Question

我有interface实施 JPARepository 并有三种方法，其中一种方法是自定义@Query。

public interface PersonRepository extends JpaRepository<Person, Long> {

  List<Person> getPersonBycountryCode(String countryCode);

  List<Person> findByCountryCodeAndCity(String string,String city);

  @Query(value = "SELECT person.firstName as firstName, person.lastName as lastName, person.countryCode as country, person.city as city,"
              + " SQRT(POWER((69.1 * (person.age - :age )) , 2 )"
              + " + POWER((53 * (person.experience - :experience )), 2)) as eligibility"
              + " FROM Person person"
              + " ORDER BY eligibility ASC")
  List<PersonDetailsDto> findPersonDetailsByEligibility(
          @Param("age") BigDecimal age,
          @Param("experience") BigDecimal experience,
          Pageable pageable
  );
}

问题是： @Query方法未返回PersonDetailsDto列表但返回字符串列表列表（ List<List<String>> ）。

PersonDetailsDto是一个POJO类，其中包含查询输出（firstName，lastName，country，city，eligibility）中描述的所有变量，以及包含所有变量作为参数的构造函数。其他两个方法确实返回Person对象列表。

有什么想法吗？

Answer 1

实际上JpaRepository<Person, Long>意味着，您只能在jpa存储库方法中使用Person作为您的dto。

对于您的解决方案，您只需在存储库中定义dto接口：

public interface PersonRepository extends JpaRepository<Person, Long> {

  List<Person> getPersonBycountryCode(String countryCode);

  List<Person> findByCountryCodeAndCity(String string,String city);

  @Query(value = "SELECT person.firstName as firstName, person.lastName as lastName, person.countryCode as country, person.city as city,"
              + " SQRT(POWER((69.1 * (person.age - :age )) , 2 )"
              + " + POWER((53 * (person.experience - :experience )), 2)) as eligibility"
              + " FROM Person person"
              + " ORDER BY eligibility ASC")
  List<PersonDetailsDto> findPersonDetailsByEligibility(
          @Param("age") BigDecimal age,
          @Param("experience") BigDecimal experience,
          Pageable pageable
  );

 //define the interface here
 public interface PersonDetailsDto{
   public String getFirstName();
   public String getLastName();
   public String getCountry();
   public String getCity();
   public Integer getEligibility();
 }

}

Answer 2

如果我没有错，JPA背后没有寻找特定字段的想法是成本（效率明智）相同，从表格的一行带来一列或所有列。但是为了解决您的问题，您可以设置{ {1}}来自Repository类的nativeQuery = true注释，如下所示：

@Query

我希望这可以帮助您解决问题。

Answer 3

您可以在查询@Query时使用new关键字。并确保您拥有PersonDetailsDto的相应构造函数，并更改包名称。

@Query(value = "SELECT new com.company.PersonDetailsDto(person.firstName, person.lastName, person.countryCode , person.city ,"
          + " SQRT(POWER((69.1 * (person.age - :age )) , 2 )"
          + " + POWER((53 * (person.experience - :experience )), 2)) "
          + " FROM Person person"
          + " ORDER BY eligibility ASC")
List<PersonDetailsDto> findPersonDetailsByEligibility(
      @Param("age") BigDecimal age,
      @Param("experience") BigDecimal experience,
      Pageable pageable
);

类似question's answer。

Answer 4

只是通过别名来称呼它，它对我来说就是这样例如：

@Query(value = "SELECT person FROM Person person"
          + " ORDER BY eligibility ASC")
List<PersonDetailsDto> findPersonDetailsByEligibility(
      @Param("age") BigDecimal age,
      @Param("experience") BigDecimal experience,
      Pageable pageable
);

Spring数据jpa，jparepository返回字符串列表代替DTO对象

4 个答案: