During my unit tests using Spock and Testcontainers, the JpaRepository is not functioning properly and is not being wired correctly. This issue persists even in non-integration tests.
As suggested in another discussion, I attempted to resolve the issue by adding the spock-spring dependency to my pom.xml file. It didn't work.
No matter the scenario, the repository consistently returns as null in all instances.
An example:
#Testcontainers
class PostgresTestContainer extends Specification {
#Autowired
private PersonRepository personRepository
#Shared
PostgreSQLContainer postgreSQLContainer = new PostgreSQLContainer("postgres:12-alpine")
.withDatabaseName("test")
.withUsername("test")
.withPassword("test")
def "waits until postgres accepts jdbc connections"() {
when: "querying the database"
def response = personRepository.findAll()
then: "result is returned"
response == 0
}
}
The database is being initialized using PostgresContainer and Testcontainers annotation. However your test infrastructure doesn't know about the database. If using spring boot, then few things are missed
Add SpringBootTest or DataJpaTest annotation on top of the class. This way the spring application context is created with the right classes and PersonRepository will be injected
Switch to
static postgresContainer = new PostgreSQLContainer("postgres:12-alpine")
#Shared
PostgreSQLContainer <?> cassandra = cassandraContainer
In order to make use of the database provided by Testcontainers, add
#DynamicPropertySource
static void registerProperties(DynamicPropertyRegistry registry) {
registry.add("spring.datasource.url", postgresContainer::getJdbcUrl);
registry.add("spring.datasource.username", postgresContainer::getUsername);
registry.add("spring.datasource.password", postgresContainer::getPassword);
}
I would suggest to separate the container setup like the following:
PostgresEnvrionement
#Testcontainers
public class PostgresEnvironment {
#Container
public static PostgreSQLContainer postgreSQLContainer = PostgresTestContainer.getInstance();
}
PostgresTestContainer
public class PostgresTestContainer extends PostgreSQLContainer<PostgresTestContainer> {
public static final String IMAGE_VERSION = "postgres:13.5";
public static final String DATABASE_NAME = "test";
private static PostgresTestContainer container;
private PostgresTestContainer() {
super(IMAGE_VERSION);
}
public static PostgresTestContainer getInstance() {
if (container == null) {
container = new PostgresTestContainer().withDatabaseName(DATABASE_NAME);
}
return container;
}
#Override
public void start() {
super.start();
System.setProperty("DB_URL", container.getJdbcUrl());
System.setProperty("DB_USERNAME", container.getUsername());
System.setProperty("DB_PASSWORD", container.getPassword());
}
#Override
public void stop() {
}
}
In your test file extend the PostgresEnvironment
#ActiveProfiles("test")
#SpringBootTest(classes = MainSpringBootApp.class, webEnvironment = SpringBootTest.WebEnvironment.RANDOM_PORT)
public class AssembleEventRepositoryIntegrationTest extends PostgresEnvrionement{
// autowire jpa
// write tests
}
in your application-test.yml file in the resources section under the test directory
spring:
datasource:
password: ${DB_USERNAME}
username: ${DB_PASSWORD}
driver-class-name: org.postgresql.Driver
url: ${DB_URL}
Also make sure that your main application.yml file that is being used once launching your application (not in for running your tests) matches the same syntax as your test profile.
As the title says, my custom properties are not registered when Application is starting. Would you mind taking a look?
My custom.yml:
bridge:
url: xxx
authentication:
auth-url: xxx
user: xxx
My BridgeProperties.java:
#Component
#PropertySource("classpath:custom.yml")
#ConfigurationProperties(
prefix = "bridge"
)
public class BridgeProperties {
#NestedConfigurationProperty
protected ApiProperties apiProperties = new ApiProperties();
}
My ApiProperties.java:
public class ApiProperties {
protected String url;
protected ApiProperties.Authentication authentication;
// getter and setter
public ApiProperties() {
}
public static class Authentication {
protected String authUrl;
protected String user;
public Authentication() {}
// getter and setter
}
My Application's entry point:
#SpringBootApplication
#PropertySources(value = {
#PropertySource("classpath:application.yml"),
#PropertySource("classpath:custom.yml")
})
public class IntegrationService {
public static void main(String... args) {
SpringApplication.run(IntegrationService.class, args);
}
}
When printing to the command line, I get null instead of the values I assigned to url, auth-url, and user in the custom.yml.
As mentioned in the Baeldung Site:
It's also worth mentioning that YAML files do not support the #PropertySource annotation, so if we need to use this annotation, it would constrain us to using a properties file.
If you really need to import additional external file in yaml format, you have to implement your own PropertySourceFactory like in this article.
For me I do not see the advantage not using the default application.yml as you can define multiple sections managed by different Properties classes.
For example, if you just put your properties in the application.yml and if you compile with a recent JDK (>=jdk 14) you can even use the record for managing your properties in a very compact manner:
What you have to do is just to add the EnableConfigurationProperties annotation:
#SpringBootApplication
#EnableConfigurationProperties(BridgeProperties.class)
public class IntegrationService {
public static void main(String... args) {
SpringApplication.run(IntegrationService.class, args);
}
}
and define a record for mapping your yaml properties:
#ConfigurationProperties(prefix = "bridge")
public final record BridgeProperties(String url, Authentication authentication) {
public record Authentication(String authUrl, String user){
}
}
Remember you can also define environment variable in profiles like application-dev.yml if you start your app with profile dev and another one e.g. for production application-prd.yml if your profile for production is named prd. You can still keep the common properties in the application.yml.
If you really want to go in the external file path, I suggest to switch to .properties as the #PropertySource supports it natively.
In my yaml file, I have config values as below:
myapp:
rest-clients:
rest-templates:
- id: myService1
username: chris
password: li
base-url: http://localhost:3000/service1
read-timeout: 2s
connect-timeout: 1s
- id: myService2
username: chris
password: li
base-url: http://localhost:3000/service1
read-timeout: 2s
connect-timeout: 1s
I want to Spring Boot 2 app register a RestTemplate for each config items.
My configuration is bean is below:
#Configuration
#AllArgsConstructor
public class MyAppRestClientsConfiguration {
private MyAppRestClientsProperties properties;
private GenericApplicationContext applicationContext;
private RestTemplateBuilder restTemplateBuilder;
#PostConstruct
public void init() {
properties.getRestTemplates().forEach(this::registerRestTemplate);
}
private void registerRestTemplate(MyAppRestTemplateConfig config) {
// do some work
applicationContext.registerBean(config.getId(), RestTemplate.class, () -> restTemplate)
}
}
The problem is that when I inject my registered RestTemplate via #Autowire, this config bean has not finished init yet. So there is no RestTemplate bean could be injected.
#Autowired
#Qualifier("myService1")
private RestTemplate client1;
The injection point has the following annotations:
- #org.springframework.beans.factory.annotation.Autowired(required=true)
- #org.springframework.beans.factory.annotation.Qualifier(value=myService1)
Is there any correct way to implement this requirement?
