我正在尝试使用Hibernate执行以下操作:
让我解释一下。
这是枚举:
public enum Position {
UNSPECIFIED(1L, "Unspecified"),
SPECIALIST(2L, "Specialist"),
NOT_SPECIALIST(3L, "Not Specialist");
private Long id;
private String name;
Position(Long id, String name) {
this.id = id;
this.name = name;
}
public Long getId() {
return id;
}
public String getName() {
return name;
}
public static Position from(Long id) {
for(Position position: values()) {
if(position.getId().equals(id))
return position;
}
throw new IllegalArgumentException("Cannot get position for ID " + id);
}
public static Position from(String name) {
for(Position position: values()) {
if(position.getName().toUpperCase().equals(name.toUpperCase()))
return position;
}
throw new IllegalArgumentException("No such position " + name);
}
}
这是一个使用枚举的类:
@Entity(name = "worker")
public class Worker {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
private String firstName;
private String lastName;
private Position position;
private String email;
}
这是数据在数据库中的显示方式:
worker
id first_name last_name position email
0 Adam Applegate Unspecified adam@company.com
1 Bob Barrett Specialist bob@company.com
所以基本上,当我调用workerRepository.findById(0);
时,我得到的工作对象具有以下值:
id --> 0
firstName --> Adam
lastName --> Applegate
position --> Position.UNSPECIFIED
email --> adam@company.com
现在,假设我使用以下值创建一个新的worker对象:
firstName --> Chad
lastName --> Carlton
position --> Position.NOT_SPECIFIED
email --> chad@company.com
调用workerRepository.save(newWorker);
后,数据库应如下所示:
id first_name last_name position email
0 Adam Applegate Unspecified adam@company.com
1 Bob Barrett Specialist bob@company.com
2 Chad Carlton Not Specialist chad@company.com
请注意,position列的值为Position.NOT_SPECIALIST.getName()。
有没有办法在休眠中执行此操作?
答案 0 :(得分:1)
我建议避免使用AttributeConverter
,然后选择简单的@Enumerated(STRING)
。
显然,这只适用于您可以站立大写并强调位置值的情况。
public enum Position
{
UNSPECIFIED,
SPECIALIST,
NOT_SPECIALIST;
}
@Entity(name = "worker")
public class Worker
{
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
private String firstName;
private String lastName;
@NotNull
@Enumerated(STRING)
@Column(nullable = false)
private Position position = Position.UNSPECIFIED;
private String email;
}
worker
id first_name last_name position email
0 Adam Applegate UNSPECIFIED adam@company.com
1 Bob Barrett SPECIALIST bob@company.com
当你已经Position.getId()
时,你还需要Position.ordinal()
,这也不清楚:Position.from(id)
相当于Position.values()[id]
。
最后,声明"标签"不是一个好习惯。在枚举中硬编码(通常在编译代码中):只需使用ResourceBundle
,您就可以为i18n做好准备。
答案 1 :(得分:0)
正如@chrylis建议的那样,我的问题的答案是使用AttributeConverter
,如下所示:
import javax.persistence.AttributeConverter;
import javax.persistence.Converter;
@Converter(autoApply = true)
public class PositionConverter implements AttributeConverter<Position, String> {
@Override
public String convertToDatabaseColumn(Position attribute) {
return attribute.getName();
}
@Override
public VolleyballPlayerPosition convertToEntityAttribute(String dbData) {
return Position.from(dbData);
}
}