设置Id（PK）生成值自动和手动 [英] set Id (PK) generation value auto and manual

问题描述

我想将用户保存到DB中，并使用IDENTITY生成类型创建的用户的ID（PK）当前场景。
eg

  @Entity 
 @Table（name =USER_PROFILES，uniqueConstraints = @UniqueConstraint（columnNames =USERNAME））
 public class UserProfiles实现java.io.Serializable {
 private Long id; 
私人字符串用户名; 
私人字符串密码; 
 
 
 
 public UserProfiles（）{
} 
 
 
 
 @Id 
 @GeneratedValue （strategy = GenerationType.IDENTITY）
 @Column（name =ID，unique = true，nullable = false，precision = 20，scale = 0）
 public Long getId（）{
返回this.id; 
} 
 
 public void setId（Long id）{
 this.id = id; 
} 
 
 @Column（name =USERNAME，unique = true，nullable = false，length = 32）
 public String getUsername（）{
 return this 。用户名; 
} 
 
 public void setUsername（String username）{
 this.username = username; 
 
 
 @Column（name =PASSWORD，nullable = false，length = 32）
 public String getPassword（）{
 return this.password; 
} 
 
 public void setPassword（String password）{
 this.password = password; 
} 
 
} 
  

但是我想<$ c在以下情况下$ c> Create Id（PK）：
1）用户明确设置 Id（PK）。
2）如果用户没有设置 Id（PK），那么它会自动分配，并且它必须是唯一的。

请给我一些可用的选项，以便我可以解决它。
Thanks。

解决方案

您可以为此定义自定义ID生成器，如 SO Answer

以下是代码的外观： -

  @Id 
 @Basic（可选= false）
 @GeneratedValue（strategy = GenerationType.IDENTITY，generator = IdOrGenerated）
 @GenericGenerator（name =IdOrGenerated，strategy =... UseIdOrGenerate）
 @Column（name =ID，unique = true，nullable = false，precision = 20 ，scale = 0）
 public Long getId（）{..} 
  

和 p>

  public class UseIdOrGenerate extends IdentityGenerator {
 @Override 
 public Serializable generate（SessionImplementor session，Object obj）throws HibernateException {
 if（obj == null）抛出新的HibernateException（new NullPointerException（））; $（）（）（）（）（） ）; 
 return id; 
} else {
 return（（EntityWithId）obj）.getId（）; // id不为空，所以使用指定的id。 
 
 
 
 
 code $ pre 
I want to persist User into DB and current scenario of ID(PK) of User created with IDENTITY generation type.
e.g.
@Entity
@Table(name = "USER_PROFILES", uniqueConstraints = @UniqueConstraint(columnNames = "USERNAME"))
public class UserProfiles implements java.io.Serializable {
private Long id;
private String username;
private String password;



public UserProfiles() {
}



@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name = "ID", unique = true, nullable = false, precision = 20, scale = 0)
public Long getId() {
    return this.id;
}

public void setId(Long id) {
    this.id = id;
}

@Column(name = "USERNAME", unique = true, nullable = false, length = 32)
public String getUsername() {
    return this.username;
}

public void setUsername(String username) {
    this.username = username;
}

@Column(name = "PASSWORD", nullable = false, length = 32)
public String getPassword() {
    return this.password;
}

public void setPassword(String password) {
    this.password = password;
}

}
but I want to Create Id(PK) in following scenarios :
1) User sets Id(PK) explicitly.
2) If User does not set Id(PK) then it is be assigned automatically and it must be unique.

Please suggest me some available options so I can resolve it.
Thanks.
解决方案
You can define your custom id generator for this purpose as pointed out in this SO Answer     

Here is how its code will look like:-  
@Id
@Basic(optional = false)
@GeneratedValue(strategy=GenerationType.IDENTITY, generator="IdOrGenerated")
@GenericGenerator(name="IdOrGenerated",strategy="....UseIdOrGenerate")
@Column(name = "ID", unique = true, nullable = false, precision = 20, scale = 0)
public Long getId(){..}
and
  public class UseIdOrGenerate extends IdentityGenerator {    
    @Override
    public Serializable generate(SessionImplementor session, Object obj) throws HibernateException {
        if (obj == null) throw new HibernateException(new NullPointerException()) ;

        if ((((EntityWithId) obj).getId()) == null) {//id is null it means generate ID
            Serializable id = super.generate(session, obj) ;
            return id;
        } else {
            return ((EntityWithId) obj).getId();//id is not null so using assigned id.

        }
    }
}


                        
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