The problem with registering new beans in a #PostConstruct annotated method is that Spring is already past that particular point in the Spring life cycle (more info on the Spring life cycle). Sometimes an annotation such as #DependsOn (already mentioned), #Order, or #Lazy might help. However, as you mentioned you'd rather not force (spring) implementation details upon projects that make use of your library, I've written a BeanFactoryPostProcessor that registers a RestTemplate bean:
#Component
public class DemoBeanFactoryPostProcessor implements BeanFactoryPostProcessor {
#Override
public void postProcessBeanFactory(ConfigurableListableBeanFactory configurableListableBeanFactory) {
GenericBeanDefinition genericBeanDefinition = new GenericBeanDefinition();
genericBeanDefinition.setBeanClass(RestTemplate.class);
HttpComponentsClientHttpRequestFactory factory = new HttpComponentsClientHttpRequestFactory();
factory.setReadTimeout(Integer.valueOf(configurableListableBeanFactory.resolveEmbeddedValue("${rest-templates[0].read-timeout}")));
factory.setConnectTimeout(Integer.valueOf(configurableListableBeanFactory.resolveEmbeddedValue("${rest-templates[0].connect-timeout}")));
// etc
ConstructorArgumentValues constructorArgumentValues = new ConstructorArgumentValues();
constructorArgumentValues.addGenericArgumentValue(factory);
genericBeanDefinition.setConstructorArgumentValues(constructorArgumentValues);
String beanId = configurableListableBeanFactory.resolveEmbeddedValue("${rest-templates[0].id}");
((DefaultListableBeanFactory) configurableListableBeanFactory).registerBeanDefinition(beanId, genericBeanDefinition);
}
}
application.yml:
rest-templates:
- id: myService1
username: chris
password: li
base-url: http://localhost:3000/service1
read-timeout: 2000
connect-timeout: 1000
- id: myService2
username: chris
password: li
base-url: http://localhost:3000/service1
read-timeout: 2000
connect-timeout: 1000
Accompanying test:
#RunWith(SpringRunner.class)
#SpringBootTest
public class DemoApplicationTests {
#Autowired
#Qualifier("myService1")
private RestTemplate restTemplate;
#Test
public void demoBeanFactoryPostProcessor_shouldRegisterBean() {
String stackOverflow =
restTemplate.getForObject("https://stackoverflow.com/questions/57122343/how-to-create-multiple-beans-same-type-in-one-spring-boot-java-config-class", String.class);
Assertions.assertThat(stackOverflow).contains("How to create multiple beans (same type) in one Spring Boot java config class (#Configuration)?");
}
}
As the BeanFactoryPostProcessor is invoked before the application context is fully set up, I had to find a different way to retrieve the application properties. I used the method ConfigurableListableBeanFactory#resolveEmbeddedValue to retrieve placeholder values instead of having them injected by an #Value annotation or environment#getProperty. Furthermore, I rewrote the property value 2s to 2000 as the HttpComponentsClientHttpRequestFactory required an int value.
You can levarage the binding mechanism that Spring Boot uses for the #ConfigurationPropeties. The corresponding beans can then be generated dynamically from a map of configurations
#Configuration
public class MyConfiguration implements BeanFactoryPostProcessor, EnvironmentAware {
private Environment environment;
#Override
public void setEnvironment(Environment environment) {
this.environment = environment;
}
#Override
public void postProcessBeanFactory(ConfigurableListableBeanFactory configurableListableBeanFactory) {
Binder.get(environment)
.bind("com.example", FooMapProps.class)
.get()
.getMap()
.forEach((name, props) -> configurableListableBeanFactory.registerSingleton(name + "FooService", new FooService(props.getGreeting())));
}
}
In this way you can create multiple FooServices with different configurations based on your YAML config
com:
example:
map:
hello:
greeting: 'hello there!'
hey:
greeting: 'hey ho :)'
The corresponding properties class looks like this
#ConfigurationProperties(prefix = "com.example")
public class FooMapProps {
private Map<String, FooProps> map;
public Map<String, FooProps> getMap() {
return map;
}
public void setMap(Map<String, FooProps> map) {
this.map = map;
}
public static class FooProps {
private String greeting;
public String getGreeting() {
return greeting;
}
public void setGreeting(String greeting) {
this.greeting= greeting;
}
}
}
In this example two FooService beans has been created with different greetings as constructor paramters. The beans can be accessed by the qualifiers 'helloFooService' and 'heyFooService'.
When testing the Hystrix fallback behavior of my Feign API, I get an error, when I expect it to succeed.
Feign interface:
This is the api to the external service.
