I notice that Spring Boot application classes can extend other classes, but that the main(String[] args) methods generally all use SpringApplication.run(Application.class, args). The examples often use different annotations above the Application class definition.
This OP asks for a simple summary of three closely related questions:
1.) What are the possible classes that a Spring Boot Application.java class can extend?
2.) What are the intended uses of each of the extension options?
3.) And does the choice of a given extension also dictate specific annotations that must be added to the class definition?
From my research, I have identified the following three extension options:
1.) Extend nothing at all, as per this example:
#Configuration
#EnableAutoConfiguration
#ComponentScan
public class Application {
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}
2.) Extend WebMvcConfigurerAdapter, as per this example:
#SpringBootApplication
#Controller
public class UiApplication extends WebMvcConfigurerAdapter {
public static void main(String[] args) {
SpringApplication.run(UiApplication.class, args);
}
}
3.) Extend SpringBootServletInitializer, as per this example:
#Configuration
#EnableAutoConfiguration
#EnableScheduling
#ComponentScan
public class Application extends SpringBootServletInitializer {
#Override
protected SpringApplicationBuilder configure(SpringApplicationBuilder application) {
return application.sources(Application.class);
}
public static void main(String... args) {
System.setProperty("spring.profiles.default", System.getProperty("spring.profiles.default", "dev"));
final ApplicationContext applicationContext = SpringApplication.run(Application.class, args);
}
}
Notice that I kept the annotations and minimal other stuff from the examples. This OP asks simply if specific annotation choices or minimal other stuff are dictated by the choice of extension.
Another one, which is not inheritance but it's composition, is to implement the CommandLineRunner interface so that you can perform some operations when the Spring Boot application starts up, like so:
#SpringBootApplication
public class DevopsbuddyApplication implements CommandLineRunner {
/** The application logger */
private static final Logger LOG = LoggerFactory.getLogger(DevopsbuddyApplication.class);
#Autowired
private UserService userService;
#Value("${webmaster.username}")
private String webmasterUsername;
#Value("${webmaster.password}")
private String webmasterPassword;
#Value("${webmaster.email}")
private String webmasterEmail;
public static void main(String[] args) {
SpringApplication.run(DevopsbuddyApplication.class, args);
}
#Override
public void run(String... args) throws Exception {
User user = UserUtils.createBasicUser(webmasterUsername, webmasterEmail);
user.setPassword(webmasterPassword);
Set<UserRole> userRoles = new HashSet<>();
userRoles.add(new UserRole(user, new Role(RolesEnum.ADMIN)));
LOG.debug("Creating user with username {}", user.getUsername());
userService.createUser(user, PlansEnum.PRO, userRoles);
LOG.info("User {} created", user.getUsername());
}
}
Not sure if this is what you're looking for though.
Related
Trying to register beans dynamically via SpringApplicationBuilder class and it's working when running the app, but when trying to execute the test and trying to verify that the beans are defined in the context, they fail for the dynamic bean. Feel like I have to use another "magical" annotation for the tests for them to properly load the dynamic beans.
This is the code used and if you run the tests you will see that both cases will fail. BarService will fail also because FooService is registered dynamically via builder, but if you would remove the dependency it will pass the BarService test.
SpringApp.java
class FooService {
}
#Component
class BarService {
private final FooService fooService;
BarService(FooService fooService) {
this.fooService = fooService;
}
}
#SpringBootApplication
public class SpringApp {
public static void main(String[] args) {
new SpringApplicationBuilder()
.sources(SpringApp.class)
.initializers((ApplicationContextInitializer<GenericApplicationContext>) context -> {
context.registerBean(FooService.class);
})
.run(args);
}
}
SpringAppTest.java
#RunWith(SpringRunner.class)
#ContextConfiguration(classes = SpringApp.class)
public class SpringAppTest {
#Autowired
ApplicationContext context;
#Test
public void barService() {
Assert.assertNotNull("The barService should not be null", context.getBean(BarService.class));
}
#Test
public void contextLoads() {
Assert.assertNotNull("The fooService should not be null", context.getBean(FooService.class));
}
}
First solution
The main error here is that the test does not have the initalization logic that the main method has. Solution is to extract the logic from the initializers method
class MyInitializer implements ApplicationContextInitializer<GenericApplicationContext> {
#Override
public void initialize(GenericApplicationContext context) {
System.out.println("Called initialize");
context.registerBean(FooService.class);
}
}
and use it in main
public static void main(String[] args) {
new SpringApplicationBuilder()
.sources(SpringApp.class)
.initializers(new MyInitializer())
.run(args);
}
and then use the MyInitializer in the test file through #ConextConfiguration
#RunWith(SpringRunner.class)
#ContextConfiguration(classes = SpringApp.class, initializers = MyInitializer.class)
public class SpringAppTest {
// ...
}
Second (better) solution
Now, this can be cumbersome as we need to reference this initializer in every test, but there is an even better solution. We can create a specific Spring file resources/META-INF/spring.factories and put inside of it a reference to the initializer:
org.springframework.context.ApplicationContextInitializer=com.acme.orders.MyInitializer
After that, we can simplify both the main method
#SpringBootApplication
public class SpringApp {
public static void main(String[] args) {
SpringApplication.run(SpringApp.class, args);
}
}
and the tests, so that they don't need to always import the initializer.
#RunWith(SpringRunner.class)
#SpringBootTest
public class SpringAppTest {
// ...
}
Now both the main run process and the tests will have access to all the beans.
I have a question, maybe simple, but I can not find out the solution.
I am using spring boot and added some annotation to the code like this:
#EnableEurekaClient
#SpringBootApplication
#EnableCaching
public class MyApplication {
public static void main(String[] args) {
SpringApplication.run(MyApplication.class, args);
}
}
But in some other environment, for example, in production environment, we want to remove EurekaClient, but I do not want to manually remove it manually for each environment, instead, I want to use environment variable or command line parameter to control the behavior. I suppose to do this way:
#EnableEurekaClient(Enabled = {EnableEureka})
#SpringBootApplication
#EnableCaching
public class MyApplication {
public static void main(String[] args) {
SpringApplication.run(MyApplication.class, args);
}
}
Then I can easily start this application without touching the code.
Can anyone tell me if this is possible? If so, how can I do it?
Thanks
You would want to work with Spring Boot Profiles. Split out the #EnableEurekaClient to another #Configuration class and also add an #Profile("eureka-client") to the class. Then when starting up the application you can set a -Dspring.profiles.active=eureka-client for the environments other than production.
Example:
#SpringBootApplication
#EnableCaching
public class MyApplication {
public static void main(String[] args) {
SpringApplication.run(MyApplication.class, args);
}
}
#Configuration
#EnableEurekaClient
#Profile("eureka-client")
public class EurekaClientConfiguration {
}
I prefer this method as you don't have to create an extra profile:
#Configuration
#EnableEurekaClient
#ConditionalOnProperty(name = "application.enabled", havingValue = "true", matchIfMissing = false)
public class EurekaClientConfiguration {
}
I have an interface like so:
public interface Animal {
void setName(String animal);
String getName();
}
and I have a Class that implements the interface:
#Component
public class Dog implements Animal {
private String name;
public void setName(String name) {
this.name= name;
}
public String getName() {
return this.name;
}
}
In another class (ProcessAnimal), I AutoWire the interface:
public class ProcessAnimal {
#Autowired
public Animal animal;
public void processAnimals() {
animal.setName("Fido");
}
}
I only have one class that implements Animal so this should work, however, I get a NullPointerException when it hits the animal.setName("Fido"); line. IntelliJ is complaining that Autowired members must be defined in valid Spring bean (#Component|#Service...) which I have... I don't understand what I'm doing wrong. I've tried to add a #Qualifier, but still it didn't work and it shouldn't be necessary since I only have one implementation.
-java
-com.example.com.AnimalProcessing
-Animal
-Animal.java
-Dog.java
-ProcessAnimal.java
-AnimalProcessingApplication.java
AnimalProcessingApplication.java
#SpringBootApplication
public class AnimalProcessingApplication {
public static void main(String[] args) {
SpringApplication.run(AnimalProcessingApplication.class, args);
run();
}
public static void run() {
ProcessAnimal processAnimal = new ProcessAnimal();
processAnimal.processAnimals();
}
}
AnimalProcessingApplication class should be one level above all other classes.
Also you are using new for creation of object instead of using Dependency Injection (autowiring).
Replace below -
ProcessAnimal processAnimal = new ProcessAnimal();
with
#Autowired
ProcessAnimal processAnimal;
Also make sure that ProcessAnimal is a bean and Animal is injected in this class using autowiring.
Animal Processing Application.java must be on root folder of all classes.
Then all components in child folders are recognized automatically.
Update:
Create a config class with #Bean method to create an instance with a Dog. Also then you can get rid of the #Component annotation of the class.
The problem here is the constructor String name which cannot be injected.
Update 2:
Don't create the instances by yourself. Let spring container create them. Remove the run method.
Following are to be done to make this program work.
1.ProcessAnimal should be made a component . Annotating the class with #Component will mark the class to be autodetected during component scan.
#Component
public class ProcessAnimal {
#Autowired
public Animal animal;
public void processAnimals() {
animal.setName("Fido");
}
}
Obtain the ProcessAnimal class from the application context. The spring will prepare the ProcessAnimal bean with all its dependencies set.
You may do this in multiple ways and following is one of those
#Component
public class CheckRedRunner implements ApplicationRunner {
#Autowired
ProcessAnimal process;
#Override
public void run(ApplicationArguments args) throws Exception {
process.processAnimals();
}
}
A bean implementing ApplicationRunner will be run when the application starts.
or else
public static void main(String[] args) {
ApplicationContext ctx = SpringApplication.run(AnimalProcessingApplication.class, args);
ProcessAnimal process = ctx.getBean(ProcessAnimal.class);
process.processAnimals();
}
Couple of observations
the package names by convention uses lower case letters
example : com.example.process.entity
Please go through the official documentation to learn the expected way of writing Spring boot application.
