Spring 数据：JPA 存储库 findAll() 返回 *Map 而不是 List？

Question

I have a Spring Data JPA repository interface that looks something like this:

@Repository
public interface DBReportRepository extends JpaRepository<TransactionModel, Long> {

    List<TransactionModel> findAll();
    List<TransactionModel> findByClientId(Long id);
}

Is there a workaround to make the same but for a Collection to be returned of type HashMap<K, V> ? I've looked through the Spring Data classes and couldn't find anything other than List<> return values.

Answer 1

I dont think you will find an easier solution as to create a simple one liner to convert your result to a map. It is simple and fast with java 8 lambdas:

Map<Long, Transaction> transactionMap = transactionList.stream()
         .collect(Collectors.toMap(Transaction::getId, Function.identity()));

Answer 2

Just had to solve something similar and Patricks answer helped but it can be improved by indicating where to add it.

To to make it appear like the JPA repo returns a map, an improvement would to wrap this up in a default method in the repository interface. Saves you having to perform a stream in all the consuming classes.

@Repository
public interface DBReportRepository extends JpaRepository<TransactionModel, Long> {

    List<TransactionModel> findAll();

    default Map<Long, TransactionModel> findAllMap() {
        return findAll().stream().collect(Collectors.toMap(TransactionModel::getId, v -> v));
    }

    List<TransactionModel> findByClientId(Long id);

    default Map<Long, TransactionModel> findByClientIdMap(Long id) {
        return findByClientId(id).stream().collect(Collectors.toMap(TransactionModel::getId, v -> v));
    }
}

Answer 3

Creating a map after having retrieved your entities is not a solution if you are trying to improve your performance. It will take longer to retrieve the whole entity and then add the selected fields into a map, than to directly retrieve a map, or a list.

Answer 4

You could get your map from a custom streamable wrapper type which is documented in Spring Data JPA reference .

@RequiredArgsConstructor(staticName = "of")
public class TransactionModels implements Streamable<TransactionModel> {
    
    private final Streamable<TransactionModel> streamable;

    @Override
    public Iterator<TransactionModel> iterator() {
        return streamable.iterator();
    }

    public Map<Long, TransactionModel> asMap() {
        return streamable.stream()
            .collect(Collectors.toMap(Transaction::getId, Function.identity()));
    }
}

@Repository
public interface DBReportRepository extends JpaRepository<TransactionModel, Long> {

    TransactionModels findAll();

    TransactionModels findByClientId(Long id);

}

The reason why you cannot get your desired Map<Long, TransactionModel> method in Spring Data is that java.util.Map<K,V> will be delegated to org.springframework.data.jpa.repository.query.JpaQueryExecution$SingleEntityExecution instead of org.springframework.data.jpa.repository.query.JpaQueryExecution$CollectionExecution by default. And it's reasonable becuase Map is not extending from Collection in Java . But in your case you would like to use Map as a "collection". Fortunately you can extend Spring Data repository with custom methods which could return any type you like.

Answer 5

Another Solution use java 8 lambdas:

Using a groupingBy method of Collectors that grouping transactionList by ID and storing results in a Map instance

Map<Long, Transaction> transactionMap = transactionList.stream().collect(Collectors.groupingBy(v->v.getId()));