带有postgresql投掷的春季应用程序“ psqlexception：错误：不存在运算符：cartare = uuid;

Question

我有一个看起来像这样的实体：

import lombok.*;
import org.hibernate.annotations.GenericGenerator;
import org.hibernate.annotations.Type;

import javax.persistence.*;
import java.util.List;
import java.util.UUID;

@Entity(name = "EnterpriseGroup")
@Data
@EqualsAndHashCode
@AllArgsConstructor
@NoArgsConstructor
@Builder
@Table(name = "enterprise_groups")
public class EnterpriseGroup {

    @Id
    @GeneratedValue(generator = "uuid4")
    @GenericGenerator(name = "UUID", strategy = "uuid4")
    @Type(type = "pg-uuid")
    @Column(columnDefinition = "CHAR(36)")
    private UUID groupId;

    @Column(unique = true)
    private String name;

    private String description;

    @ManyToMany(fetch = FetchType.EAGER, cascade = {CascadeType.PERSIST})
    @JoinTable(
            name = "groups_roles",
            joinColumns = {@JoinColumn(name = "groupId")},
            inverseJoinColumns = {@JoinColumn(name = "roleId")})
    private List<UserRole> roles;
}

我有一个默认的JPA存储库：

import com.fadata.security.domain.entities.EnterpriseGroup;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;

import java.util.UUID;

@Repository
public interface EnterpriseGroupRepository extends JpaRepository<EnterpriseGroup, UUID> {
}

当我尝试调用 repository.findall（）方法时，我会得到以下例外：

o.h.engine.jdbc.spi.SqlExceptionHelper: could not extract ResultSet [n/a]
org.postgresql.util.PSQLException: ERROR: operator does not exist: character = uuid

Hint: No operator matches the given name and argument types. You might need to add explicit type casts.

我尝试更改UUID类型和发电策略，列定义和名称，到目前为止什么都没有。我尝试通过在存储库方法上放置断点来调试它，但是它永远不会受到打击，这意味着在该方法击中该方法之前有某种验证，这也是此问题的起源。我敢肯定，我通过有效的UUID，因为如果我通过无效的UUID格式，就会抛出适当的例外。然而，我得到的例外使我认为我通过请求转换并击中实际应用程序的UUID的方式存在某种问题。欢迎任何想法！谢谢！

Answer 1

尝试一下 uuidchartype

@Id
@Type(type = "org.hibernate.type.UUIDCharType")
@GeneratedValue(generator = "UUID")
@GenericGenerator(
        name = "UUID",
        strategy = "org.hibernate.id.UUIDGenerator"
)
@Column(name = "groupId", updatable = false, nullable = false)
private UUID groupId;

注意：不需要通过uuid.randomuuid（）