3 回答

TA貢獻(xiàn)1802條經(jīng)驗(yàn) 獲得超10個(gè)贊
您需要像這樣設(shè)置適當(dāng)?shù)?Hibernate?Generator
@Id
@GeneratedValue(generator = “UUID”)
@GenericGenerator(
? ? name = “UUID”,
? ? strategy = “org.hibernate.id.UUIDGenerator”,
)
private UUID id;

TA貢獻(xiàn)1872條經(jīng)驗(yàn) 獲得超4個(gè)贊
import java.util.UUID;
@Table(name="address_book")
@Entity??
public class AddressBook {??
@Id
@GeneratedValue(generator = "uuid2")
@GenericGenerator(name = "uuid2", strategy = "uuid2")
@Column(name = "name", updatable = false, nullable = false, columnDefinition = "VARCHAR(36)")
@Type(type = "uuid-char")
private String name;
.
.
.
}
這肯定會(huì)奏效。

TA貢獻(xiàn)1860條經(jīng)驗(yàn) 獲得超8個(gè)贊
你真的是說BLOB嗎?一個(gè) UUID 是 128 位,當(dāng)然不符合 LOB 列的條件。例如,您可以在 MySQL 中使用 Binary(16) 列。
現(xiàn)在回答你的問題,你需要告訴hibernate如何生成ID值。
@Id
@GeneratedValue(generator = "hibernate-uuid")
@GenericGenerator(name = "hibernate-uuid", strategy = "uuid2")
@Column(name = "id", columnDefinition = "BINARY(16)")
protected UUID id;
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