In Spring Boot, a main class needs to be specified, which is the entry point to the app. Typically this is a simple class with a standard main method, as follows;
#SpringBootApplication
public class MySpringApplication {
public static void main(String [] args) {
SpringApplication.run(MySpringApplication.class, args);
}
}
This class is then specified as the main entry point when the application runs.
However, I want to run my code using a different main class using config to define this, And without using a different jar!! (I know rebuilding the jar will enable me to specify an alternative main class, but this effectively gives me two apps, not one! So, how can I do this to utilise one jar with two main classes and select the one to use via the Spring application.yml file?
I found an answer - use the CommandLineRunner interface...
So now I have two classes;
public class ApplicationStartupRunner1 implements CommandLineRunner {
#Override
public void run(String... args) throws Exception {
//implement behaviour 1
}
and
public class ApplicationStartupRunner2 implements CommandLineRunner {
#Override
public void run(String... args) throws Exception {
//implement behaviour 2
}
and how to switch between them in config..
#Configuration
public class AppConfig {
#Value("${app.runner}")
private int runner;
#Bean
CommandLineRunner getCommandLineRunner() {
CommandLineRunner clRunner = null;
if (runner == 1) {
clRunner = new ApplicationStartupRunner1();
} (else if runner == 2) {
clRunner = new ApplicationStartupRunner2();
} else {
//handle this case..
}
return clRunner;
}
}
and finally in the application.properties file, use
app.runner=1
I would stick with the original pattern of having just one main class and main method, however within that method you could configure where you want to go. I.e. rather than having 2 main methods and configuring which gets called make these 2 methods just normal methods, and create one main method which uses the config to determine which of your two methods get run.
The answer of jonny.l is fine.
Another very similar, but more manual/DIY solution is to get the ApplicationContext, from which you can get all other things:
#SpringBootApplication
public class App {
public static void main(String[] args) {
ConfigurableApplicationContext context = SpringApplication.run(App.class);
ConfigurableEnvironment env = context.getBean(ConfigurableEnvironment.class);
String mainApp = env.getProperty("app.runner");
if (mainApp.equals("Main1")) {
Main1 main1 = context.getBean(Main1.class);
main1.run();
} else if (mainApp.equals("Main2")) {
Main2 main2 = context.getBean(Main2.class);
main2.run();
}
}
}
#Service #Lazy
public class Main1 {
public void run() {}
}
#Service #Lazy
public class Main2 {
public void run() {}
}
#Lazyis used to prevent those beans from loading unnecessarily automatically.
If I am developing a rather simple Spring Boot console-based application, I am unsure about the placement of the main execution code. Should I place it in the public static void main(String[] args) method, or have the main application class implement the CommandLineRunner interface and place the code in the run(String... args) method?
I will use an example as the context. Say I have the following [rudimentary] application (coded to interfaces, Spring style):
Application.java
public class Application {
#Autowired
private GreeterService greeterService;
public static void main(String[] args) {
// ******
// *** Where do I place the following line of code
// *** in a Spring Boot version of this application?
// ******
System.out.println(greeterService.greet(args));
}
}
GreeterService.java (interface)
public interface GreeterService {
String greet(String[] tokens);
}
GreeterServiceImpl.java (implementation class)
#Service
public class GreeterServiceImpl implements GreeterService {
public String greet(String[] tokens) {
String defaultMessage = "hello world";
if (args == null || args.length == 0) {
return defaultMessage;
}
StringBuilder message = new StringBuilder();
for (String token : tokens) {
if (token == null) continue;
message.append(token).append('-');
}
return message.length() > 0 ? message.toString() : defaultMessage;
}
}
The equivalent Spring Boot version of Application.java would be something along the lines:
GreeterServiceImpl.java (implementation class)
#EnableAutoConfiguration
public class Application
// *** Should I bother to implement this interface for this simple app?
implements CommandLineRunner {
#Autowired
private GreeterService greeterService;
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
System.out.println(greeterService.greet(args)); // here?
}
// Only if I implement the CommandLineRunner interface...
public void run(String... args) throws Exception {
System.out.println(greeterService.greet(args)); // or here?
}
}
You should have a standard loader:
#SpringBootApplication
public class MyDemoApplication {
public static void main(String[] args) {
SpringApplication.run(MyDemoApplication.class, args);
}
}
and implement a CommandLineRunner interface with #Component annotation
#Component
public class MyRunner implements CommandLineRunner {
#Override
public void run(String... args) throws Exception {
}
}
#EnableAutoConfiguration will do the usual SpringBoot magic.
UPDATE:
As #jeton suggests the latest Springboot implements a straight:
spring.main.web-application-type=none
spring.main.banner-mode=off
See docs at 72.2