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Can't inject custom repository

By : Yann
Date : October 18 2020, 06:10 AM
To fix this issue Composition over inheritance, that is wrapping a repository, and using it as a provider can help here:
code :
export class EnqueuedRepository {
    @Inject() readonly cache: RequestScopedCache;

        @InjectRepository(Enqueued) private readonly enqueuedRepository: Repository<Enqueued>
    ) {

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How do I inject a repository into a custom MembershipProvider using AutoFac

By : Александр Крылосов
Date : March 29 2020, 07:55 AM
will be helpful for those in need No. It's not possible.
The DependencyResolver is not used for the providers (roles/membership) etc.

How do you inject a Symfony repository into a custom user provider?

By : Yohanes Kuma
Date : March 29 2020, 07:55 AM
I wish this help you The problem is that the method is_approved() is static. You cannot use $this in a static context. You need to declare your method non-static.

How to inject configuration in a custom Spring Data JPA repository

By : Haiying Tian
Date : March 29 2020, 07:55 AM
I wish this helpful for you I came across the exact same problem and could not find any way around defining my own JpaRepositoryFactoryBean class. It seems that the dependencies of the custom repository base class are not automatically injected the way they are in a standard bean (see here and here). Also, when creating instances of repository interfaces, the default JpaRepositoryFactory only passes instances of JpaEntityInformation and EntityManager to the class constructor (see here). This, as far as I can tell, effectively prevents you from including additional dependencies for classes extending SimpleJpaRepository.
I ended up defining the custom factory the following way:
code :
@ConfigurationProperties(prefix = "spring.jpa.properties.hibernate.jdbc")
public class RepositoryConfiguration {
    private int batchSize;

public class MyCustomRepositoryFactoryBean<R extends JpaRepository<T, I>, T, I extends Serializable> extends JpaRepositoryFactoryBean<R, T, I> {

    private RepositoryConfiguration repositoryConfiguration;

    public MyCustomRepositoryFactoryBean(Class<? extends R> repositoryInterface, RepositoryConfiguration repositoryConfiguration) {
        this.repositoryConfiguration = repositoryConfiguration;

    protected RepositoryFactorySupport createRepositoryFactory(EntityManager entityManager) {
        return new MyCustomRepositoryFactory(entityManager, repositoryConfiguration);

    private static class MyCustomRepositoryFactory<T, I extends Serializable> extends JpaRepositoryFactory {

        private RepositoryConfiguration repositoryConfiguration;

        MyCustomRepositoryFactory(EntityManager entityManager, RepositoryConfiguration repositoryConfiguration) {
            this.repositoryConfiguration = repositoryConfiguration;

        protected SimpleJpaRepository<?, ?> getTargetRepository(RepositoryInformation information, 
                EntityManager entityManager) {

            JpaEntityInformation<T, ?> entityInformation = 
                    (JpaEntityInformation<T, ?>) getEntityInformation(information.getDomainType());

            return new MyCustomRepositoryImpl<T, I>(

        protected Class<?> getRepositoryBaseClass(RepositoryMetadata metadata) {
            return MyCustomRepositoryImpl.class;
        repositoryFactoryBeanClass = MyCustomRepositoryFactoryBean.class

Automapper Custom Resolver - Inject Repository into constructor

By : user3928689
Date : March 29 2020, 07:55 AM
this one helps. Found the solution shorlty after...i was forgetting to add my resolvers as an IoC container.
Works great now!

Inject repository to custom membership provider with Ninject

By : Jacky Hicks
Date : March 29 2020, 07:55 AM
wish of those help You could use the approach @Remo Gloor outlines in his blog post on provider injection. It involves 3 steps:
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