Spring Boot JPA 本机查询枚举值

Question

我正在尝试编写一个本机查询来从基于 EnumType 实体的表中进行搜索。此 ENUM MealType 是 @Table Meal 的一部分。

@Column(name = "meal_type")
@Enumerated(EnumType.STRING)
private MealType mealType;

现在，我的查询是：

@Repository
public interface MealRepository extends JpaRepository<Meal, Long> {
@Query(value ="select * from meal m where m.meal_type = ?1", nativeQuery = true)
List<Meal> findMealByType(MealType mealType);

}

但是当我对其进行测试时，我不断收到org.springframework.orm.jpa.JpaSystemException: could not extract ResultSet; nested exception is org.hibernate.exception.GenericJDBCException: could not extract ResultSet

除此之外，我还尝试使用 MealType 作为参数重写查询：

  @Query(value ="select * from meal m where m.meal_type in :meal_type ", nativeQuery = true)
List<Meal> findMealByType(@Param("meal_type") MealType mealType);

但它导致了另一种错误

InvalidDataAccessResourceUsageException: could not prepare statement; SQL [select * from meal m where m.meal_type in ? ]; nested exception is org.hibernate.exception.SQLGrammarException: could not prepare statement

我预计其他地方会出现问题，但是基于 ID 搜索的相同自定义查询可以正常工作。

Answer 1

您不能使用枚举和 SQL。您必须将参数作为字符串传递：

@Repository
public interface MealRepository extends JpaRepository<Meal, Long> {

    @Query(value ="select * from meal m where m.meal_type = ?1", nativeQuery = true)
    List<Meal> findMealByType(String mealType);

Answer 2

@Repository
public interface MealRepository extends JpaRepository<Meal, Long> {

    @Query(value ="select * from meal m where m.meal_type = :mealType", nativeQuery = true)
    List<Meal> findMealByType(@Param("mealType")MealType mealType);
}