枚举值的 Spring Boot JPA 本机查询

Vla*_*yan 2 java spring jpa spring-data-jpa

我正在尝试编写本机查询以从基于 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;
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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);
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}

但是当我对其进行测试时,我不断得到 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 作为参数重新编写查询:

  @Query(value ="select * from meal m where m.meal_type in :meal_type ", nativeQuery = true)
List<Meal> findMealByType(@Param("meal_type") MealType mealType);
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但它引起了不同类型的错误

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一些问题,但是基于 ID 搜索的相同自定义查询可以正常工作。

Sim*_*lli 6

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);
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  • 你实际上可以::#{#mealType.name()}参见/sf/ask/3112227611/ (2认同)