在 JPA (Hibernate) 中保存非持久对象 ID
我有一个静态类，例如：
public class Sex{
private final int code;
private final String status;
public static final int TOTAL = 3;
private Sex(int c, String s) {
code = c;
status = s;
}
public static final Sex UNDEFINED = new Sex(1, "UNDEFINED");
public static final Sex MALE = new Sex(2, "MALE");
public static final Sex FEMALE = new Sex(3, "FEMALE");
private static final Sex[] list = { UNDEFINED, MALE, FEMALE };
public int getCode() {
return code;
}
public String getStatus() {
return status;
}
public static Sex fromInt(int c) {
if (c < 1 || c > TOTAL)
throw new RuntimeException("Unknown code for fromInt in Sex");
return list[c-1];
}
public static List getSexList() {
return Arrays.asList(list);
}
}
而且，我有一个实体类
@Entity
@Table(name="person")
public class Person{
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private long id;// setter/getter omitted
private Sex sex;
public final Sex getSex() {
return this.sex;
}
public final void setSex(final Sex argSex) {
this.sex = argSex;
}
}
，我想将 sex_id 保存在数据库的 person 表中。但 setter/getter 应该按照指定的那样，因为我想将代码编写为 -
Person person = new Person();
person.setSex(Sex.MALE);
Dao.savePerson(person);
How to annotate Sex with JPA?
I've a static class like:
public class Sex{
private final int code;
private final String status;
public static final int TOTAL = 3;
private Sex(int c, String s) {
code = c;
status = s;
}
public static final Sex UNDEFINED = new Sex(1, "UNDEFINED");
public static final Sex MALE = new Sex(2, "MALE");
public static final Sex FEMALE = new Sex(3, "FEMALE");
private static final Sex[] list = { UNDEFINED, MALE, FEMALE };
public int getCode() {
return code;
}
public String getStatus() {
return status;
}
public static Sex fromInt(int c) {
if (c < 1 || c > TOTAL)
throw new RuntimeException("Unknown code for fromInt in Sex");
return list[c-1];
}
public static List getSexList() {
return Arrays.asList(list);
}
}
And, I've an entity class
@Entity
@Table(name="person")
public class Person{
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private long id;// setter/getter omitted
private Sex sex;
public final Sex getSex() {
return this.sex;
}
public final void setSex(final Sex argSex) {
this.sex = argSex;
}
}
I want to save sex_id in database in person table. But setter/getter should be as specified, because I want write my code as -
Person person = new Person();
person.setSex(Sex.MALE);
Dao.savePerson(person);
How to annotate Sex with JPA?
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评论（2）
为什么不使用枚举器 Sex 来代替，然后使用
@Enumerated来代替？Why not to use enumerator Sex instead and then use
@Enumeratedinstead?既然您不想在数据库中创建新的
Sex实例，为什么不使用enum而不是class来Sex实例？代码>性别？如果这样做，您只需在
Person类中使用@Enumerated(EnumType.STRING)注释您的Sex属性即可。完整示例此处（或者直接 Google 一下，您会发现很多）Since you don't want to create new
Sexinstance in your database, why don't you use anenuminstead of aclassforSex?If you do so, you will just have to annotate you
Sexattribute with@Enumerated(EnumType.STRING)in thePersonclass. Full example here (or just Google it and you will find plenty)