JPA：在使用findAll和規范時如何按欄位自定義順序排序

我正在將 Spring Tool Suite 4 (4.12.0.RELEASE) 與 Java 8 一起使用，我有以下代碼可以從過濾、分頁和排序的表中獲取結果：

// 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);

問題是添加自定義排序。我需要按狀態欄位排序，它是一個列舉器，但按特定順序排序，而不是撰寫列舉器的順序。

在使用 @Query 和 nativeQuery=true 的存盤庫上使用本機查詢時，自定義排序在不同的情況下作業，但在這種情況下，當我想使用帶有過濾規范的 findAll 方法時，自定義排序不適用。

這是物體類：

@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!

我怎么能解決這個問題？謝謝

uj5u.com熱心網友回復：

由于 Spring 抱怨您沒有排序欄位，您可以通過使用FIELDa@Formula中的函式創建它來解決此問題BestPracticeAdminList

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

并使用它進行排序

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

標籤：爪哇 排序 jpa

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