JPA Criteria order by enum set

Question

我想使用JPA Criteria

按枚举顺序排序

这是我的枚举：

    public enum UserRole {
    SYS_ADMIN,
    CALIBRATOR_ADMIN,
    PROVIDER_ADMIN,
    }

实体：

@Entity
@NoArgsConstructor(access = AccessLevel.PROTECTED)
@Table(name = "USER")
public class User {

    @Id
    private String username;
    private String password;
    private String firstName;
    private String lastName;

@Embedded
    private Address address;

    @ManyToOne
    @JoinColumn(name = "organizationId")
    @JsonManagedReference
    private Organization organization;

    @ElementCollection
    @JoinTable(name = "USER_ROLE", joinColumns = @JoinColumn(name = "username"))
    @Enumerated(EnumType.STRING)
    @Column(name = "value", length = 30)
    private Set<UserRole> userRoles = new HashSet<>();

以及我如何尝试进行搜索查询：

public static CriteriaQuery<User> buildSearchQuery(String userName, String role, String firstName, String lastName, EntityManager em, String fieldToSort) {

        CriteriaBuilder cb = em.getCriteriaBuilder();
        CriteriaQuery<User> criteriaQuery = cb.createQuery(User.class);
        Root<User> root = criteriaQuery.from(User.class);

        Predicate predicate = ProviderEmployeeQuary.buildPredicate(root, cb, userName,role, firstName, lastName);

if (fieldToSort.length() > 0) {
            if (fieldToSort.substring(0, 1).equals("-")) {
                if (fieldToSort.substring(1, fieldToSort.length()).equals("userRoles")) {   
                    criteriaQuery.orderBy(cb.desc(root.<Set<UserRole>>get(fieldToSort.substring(1, fieldToSort.length()))));
                } else {
                    criteriaQuery.orderBy(cb.desc(root.get(fieldToSort.substring(1, fieldToSort.length()))));
                }
            } else {
                if (fieldToSort.equals("userRoles")) {

                    criteriaQuery.orderBy(cb.asc(root.<Set<UserRole>>get(fieldToSort)));

                } else {
                    criteriaQuery.orderBy(cb.asc(root.get(fieldToSort)));
                }
            }
        }
        criteriaQuery.select(root).distinct(true);
        criteriaQuery.where(predicate);
        return criteriaQuery;
    }

执行此代码时出错：

  

com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException：你有一个   SQL语法错误;查看与您的手册相对应的手册   MySQL服务器版本的正确语法使用附近＆＃39; asc limit 5＆＃39;在   第1行

生成的JPQL查询：

  

从User选择distinct generatedAlias0作为generatedAlias0 inner   将generatedAlias0.organization加入generateAlias1其中（   generatedAlias1.id = 11L）和（1 = 1）order by   generatedAlias0.userRoles desc