JPA：在使用 findAll 和规范时如何按字段自定义顺序排序

Question

我正在将 Spring Tool Suite 4 (4.12.0.RELEASE) 与 Java 8 一起使用，并且我有以下代码可以从过滤、分页和排序的表中获取结果：

// Define pageable
    Sort sort = JpaSort.unsafe(Sort.Direction.ASC, Arrays.asList("FIELD(Status, '0','6','1','3','7')"));
    Pageable pageRequest = PageRequest.of(searchCriteria.getPageNo(), searchCriteria.getPageSize(), sort);
    
    // Initialise specification
    Specification<BestPracticeAdminList> spec = Specification.where(null);

    // If search criteria contains searchText field, add it to the specification
    if(searchCriteria.getSearchText() != null)
        spec = spec.and(BestPracticeAdminListFilter.titleLike(searchCriteria.getSearchText()));
    
    // If search criteria contains filters field, inspect filters field
    if(searchCriteria.getFilters() != null) {
        
        // If search criteria contains subjectID field, add it to the specification
        if(searchCriteria.getFilters().getSubjectID() != null)
            spec = spec.and(BestPracticeAdminListFilter.hasAlternativeSubjectID(searchCriteria.getFilters().getSubjectID()));
        
        // If search criteria contains categoryID field, add it to the specification
        if(searchCriteria.getFilters().getCategoryID() != null)
            spec = spec.and(BestPracticeAdminListFilter.hasAlternativeSubjectCategoryID(searchCriteria.getFilters().getCategoryID()));
        
        // If search criteria contains status field, add it to the specification
        if(searchCriteria.getFilters().getStatus() != null)
            spec = spec.and(BestPracticeAdminListFilter.hasStatus(searchCriteria.getFilters().getStatus()));
        
    }
    
    // Get results
    Page<BestPracticeAdminList> pagedResults = bestPracticeAdminListRepository.findAll(spec, pageRequest);

问题是添加自定义排序。 我需要按状态字段排序，它是一个枚举器，但按特定顺序排序，而不是编写枚举器的顺序。

在使用 @Query 和 nativeQuery=true 的存储库上使用本机查询时，自定义排序在不同的情况下工作，但在这种情况下，当我想使用带有过滤规范的 findAll 方法时，自定义排序不适用。

这是实体类：

@Entity
@Table
public class BestPracticeAdminList implements Serializable {

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name = "BestPracticeAdminListID")
private int bestPracticeAdminListID;

@Column(name = "Title")
private String title;

@Column(name = "BestPracticeID")
private int bestPracticeID;

@Column(name = "AlternativeSubjectID")
private int alternativeSubjectID;

@Column(name = "AlternativeSubjectCategoryID")
private int alternativeSubjectCategoryID;

@Column(name = "Status")
@Enumerated(EnumType.ORDINAL)
private LookupReportingProvisionWorkflowStatus status;

@Column(name = "IsPreviousVersionPublished")
private Boolean isPreviousVersionPublished; 

public BestPracticeAdminList() {
}

目前，我得到的例外是：

 org.springframework.data.mapping.PropertyReferenceException: No property fIELD(Status, '0','6','1','3','7') found for type BestPracticeAdminList!

我怎么能解决这个问题？ 谢谢

Answer 1

由于 Spring 抱怨您没有排序字段，您可以通过在@Formula中使用FIELD函数创建它来解决此BestPracticeAdminList

@Formula("FIELD(Status, '0','6','1','3','7')")
private String sortingStatus;

并使用它进行排序

Sort sort = JpaSort.unsafe(Sort.Direction.ASC, Arrays.asList("sortingStatus"));