No property found for type... custom Spring Data repository

The problem here is that you are creating FilterRepositoryImpl but you are using it in UserRepository. You need to create UserRepositoryImpl to make this work.

Read this doc for more detail

Basically

public interface UserRepositoryCustom {
    List<User> filterBy(String role);
}

public class UserRepositoryImpl implements UserRepositoryCustom {
...
}

public interface UserRepository extends JpaRepository<User, String>, UserRepositoryCustom {
...
}

Spring Data 2.x update
This answer was written for Spring 1.x. As Matt Forsythe pointed out, the naming expectations changed with Spring Data 2.0. The implementation changed from the-final-repository-interface-name-with-an-additional-Impl-suffix to the-custom-interface-name-with-an-additional-Impl-suffix.

So in this case, the name of the implementation would be: UserRepositoryCustomImpl.

Another way this error can happen if the impl class for FilterRepositoryCustom isn't picked up in your spring configuration:

@EnableJpaRepositories(basePackageClasses = {RepoPackageMarker.class, FilterRepositoryCustomImpl.class})

Is it a must that the customMethod() in the CustomRepository can only have parameters defined that are either

1.Entity class name - customMethod(User user),

2.Entity class attributes - customMethod(String firstName), here firstName is an attribute of User Entity class.

Can I not have something like customMethod(CustomCriteria criteria), the criteria class contain the various attributes that are used to construct a dynamic query.

e.g. getStatusByCriteria(CustomCriteria criteria), CustomCriteria is a simple pojo annotated with @Component so that spring identifies it.

When I tried this I get an error:

org.springframework.data.mapping.PropertyReferenceException: No property criteria found for type UserRepository!

I had the same problem. Please check if your packages structure looks like this

custom
   impl
      - FilterRepositoryCustomImpl.class
- FilterRepositoryCustom.class      

Because when I try to use my custom repo it doesn't see the implementation. (implementation should be in the same package or in sub-packages for Spring to see it)

Maybe it helps somebody (ノ^∇^)

for me the keys was

@EnableJpaRepositories(repositoryBaseClass = MyRepositoryImpl.class) @NoRepositoryBean annotation on interface

follow the docs for your spring ver https://docs.spring.io/spring-data/data-commons/docs/2.3.4.RELEASE/reference/html/#repositories

so as result i have:

@NoRepositoryBean
public interface SliceRepository<T, ID> extends JpaRepository<T, ID> {..}

public class SliceRepositoryImpl<T, ID> extends SimpleJpaRepository<T, ID> implements SliceRepository<T, ID> {..}


@EnableJpaRepositories(
        entityManagerFactoryRef = "entityManagerFactory",
        basePackages = {"com.some.servicename.repository"},
        repositoryBaseClass = SliceRepositoryImpl.class
)

I also had this error. It occured because I had configured elasticSearch repository with package

@EnableElasticsearchRepositories("some.package")

Next I moved MyElasticSearchRepository from package 'some.package' to another, but didn't change configuration

I had the same problem in a project of mine. I solved the problem by adding a line in my pom.xml

<plugin>
    <groupId>org.apache.maven.plugins</groupId>
    <artifactId>maven-compiler-plugin</artifactId>
    <configuration>
        <includes>
            <include>com/my/package/entities/*.java</include>
            <include>com/my/package/repositories/*.java</include>
            <include>com/my/package/repositories/impl/*.java</include> <!-- add this -->
        </includes>
    </configuration>
</plugin>

The most important part of the class name that corresponds to the fragment interface is the Impl postfix.

Old way:

Entity aThing = repository.findOne(1L);

New way:

Optional<Entity> aThing = repository.findById(1L);