Question

我正在使用jpa和hibernate。有两个类：

SRTP_KEY_LEN = 16
SRTP_SALT_LEN = 14

def get_srtp_key_salt(src, idx):
    key_start = idx * SRTP_KEY_LEN
    salt_start = 2 * SRTP_KEY_LEN + idx * SRTP_SALT_LEN
    return (
        src[key_start:key_start + SRTP_KEY_LEN] +
        src[salt_start:salt_start + SRTP_SALT_LEN]
    )

material = connection.export_keying_material(
    b'EXTRACTOR-dtls_srtp',
    2 * (SRTP_KEY_LEN + SRTP_SALT_LEN))

if is_server:
    srtp_tx_key = get_srtp_key_salt(material, 1)
    srtp_rx_key = get_srtp_key_salt(material, 0)
else:
    srtp_tx_key = get_srtp_key_salt(material, 0)
    srtp_rx_key = get_srtp_key_salt(material, 1)

当我尝试进行查询时，我得到一个例外。这是我的疑问：

@Entity(name = "constant")
@Table(name = "CONSTANT")
public class Constant extends BaseEntity {

   @Id
   @GeneratedValue(strategy = GenerationType.AUTO, generator = "CONSTANT_SEQ")
   private Long constantID;

   private String title;

}

@Entity(name = "constantValue")
@Table(name = "CONSTANT_VALUE")
public class ConstantValue extends BaseEntity {

   @Id
   @SequenceGenerator(sequenceName = "CONSTANT_VALUE_SEQ", name = "CONSTANT_VALUE_SEQ", allocationSize = 1, initialValue = 1)
   @GeneratedValue(strategy = GenerationType.AUTO, generator = "CONSTANT_VALUE_SEQ")

   private Long constantValueID;

   @ManyToOne(fetch = FetchType.EAGER)
   @JoinColumn(name = "CONSTANT_ID")
   private Constant constant;
}

如何在查询中使用o.constant.title？

这是例外：

     select o from constantValue o where o.constant.constantID=4 and (upper(o.constant.title) like upper(:constant.title) )

在jpa中嵌套查询@ManyToOne

0 个答案: