I know I can inject the value from properties file with the following syntax:
#Scheduled(fixedRate=${myRate})
public void getSchedule(){
System.out.println("in scheduled job");
}
However I can't guess how to accomplish the same if the configuration is in YAML file.
Thanks in advance,
In my application.properties (YAML) I put this
console:
fetchMetrics: 5000
Then in my simple Task class I push the definition :
#Scheduled(fixedRateString ="${console.fetchMetrics}", initialDelay=1000)
public void fetchMetrics() {
logger.info("What's up ?");
}
Please notice that fixedRate expects a long and you want to inject a placeholder, you will need fixedRateString
I find it easy once done for my project.
Change fixedRate to fixedRateString and put the property key in double quotes like this:
#Scheduled(fixedRateString="${myRate}")
public void getSchedule() {
System.out.println("Scheduled job");
}
In my application I use the annotation PropertySource on my config class:
#PropertySource("application-${spring.profiles.active}.yml")
spring.profiles.active returns the active profile (dev, test, etc). My properties file name is application-dev.yml
The annotation #Scheduled works with property injection.
Dont forget the annotation with prefix configuration on your class.
Related
I have my custom properties defined in application.properties file located in src/main/resources folder. I want to add test cases for my application hence I added application-integration.properties in the src/test/resources folder as mentioned here. The test case is added as below
#SpringBootTest
#ActiveProfiles("integration")
public class ServiceATest {
#Value("${app.api.url}")
private String apiUrl;
#Test
public void testService() {
Assertions.assertNotNull(apiUrl);
}
}
The above test case fails. I do not understand why the properties are not read from application-integration.properties and nor from application.properties. The latter is not expected though still just for the understanding. Any idea what could be going wrong?
Thanks.
I have created a myApp.properties in resources folder location and mentioned the server.port in this file.
myApp.properties
myApp.server.port=8020
Now I want to read load this property into my application. But I have to read this before I actually a server.
Here I am trying to do like this
#SpringBootApplication
#ComponentScan(basePackages = {"com.myorg.myapp" })
#EnableConfigurationProperties
#PropertySource("classpath:myApp.properties")
#Component
public class MyAppApplication {
#Value("${myApp.server.port}")
private static String serverPort;
public static void main(String[] args) throws Exception{
try {
SpringApplication appCtxt = new SpringApplication(MyAppApplication.class);
appCtxt.setDefaultProperties(Collections
.singletonMap("server.port", serverPort));
appCtxt.run(args);
} catch (Exception e) {
e.printStackTrace();
}
}
But serverPort is coming as null.
I also tried to create a separate Config file like this but it can't be accessed in static main
#Configuration
#PropertySource("myApp.properties")
#ConfigurationProperties
public class MyAppConfig {
#Value("${myApp.server.port}")
private String serverPort;
/**
* #return the serverPort
*/
public String getServerPort() {
return serverPort;
}
}
Any suggestion would be helpful.
Spring boot injects properties during the initialization of the application context.
This happens (gets triggered) in the line:
appCtxt.run(args);
But you try to access the property before this line - that why it doesn't work.
So bottom line, using "#Value" in the main method doesn't work and it shouldn't.
Now from the code snippet, it looks like you could merely follow the "standards" of spring boot and create the file application.properties with:
server.port=1234
The process of starting the embedded web server in spring boot honors this property and bottom line it will have the same effect and Tomcat will be started on port 1234
Update 1
Based on OP's comment:
So, how can I have multiple application.properties.
In the Spring Boot's documentation it is written that application.properties are resolved from the classpath. So you can try the following assuming you have different modules A,B,C and web app D:
Create src/main/resources/application.properties inside each of 4 modules and pack everything together. The configuration values will be merged (hopefully they won't clash)
If you insist on naming properties A.properties, B.properties and C.properties for each of non-web modules, you can do the following (I'll show for module A, but B and C can do the same).
#Configuration
#PropertySource("classpath:A.properties")
public class AConfiguration {
}
Create in Module A: src/main/resources/A.properties
If you need to load the AConfiguration automatically - make the module A starter (using autoconfig feature of spring-boot):
Create src/resources/META-INF/spring.factories file with the following content:
org.springframework.boot.autoconfigure.EnableAutoConfiguration=\
<package_of_AConfiguration>.AConfiguration
Also this has been the requirement to separate C from entire bundle where it might run as bundle for some and as a separate for some others
Although I haven't totally understood the requirement, but you can use #ConditionalOnProperty for configuration CConfiguration (that will be created just like AConfiguration.java in my previous example) but this times for module C.
If the conditional is met, configuration will run and load some beans / load its own properties or whatever. All in all conditionals (and in particular Profiles in spring) can help to reach the desired flexibility.
By default, the application.properties file can be used to store property pairs, though you can also define any number of additional property files.
If you save myApp.server.port=8020 in application.properties, it will work fine.
To register a custome property file, you can annotate a #Configuration class with the additional #PropertySource annotation:
#Configuration
#PropertySource("classpath:custom.properties")
#PropertySource("classpath:another.properties")
public class ConfigClass {
// Configuration
}
make sure, your class path is correct.
Normally, we can use a cron expression defined as "cron.expression" in the default property file, as follows:
#Scheduled(cron = "${cron.expression}")
public void demoServiceMethod(){
}
But I wish to define a property file for this class itself, and use the "cron.expression" property from this file. How can I do that?
P.S: I am using Java 1.7
Add to your class PropertySource
#PropertySource("classpath:other.properties")
Or using Configuration
#Configuration
#PropertySources(value = {#PropertySource("classpath:/datasource.properties")})
I'm tying to do a very minimal programmatic/annotation based configuration of Spring, to do some command line stuff and I want to be able to inject value of some bean values from System properties.
I'm using the #Value like this:
#Value("${MigrateDb.task:default}")
private String task;
It's sort of working, but it's not evaluating the value definition, I'm just getting "${MigrateDb.task:default}" in the actual field, instead of Spring evaluating it and giving me the value of the Migrate.db.task system property (or default).
What do I need to add to my Configuration class to enable this behaviour?
try using it this way:
#Value("${MigrateDb.task:default}")
private String task;
XML Config:
<context:property-placeholder
location="your.filelocation.properties" />`
Java Config :
#Bean
public static PropertyPlaceholderConfigurer propertyPlaceholderConfigurer() {
PropertyPlaceholderConfigurer propertyPlaceholderConfigurer = new PropertyPlaceholderConfigurer();
propertyPlaceholderConfigurer.setLocation(new ClassPathResource("file.properties"));
return propertyPlaceholderConfigurer;
}
From ShadowRay's answer, the minimum code to enable the requested behaviour is:
#Bean
public static PropertyPlaceholderConfigurer propertyPlaceholderConfigurer(){
return new PropertyPlaceholderConfigurer();
}
Method should be static as per: https://stackoverflow.com/a/14943106/924597
I've defined an application name using the bootstrap.yml file in my spring boot application.
spring:
application:
name: abc
How can i get this application name during runtime/programmatically ?
You should be able to use the #Value annotation to access any property you set in a properties/YAML file:
#Value("${spring.application.name}")
private String appName;
#Autowired
private ApplicationContext applicationContext;
...
this.applicationContext.getId();
Please, find this:
# IDENTITY (ContextIdApplicationContextInitializer)
spring.application.name=
spring.application.index=
In Spring Boot Reference Manual.
And follow with source code for that ContextIdApplicationContextInitializer class:
#Override
public void initialize(ConfigurableApplicationContext applicationContext) {
applicationContext.setId(getApplicationId(applicationContext.getEnvironment()));
}
Where the default behavior is with this:
/**
* Placeholder pattern to resolve for application name
*/
private static final String NAME_PATTERN = "${vcap.application.name:${spring.application.name:${spring.config.name:application}}}";
Since the #Value annotation is discouraged in Spring Boot when referencing configuration properties, and because applicationContext.getId(); doesn't always return the value of spring.application.name another way is to get the value from the Environment directly
private final Environment environment;
...
public MyBean(final Environment environment) {
this.environment = environment;
}
...
private getApplicationName() {
return this.environment.get("spring.application.name");
}
Another possible way would be to create your own ConfigurationProperties class to get access to the value.
I'm not saying these are the best ways, and I hope/wish that there is a better way, but it is a way.
Note! If your using a SpringBootTest, you need to suplly the properties/yml.
Otherwise, the environment/appcontext does not load the config files.
The, your app name is not set.
Like so:
#PropertySource("classpath:application.properties")
#RunWith(SpringRunner.class)
#SpringBootTest
....
This post is aged but I hate unanswered questions.
So use the following snippet:
#Value("${spring.application.name [: defaultValue]}")
private String appName;
What is between [] is optional.
So I found a really ugly way to do this, but it works so I'm not searching further. Maybe this will help someone.
The basic premise is that spring Environment stores the value inside a propertySource.. It appears that bootstrap config is stored in the ResourcePropertySource and so you can get it from that. For me it is currently throwing an exception, but then I can get the value out of the exception, so I haven't looked any further:
try {
this.environment.getProperty("name", ResourcePropertySource.class);
} catch (ConversionFailedException e) {
String res = (String)e.getValue();
}
And then you can just do this for every property you are interested in.
Like I said ugly, but it works.