#FeignClient(name = "book", fallback = BookAPI.BookAPIFallback.class)
public interface BookAPI {
#RequestMapping("/")
Map<String, String> getBook();
#Component
class BookAPIFallback implements BookAPI {
#Override
#RequestMapping("/")
public Map<String, String> getBook() {
Map<String, String> fallbackmap = new HashMap<>();
fallbackmap.put("book", "fallback book");
return fallbackmap;
}
}
}
Test class
This test exists just to verify fallback behavior:
#RunWith(SpringRunner.class)
#SpringBootTest(webEnvironment = NONE)
public class BookServiceClientTest {
#MockBean
RestTemplate restTemplate;// <---- #LoadBalanced bean
#Autowired
private BookServiceClient bookServiceClient;
#Before
public void setup() {
when(restTemplate.getForObject(anyString(), any()))
.thenThrow(new RuntimeException("created a mock failure"));
}
#Test
public void fallbackTest() {
assertThat(bookServiceClient.getBook())
.isEqualTo(new BookAPI.BookAPIFallback().getBook().get("book")); // <--- I thought this should work
}
}
config files
application.yml
These files show configuration that might be relevant:
feign:
hystrix:
enabled: true
test/application.yml
eureka:
client:
enabled: false
The Question
Everything works fine when running the apps.
But when running this test, I get the below error.
Naturally, it's a test, so I'm trying to bypass the lookup anyway.
java.lang.RuntimeException: com.netflix.client.ClientException: Load balancer does not have available server for client: book
at org.springframework.cloud.netflix.feign.ribbon.LoadBalancerFeignClient.execute(LoadBalancerFeignClient.java:71)
at feign.SynchronousMethodHandler.executeAndDecode(SynchronousMethodHandler.java:97)
What am I missing?
Addendums
Application class
#SpringBootApplication
#EnableCircuitBreaker
#EnableDiscoveryClient
#EnableFeignClients
public class LibraryApplication {
public static void main(String[] args) {
SpringApplication.run(LibraryApplication.class, args);
}
}
LibraryController
#Controller
public class LibraryController {
private final BookServiceClient bookService;
public LibraryController(BookServiceClient bookServiceClient) {
this.bookService = bookServiceClient;
}
#GetMapping("/")
String getLibrary(Model model) {
model.addAttribute("msg", "Welcome to the Library");
model.addAttribute("book", bookService.getBook());
return "library";
}
}
There are no other classes.
so! I was able to recreate the issue, thanks for adding more code, had to play about with it a tad as I was unsure what the BookClientService looked like and it wouldn't make sense for it to implement the BookAPI as that would be an internal call e.g. in your application and not an external API call with Feign.
Anyway,
I pushed my version of what you provided here.
https://github.com/Flaw101/feign-testing
The issue was resolved when I renamed the second application.yml which lives in the src/test/resources folder to application-test.yml which will merge the properties.
The issue was caused by the fact the second property source, the testing one, overrides the initial application.yml and disables hystrix, because Hystrix is disabled there is no fallback to go to and it throws the root cause of what would cause the fallback, a lack of a server to call to for the Book API. Renaming it to application-test will always be loaded into spring test contexts. You could resolve it with the use of inlined properties or profiles.
I've added another test disabling feign /w hystrix within the test which re-creates the error you are recieving.
I'm wondering what the best way to load initial database data before the application starts? What I'm looking for is something that will fill my H2 database with data.
For example, I have a domain model "User" I can access users by going to /users but initially there won't be any users in the database so I have to create them. Is there anyway to fill the database with data automatically?
At the moment I have a Bean that gets instantiated by the container and creates users for me.
Example:
#Component
public class DataLoader {
private UserRepository userRepository;
#Autowired
public DataLoader(UserRepository userRepository) {
this.userRepository = userRepository;
LoadUsers();
}
private void LoadUsers() {
userRepository.save(new User("lala", "lala", "lala"));
}
}
But I very much doubt that is the best way of doing it. Or is it?
