How can I add columns dynamically to a Spring Data JPA @Query

Question

Consider the following method on a Spring Data JPA interface:

@Query(\"select distinct :columnName from Item i\")
List findByName(@Param(\"columnName         


        

Answer 1

First of all you must implement custom Spring Data repository by adding interface:

public interface ItemCustomRepository {

    List<Item> findBy(String columnName, String columnValue);

}

then you must extend your current Spring Data repository interface with newly created i.e.:

public interface ItemRepository extends JpaRepository<Item, Long>, ItemCustomRepository, QueryDslPredicateExecutor {

}

and then you must implement your interface using Query DSL dynamic expression feature (the name ItemRepositoryImpl is crucial - it will let you use original Spring Data repository implementation):

public class ItemRepositoryImpl implements ItemCustomRepository {

    @Autowired
    private ItemRepository itemRepository;

    public List<Item> findBy(final String columnName, final String columnValue) {
        Path<Item> item = Expressions.path(Item.class, "item");
        Path<String> itemColumnName = Expressions.path(String.class, item, columnName);
        Expression<String> itemValueExpression = Expressions.constant(columnValue);
        BooleanExpression fieldEqualsExpression = Expressions.predicate(Ops.EQ, itemColumnName, itemValueExpression);

        return itemRepository.findAll(fieldEqualsExpression);
    }

}

Answer 2

You can't. You'll have to implement such a method by yourself. And you won't be able to use parameters: you'll have to use String concatenation or the criteria API. What you'll pass won't be a column name but a field/property name. And it won't return a List<Item>, since you only select one field.

Answer 3

You can use QueryDSL support built into Spring Data. See this tutorial to get started.