CODEWITHSUNDEEP
springspring-bootspring-data
Héctor
Héctor

Reputation: 26054

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

I'm trying to implement a custom Spring repository. I have the interface:

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

the implementation:

public class FilterRepositoryImpl implements FilterRepositoryCustom {
...
}

and the "main" repository, extending my custom repository:

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

I'm using Spring Boot and, according to the docs:

By default, Spring Boot will enable JPA repository support and look in the package (and its subpackages) where @SpringBootApplication is located.

When I run my application, I get this error:

org.springframework.data.mapping.PropertyReferenceException: No property filterBy found for type User!

Upvotes: 66

Views: 119420

Answers (10)

alierdogan7
alierdogan7

Reputation: 840

The new naming convention is explained in the documentation:

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

So, for adding a new function someCustomMethod to CustomizedUserRepository, you need to add a new class named CustomizedUserRepositoryImpl which implements CustomizedUserRepository interface.

interface CustomizedUserRepository {
  void someCustomMethod(User user);
}

class CustomizedUserRepositoryImpl implements CustomizedUserRepository {
  public void someCustomMethod(User user) {
    // Your custom implementation
  }
}

Upvotes: 0

Oleksandr Bondarchuk
Oleksandr Bondarchuk

Reputation: 1053

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

Upvotes: 1

pureevill
pureevill

Reputation: 99

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

Upvotes: 3

Alex Xela
Alex Xela

Reputation: 11

for me the keys was

  1. @EnableJpaRepositories(repositoryBaseClass = MyRepositoryImpl.class)
  2. @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
)

Upvotes: 1

Andrew Mikolyk
Andrew Mikolyk

Reputation: 181

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^∇^)

Upvotes: 7

Java Man
Java Man

Reputation: 1

Old way:

Entity aThing = repository.findOne(1L);

New way:

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

Upvotes: -5

A0__oN
A0__oN

Reputation: 9100

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.

Upvotes: 105

user64141
user64141

Reputation: 5331

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})

Upvotes: 6

Vivek Vhatkar
Vivek Vhatkar

Reputation: 41

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!

Upvotes: 3

Bruno 82
Bruno 82

Reputation: 519

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>

Upvotes: 0

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