would be at
src/main/resources/application.properties
Production
src/main/resources/application-prod.properties
Development
src/main/resources/application-dev.properties
Testing
src/main/resources/application-test.properties
And then just use the profile name as your environment variable
java -jar -Dspring.profiles.active=prod demo-0.0.1-SNAPSHOT.jar
Actually, you can just use a placeholder in #PropertySource annotation.
See documentation:
Any ${...} placeholders present in a #PropertySource resource location will be resolved against the set of property sources already registered against the environment.
Assuming that placeholder is present in one of the property sources already registered, e.g. system properties or environment variables, the placeholder will be resolved to the corresponding value.
I've made a simple example, it receives a 'property.environment' value to choose, which .properties file should be used as property source. I have two resource files in my classpath - application-test.properties and application-dev.properties, each one contains a 'test.property' value ('test-env' and 'dev-env' respectively).
Property configuration:
#Configuration
#PropertySource("classpath:/config/application-${property.environment}.properties")
public class PropertyConfig {
#Bean
public static PropertySourcesPlaceholderConfigurer propertySourcesPlaceholderConfigurer() {
PropertySourcesPlaceholderConfigurer propertySourcesPlaceholderConfigurer = new PropertySourcesPlaceholderConfigurer();
return propertySourcesPlaceholderConfigurer;
}
}
Component with #Value
#Component
public class TestService {
#Value("${test.property}")
String testProperty;
#PostConstruct
void init() {
System.out.println("---------------------------------------------------------");
System.out.println("Running in " + testProperty + " environment");
System.out.println("---------------------------------------------------------");
}
}
Build command line example (it runs tests with test environment properties)
mvn clean install -DargLine="-Dproperty.environment=test"
Output
---------------------------------------------------------
Running in test-env environment
---------------------------------------------------------
Run command line example
java -jar -Dproperty.environment=dev PATH_TO_YOUR_JAR.jar
Output
---------------------------------------------------------
Running in dev-env environment
---------------------------------------------------------
Don't hard code based on different environment, in spring boot you can able to maintain properties specific environment easily. Refer https://spapas.github.io/2016/03/31/spring-boot-settings/
I would try to take advantage of the profile mechanism already in place in Spring. You basically have done the job yourself already, the only thing you need to change is to have different configurations for "test" and "production" profiles. I prefer to keep everything related to test away from production code (allowing me to place the TestConfig class below in the test source path), so I would probably do something like this:
#Configuration
#Profile("!test")
#PropertySource(value = "classpath:/environment/production.properties")
#Import(AppConfig.class)
public class ProductionConfig
{
// Your production-specific config goes here
}
#Configuration
#Profile("test")
#PropertySource(value = "classpath:/environment/development.properties")
#Import(AppConfig.class)
public class TestConfig
{
// Your test-specific config goes here
}
#Configuration
public class AppConfig
{
// Needed for spring to handle ${property:default} syntax
#Bean
public static PropertySourcesPlaceholderConfigurer propertyConfigIn() {
return new PropertySourcesPlaceholderConfigurer();
}
}
If you prefer to have one config for both cases, you can let the AppConfig import the TestConfig and the ProductionConfig instead, but that will put test code in to production...
Good luck with your project!
How to create project architecture to support multiple envionment. Each environment will have different datasource from different property file like(dev-propertfile,test-propertyFil,Production-propertyfile) with help of spring's
org.springframework.core.env.Environment;
I'll give step by step procedure for Spring boot applications.
Inside /src/main/resources/application.properties mention spring.profiles.active=dev (or Prod)
Create /src/main/resources/application-dev.properties and give your custom dev configurations here.
Create /src/main/resources/application-prod.properties and give your custom prod configurations here.
Run.
Put property file in same location as application.property and follow
the naming convention application-{profile}.properties like
application-dev.properties,application-test.properties,
application-prod.properties
And in application.properties set spring.profiles.active=dev,test etc
For Spring Boot applications it will work easily even by using a YAML File
spring:
profiles: dev
property: this is a dev env
---
spring:
profiles: prod
property: this is a production env
---
However, for a Spring MVC application, it needs more work. Have a look at this link
Basically, it involves 2 steps
Get the Spring Profile within servlet context
If you have set the profile on the server and want it to retrieve it within your application you can use System.getProperty or System.getenv methods.
Here is the code which fetches the profile and defaults it to a local profile, if no profile has been found.
private static final String SPRING_PROFILES_ACTIVE = "SPRING_PROFILES_ACTIVE";
String profile;
/**
* In local system getProperty() returns the profile correctly, however in docker getenv() return profile correctly
* */
protected void setSpringProfile(ServletContext servletContext) {
if(null!= System.getenv(SPRING_PROFILES_ACTIVE)){
profile=System.getenv(SPRING_PROFILES_ACTIVE);
}else if(null!= System.getProperty(SPRING_PROFILES_ACTIVE)){
profile=System.getProperty(SPRING_PROFILES_ACTIVE);
}else{
profile="local";
}
log.info("***** Profile configured is ****** "+ profile);
servletContext.setInitParameter("spring.profiles.active", profile);
}
To access the application-dev.properties, say now you will need to use
#Profile("dev") at the class level
The following code will fetch the application-dev.properties and common.properties
#Configuration
#Profile("dev")
public class DevPropertyReader {
#Bean
public static PropertyPlaceholderConfigurer properties() {
PropertyPlaceholderConfigurer ppc = new PropertyPlaceholderConfigurer();
Resource[] resources = new ClassPathResource[] { new ClassPathResource("properties/common.properties"), new ClassPathResource("properties/application-dev.properties") };
ppc.setLocations(resources);
ppc.setIgnoreUnresolvablePlaceholders(true);
return ppc;
}
}
For accessing say application-prod.properties you have to use #Profile("prod") at the class level. More details can be found here
Take a look at Spring Profile. You will define a set of profiles configurations, like Test, Dev, Production. And then, when you launch the application, you can define wich profile it should use.
Here are some tutorials of how to use.
And this guys had the same problem as yours: How to config #ComponentScan dynamic?
We wanted a way to load different properties from application-<your_env>.properties file depending on the environment (spring profile) in a Spring MVC project, so we implemented a configuration class something like this.
#Configuration
#PropertySource({ "classpath:application-${envTarget:dev}.properties" })
#Data
public class EnvironmentConfig {
#Value("${files.s3.accessId:}")
String s3AccessId;
#Value("${files.s3.accessToken:}")
String s3AccessToken;
.
.
.
}
Then we loaded the EnvironmentConfig in the class where we needed to use it.
While running the application, you just need to pass the -DenvTarget=<your_env>, and it will pick up the application-<your_env>.properties file from src/resources folder of the project.
In the above code, it will load values from application-dev.properties files when no envTarget is specified.
Thanks to Karthikeyan Muthurangam for suggesting this clever solution.