You can create a data.sql file in your src/main/resources folder and it will be automatically executed on startup. In this file you can add some insert statements, eg.:
INSERT INTO users (username, firstname, lastname) VALUES
('lala', 'lala', 'lala'),
('lolo', 'lolo', 'lolo');
Similarly, you can create a schema.sql file (or schema-h2.sql) as well to create your schema:
CREATE TABLE task (
id INTEGER PRIMARY KEY,
description VARCHAR(64) NOT NULL,
completed BIT NOT NULL);
Though normally you shouldn't have to do this since Spring boot already configures Hibernate to create your schema based on your entities for an in memory database. If you really want to use schema.sql you'll have to disable this feature by adding this to your application.properties:
spring.jpa.hibernate.ddl-auto=none
More information can be found at the documentation about Database initialization.
If you're using Spring Boot 2, database initialization only works for embedded databases (H2, HSQLDB, ...). If you want to use it for other databases as well, you need to change the initialization mode property:
spring.sql.init.mode=always # Spring Boot >=v2.5.0
spring.datasource.initialization-mode=always # Spring Boot <v2.5.0
If you're using multiple database vendors, you can name your file data-h2.sql or data-mysql.sql depending on which database platform you want to use.
To make that work, you'll have to configure the datasource platform property:
spring.sql.init.platform=h2 # Spring Boot >=v2.5.0
spring.datasource.platform=h2 # Spring Boot <v2.5.0
If I just want to insert simple test data I often implement a ApplicationRunner. Implementations of this interface are run at application startup and can use e.g. a autowired repository to insert some test data.
I think such an implementation would be slightly more explicit than yours because the interface implies that your implementation contains something you would like to do directly after your application is ready.
Your implementation would look sth. like this:
#Component
public class DataLoader implements ApplicationRunner {
private UserRepository userRepository;
#Autowired
public DataLoader(UserRepository userRepository) {
this.userRepository = userRepository;
}
public void run(ApplicationArguments args) {
userRepository.save(new User("lala", "lala", "lala"));
}
}
You can add a spring.datasource.data property to application.properties listing the sql files you want to run. Like this:
spring.datasource.data=classpath:accounts.sql, classpath:books.sql, classpath:reviews.sql
//or (depending on SB version)
spring.sql.init.data-locations=classpath:accounts.sql, classpath:books.sql, file:reviews.sql
The sql insert statements in each of these files will then be run, allowing you to keep things tidy.
If you put the files in the classpath, for example in src/main/resources they will be applied. Or replace classpath: with file: and use an absolute path to the file
If you want to run DDL type SQL then use:
spring.datasource.schema=classpath:create_account_table.sql
// depending on spring version
spring.sql.init.schema-locations=classpath:create_account_table.sql
Edit: these solutions are great to get you up and running quickly, however for a more production ready solution it would be worth looking at a framework such as flyway, or liquibase. These frameworks integrate well with spring, and provide a quick, consistent, version-controlled means of initialising schema, and standing-data.
There are multiple ways how to achieve this. I prefer to use one of following options:
Option 1: Initializing with CommandLineRunner bean:
#Bean
public CommandLineRunner loadData(CustomerRepository repository) {
return (args) -> {
// save a couple of customers
repository.save(new Customer("Jack", "Bauer"));
repository.save(new Customer("Chloe", "O'Brian"));
repository.save(new Customer("Kim", "Bauer"));
repository.save(new Customer("David", "Palmer"));
repository.save(new Customer("Michelle", "Dessler"));
// fetch all customers
log.info("Customers found with findAll():");
log.info("-------------------------------");
for (Customer customer : repository.findAll()) {
log.info(customer.toString());
}
log.info("");
// fetch an individual customer by ID
Customer customer = repository.findOne(1L);
log.info("Customer found with findOne(1L):");
log.info("--------------------------------");
log.info(customer.toString());
log.info("");
// fetch customers by last name
log.info("Customer found with findByLastNameStartsWithIgnoreCase('Bauer'):");
log.info("--------------------------------------------");
for (Customer bauer : repository
.findByLastNameStartsWithIgnoreCase("Bauer")) {
log.info(bauer.toString());
}
log.info("");
}
}
Option 2: Initializing with schema and data SQL scripts
Prerequisites:
application.properties
spring.jpa.hibernate.ddl-auto=none
Explanation:
Without ddl-auto SQL scripts will be ignored by
Hibernate and trigger default behavior - scanning project for
#Entity and/or #Table annotated classes.
Then, in your MyApplication class paste this:
#Bean(name = "dataSource")
public DriverManagerDataSource dataSource() {
DriverManagerDataSource dataSource = new DriverManagerDataSource();
dataSource.setDriverClassName("org.h2.Driver");
dataSource.setUrl("jdbc:h2:~/myDB;MV_STORE=false");
dataSource.setUsername("sa");
dataSource.setPassword("");
// schema init
Resource initSchema = new ClassPathResource("scripts/schema-h2.sql");
Resource initData = new ClassPathResource("scripts/data-h2.sql");
DatabasePopulator databasePopulator = new ResourceDatabasePopulator(initSchema, initData);
DatabasePopulatorUtils.execute(databasePopulator, dataSource);
return dataSource;
}
Where scripts folder is located under resources folder (IntelliJ Idea)
Hope it helps someone
Update 04-2021: Both options are great to combine with Spring Profiles as this will help you to avoid creating additional config files making your life as the developer easy.
You can use something like this:
#SpringBootApplication
public class Application {
#Autowired
private UserRepository userRepository;
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
#Bean
InitializingBean sendDatabase() {
return () -> {
userRepository.save(new User("John"));
userRepository.save(new User("Rambo"));
};
}
}
In Spring Boot 2 data.sql was not working with me as in spring boot 1.5
import.sql
In addition, a file named import.sql in the root of the classpath is executed on startup if Hibernate creates the schema from scratch (that is, if the ddl-auto property is set to create or create-drop).
Note very important if you insert Keys cannot be duplicated do not use ddl-auto property is set to update because with each restart will insert same data again
For more information you vist spring websit
https://docs.spring.io/spring-boot/docs/current/reference/html/howto-database-initialization.html
Spring Boot allows you to use a simple script to initialize your database, using Spring Batch.
Still, if you want to use something a bit more elaborated to manage DB versions and so on, Spring Boot integrates well with Flyway.
See also:
Spring Boot Database initialization
You can simply create a import.sql file in src/main/resources and Hibernate will execute it when the schema is created.
If you came here and nothing seems to work for you, then it might be the case that you are affected from some changes that were introduced with Spring Boot 2.5 and onwards.
Here is the total set of properties which I use for postgresql.
spring:
sql.init.mode: always <-----------------
datasource:
url: jdbc:postgresql://localhost:5432/products
username:
password:
jpa:
defer-datasource-initialization: true <------------------
hibernate:
ddl-auto: create-drop <----------------
database-platform: org.hibernate.dialect.PostgreSQLDialect
I have also marked with <--- the relevant properties for the current topic in order to achieve the following.
ORM vendor will create database schema for you from Java Entities model.
After database schema is created, initial data will be loaded to database from the file data.sql
Ps: Don't forget to add the file with initial data, data.sql under src/main/resources
Also as reference: Spring Boot 2.5 release notes
Here is the way I got that:
#Component
public class ApplicationStartup implements ApplicationListener<ApplicationReadyEvent> {
/**
* This event is executed as late as conceivably possible to indicate that
* the application is ready to service requests.
*/
#Autowired
private MovieRepositoryImpl movieRepository;
#Override
public void onApplicationEvent(final ApplicationReadyEvent event) {
seedData();
}
private void seedData() {
movieRepository.save(new Movie("Example"));
// ... add more code
}
}
Thanks to the author of this article:
http://blog.netgloo.com/2014/11/13/run-code-at-spring-boot-startup/
I solved similar problem this way:
#Component
public class DataLoader {
#Autowired
private UserRepository userRepository;
//method invoked during the startup
#PostConstruct
public void loadData() {
userRepository.save(new User("user"));
}
//method invoked during the shutdown
#PreDestroy
public void removeData() {
userRepository.deleteAll();
}
}
You're almost there!
