java - 在rest API中应用@OrderBy注释而不是排序

Question

我正在尝试在 Rest API 中应用 @OrderBy 注释，我正在尝试按 styleName 对输出进行排序，但是，它不是按styleName 并且仍然按 id 顺序显示所有这些。我正在 postman 中测试这个。我如何错误地实现 @OrderBy 注释。

我用于@OrderBy的导入语句是

import javax.persistence.OrderBy;

我知道我可以创建一个像这样的新查询

@NamedQuery(name = "Styles.findAll", query = "SELECT s FROM Styles s ORDER BY s.name")

但是，我想使用@OrderBy注释

我将@Orderby注释放在Controller方法中，即 Controller

  @GetMapping(produces=MediaType.APPLICATION_JSON_VALUE)
  @OrderBy("Styles.styleName DESC")
    public List<Styles> GetAllStyles() {

        return service.getAllStyles();
    }

是否可以执行类似于服务和 Controller 的操作，而不是过滤数据，而是可以按列名称排序

服务是

  public List<Styles> getAllStyles() {
        EntityManager em = DBUtil.getEMF().createEntityManager();

        List<Styles> list = null;

        try {
        list = em.createNamedQuery("Styles.findAll", Styles.class)
                .getResultList();
        if (list == null || list.isEmpty()) {
            list = null;
        }

    } finally {
        em.close();
    }
    return list;

}

我使用的JPA是

@Entity
@Table(name = "styles")
@NamedQueries({
    @NamedQuery(name = "Styles.findAll", query = "SELECT s FROM Styles s"),
    @NamedQuery(name = "Styles.findById", query = "SELECT s FROM Styles s WHERE s.id = :id"),
    @NamedQuery(name = "Styles.findByCatId", query = "SELECT s FROM Styles s WHERE s.catId = :catId"),
    @NamedQuery(name = "Styles.findByStyleName", query = "SELECT s FROM Styles s WHERE s.styleName = :styleName"),
    @NamedQuery(name = "Styles.findByLastMod", query = "SELECT s FROM Styles s WHERE s.lastMod = :lastMod")})
public class Styles implements Serializable {

    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    @Column(name = "id")
    private Integer id;
    @Basic(optional = false)
    @NotNull
    @Column(name = "cat_id")
    private int catId;
    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 255)
    @Column(name = "style_name")
    private String styleName;
    @Basic(optional = false)
    @NotNull
    @Column(name = "last_mod")
    @Temporal(TemporalType.TIMESTAMP)
    private Date lastMod;

    public Styles() {
    }

    public Styles(Integer id) {
        this.id = id;
    }

    public Styles(Integer id, int catId, String styleName, Date lastMod) {
        this.id = id;
        this.catId = catId;
        this.styleName = styleName;
        this.lastMod = lastMod;
    }

    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public int getCatId() {
        return catId;
    }

    public void setCatId(int catId) {
        this.catId = catId;
    }

    public String getStyleName() {
        return styleName;
    }

    public void setStyleName(String styleName) {
        this.styleName = styleName;
    }

    public Date getLastMod() {
        return lastMod;
    }

    public void setLastMod(Date lastMod) {
        this.lastMod = lastMod;
    }

    @Override
    public int hashCode() {
        int hash = 0;
        hash += (id != null ? id.hashCode() : 0);
        return hash;
    }

    @Override
    public boolean equals(Object object) {
        // TODO: Warning - this method won't work in the case the id fields are not set
        if (!(object instanceof Styles)) {
            return false;
        }
        Styles other = (Styles) object;
        if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
            return false;
        }
        return true;
    }

    @Override
    public String toString() {
        return "Service.Styles[ id=" + id + " ]";
    }

Answer 1

由于您使用的是 javax.persistence.OrderBy，并遵循其文档，因此应该将其应用于 Entity 或 JPA(特别是 Spring)以使其正常工作