#Component
public class DataLoader implements CommandLineRunner {
private UserRepository userRepository;
public DataLoader(UserRepository userRepository) {
this.userRepository = userRepository;
}
#Override
public void run(String... args) throws Exception {
LoadUsers()
}
private void LoadUsers() {
userRepository.save(new User("lala", "lala", "lala"));
}
}
you can register and event listener to achieve that like below:
#EventListener
public void seed(ContextRefreshedEvent event) {
userRepository.save(new User("lala", "lala", "lala"));
}
When the ContextRefreshEvent is fired, we get access to all autowired beans in the application — including models and repositories.
If someone are struggling in make this to work even following the accepted answer, for me only work adding in my src/test/resources/application.yml the H2 datasource details:
spring:
datasource:
platform: h2
url: jdbc:h2:mem:test;DB_CLOSE_DELAY=-1
driver-class-name: org.h2.Driver
username: sa
password:
If you want to insert only few rows and u have JPA Setup. You can use below
#SpringBootApplication
#Slf4j
public class HospitalManagementApplication {
public static void main(String[] args) {
SpringApplication.run(HospitalManagementApplication.class, args);
}
#Bean
ApplicationRunner init(PatientRepository repository) {
return (ApplicationArguments args) -> dataSetup(repository);
}
public void dataSetup(PatientRepository repository){
//inserts
}
You can use the below code. In the following code a database insertion occurs during the startup of the spring boot application.
#SpringBootApplication
public class Application implements CommandLineRunner {
#Autowired
private IService<Car> service;
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
#Override
public void run(String... args) throws Exception {
for(int i=1; i<=1000; i++) {
Car car = new Car();
car.setName("Car Name "+i);
book.setPrice(50 + i);
service.saveOrUpdate(car);
}
}
}
This will also work.
#Bean
CommandLineRunner init (StudentRepo studentRepo){
return args -> {
// Adding two students objects
List<String> names = Arrays.asList("udara", "sampath");
names.forEach(name -> studentRepo.save(new Student(name)));
};
}
The most compact (for dynamic data) put #mathias-dpunkt solution into MainApp (with Lombok #AllArgsConstructor):
#SpringBootApplication
#AllArgsConstructor
public class RestaurantVotingApplication implements ApplicationRunner {
private final VoteRepository voteRepository;
private final UserRepository userRepository;
public static void main(String[] args) {
SpringApplication.run(RestaurantVotingApplication.class, args);
}
#Override
public void run(ApplicationArguments args) {
voteRepository.save(new Vote(userRepository.getOne(1), LocalDate.now(), LocalTime.now()));
}
}
One possibility is using incorrect JDBC URL. make sure it is jdbc:h2:mem:testdb
I created a library that facilitates initial/demo data loading in a Spring Boot application. You can find it at https://github.com/piotrpolak/spring-boot-data-fixtures
Once the data fixtures starter is on the classpath, it will automatically try to load DICTIONARY data upon application startup (this behavior can be controlled by properties) - all you need to do is to register a bean implementing DataFixture.
I find loading initial data by code superior to loading it using SQL scripts:
the logic of your fixtures lives close to your application logic/domain model and it is subject to refactoring as your domain evolves
you benefit from incremental demo data updates - imagine a QA environment with some user data (that needs not to be lost after application deploy) but at the same time you want to add data for the new features you developed
Example data fixture:
/**
* You can have as many fixture classes as you want.
* #Order annotation is respected for the fixtures belonging to the same set.
* You can make your demo database to be incrementally updated with fresh data
* each time the application is redeployed - all you need to do is to write
* a good condition in `canBeLoaded()` method.
*/
#Component
public class InitialDataFixture implements DataFixture {
private final LanguageRepository languageRepository;
// ...
#Override
public DataFixtureSet getSet() {
return DataFixtureSet.DICTIONARY;
}
/**
* We want to make sure the fixture is applied once and once only.
* A more sophisticated condition can be used to create incremental demo data
* over time without the need to reset the QA database (for example).
*/
#Override
public boolean canBeLoaded() {
return languageRepository.size() == 0;
}
/**
* The actual application of the fixture.
* Assuming that data fixtures are registered as beans, this method can call
* other services and/or repositories.
*/
#Override
public void load() {
languageRepository.saveAll(Arrays.asList(
new Language("en-US"), new Language("pl-PL")));
}
}
The concept is inspired by the Symfony Doctrine Data Fixtures bundle.
For those using MysqlDriver, I tried using Init attribute of #bean annotation and it works.
After created the Schema and Data sql file in the path of resources\Scripts
Add the line in application.properties
spring.jpa.hibernate.ddl-auto=none
Edit the Application content:
package com.spring_mvaen.demo;
import org.springframework.boot.CommandLineRunner;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.context.annotation.Bean;
import org.springframework.core.io.ClassPathResource;
import org.springframework.core.io.Resource;
import org.springframework.jdbc.datasource.DriverManagerDataSource;
import org.springframework.jdbc.datasource.init.DatabasePopulator;
import org.springframework.jdbc.datasource.init.DatabasePopulatorUtils;
import org.springframework.jdbc.datasource.init.ResourceDatabasePopulator;
#SpringBootApplication
public class DemoApplication implements CommandLineRunner {
public static void main(String[] args) {
SpringApplication.run(DemoApplication.class, args);
}
#Override
public void run(String... arg0) throws Exception {
System.out.println("Hello world from Command Line Runner");
}
#Bean(name = "dataSource")
public DriverManagerDataSource dataSource() {
DriverManagerDataSource dataSource = new DriverManagerDataSource();
dataSource.setDriverClassName("com.mysql.cj.jdbc.Driver");
dataSource.setUrl("jdbc:mysql://localhost:3306/db_spring_rest?useUnicode=true&useLegacyDatetimeCode=fa lse&serverTimezone=UTC&createDatabaseIfNotExist=true&allowPublicKeyRetrieval=true&useSSL=false");
dataSource.setUsername("root");
dataSource.setPassword("root");
// schema init
Resource initSchema = new ClassPathResource("scripts/schema.sql");
Resource initData = new ClassPathResource("scripts/data.sql");
DatabasePopulator databasePopulator = new ResourceDatabasePopulator(initSchema, initData);
DatabasePopulatorUtils.execute(databasePopulator, dataSource);
return dataSource;
}
}
If you want to do insert quick some queries, you can do with h2 data.sql queries as well
application.properties include:
spring.jpa.show-sql=true
spring.h2.console.enabled=true
spring.datasource.url=jdbc:h2:mem:testdb
#This directs the data.sql file and help it to run
spring.sql.init.data-locations=classpath:data.sql
spring.jpa.defer-datasource-initialization=true
data.sql file include:
INSERT INTO todo (id, username, description, target_date, is_done) VALUES (10001, 'lighteducation', 'Learn dance', CURRENT_DATE ,false);
INSERT INTO todo (id, username, description, target_date, is_done) VALUES (10002, 'lighteducation', 'Learn Angular14', CURRENT_DATE, false);
INSERT INTO todo (id, username, description, target_date, is_done) VALUES (10003, 'lighteducation', 'Learn Microservices', CURRENT_DATE,false);
P.S.: data.sql file should be inside src/main/resources
Your #Entity include
#Getter
#Setter
#AllArgsConstructor
#ToString
#Entity
public class Todo {
#Id
#GeneratedValue
private Long id;
private String username;
private String description;
private Date targetDate;
private boolean isDone;
protected Todo() {
}
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Todo todo = (Todo) o;
return id == todo.id;
}
#Override
public int hashCode() {
return Objects.hash(id);
}
}
That is it basically. it will be in memory, it means when you restart application data will arese and will be again same as queries show.
But it is easy for quick check
also you can access the path with http://localhost:8080/h2-console/ or you can edit the path from .properties file