Hibernate:必须在调用 save() 之前手动分配此类的 id
Posted
技术标签:
【中文标题】Hibernate:必须在调用 save() 之前手动分配此类的 id【英文标题】:Hibernate: ids for this class must be manually assigned before calling save() 【发布时间】:2013-04-15 01:22:50 【问题描述】:我在使用 Hibernate 和 oneToMany
映射时遇到了一些问题。
这是我的功能:
Location location = new Location();
location.setDateTime(new Date());
location.setLatitude(lat);
location.setLongitude(lon);
location = this.locationDao.save(location);
merchant = new Merchant();
merchant.setAddress(address);
merchant.setCity(city);
merchant.setCountry(country);
merchant.setLocation(location);
merchant.setName(name);
merchant.setOrganization(organization);
merchant.setPublicId(publicId);
merchant.setZipCode(zipCode);
merchant.setApplication(this.applicationDAO.findByPublicId(applicationPublicId));
merchant = this.merchantDao.save(merchant);
return merchant;
这是我的两个实体:
位置
import java.io.Serializable;
import java.util.Date;
import java.util.List;
import javax.persistence.*;
import javax.validation.constraints.NotNull;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlTransient;
@Entity
@Table(name = "location")
@XmlRootElement
public class Location implements Serializable
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Basic(optional = false)
@Column(name = "id", nullable = false)
private Long id;
@Basic(optional = false)
@NotNull
@Column(name = "latitude", nullable = false)
private double latitude;
@Basic(optional = false)
@NotNull
@Column(name = "longitude", nullable = false)
private double longitude;
@Basic(optional = false)
@NotNull
@Column(name = "date_time", nullable = false)
@Temporal(TemporalType.TIMESTAMP)
private Date dateTime;
@OneToMany(cascade = CascadeType.ALL, mappedBy = "location")
private List<Merchant> retailerList;
@OneToMany(cascade = CascadeType.ALL, mappedBy = "location")
private List<MobileUser> userList;
public Location()
public Location(Long id)
this.id = id;
public Location(Long id, double latitude, double longitude, Date dateTime)
this.id = id;
this.latitude = latitude;
this.longitude = longitude;
this.dateTime = dateTime;
public Long getId()
return id;
//getters and setters
@XmlTransient
public List<Merchant> getRetailerList()
return retailerList;
public void setRetailerList(List<Merchant> retailerList)
this.retailerList = retailerList;
@XmlTransient
public List<MobileUser> getUserList()
return userList;
public void setUserList(List<MobileUser> userList)
this.userList = userList;
@Override
public int hashCode()
int hash = 0;
hash += (id != null ? id.hashCode() : 0);
return hash;
@Override
public boolean equals(Object object)
// TODO: Warning - this method won't work in the case the id fields are not set
if (!(object instanceof Location))
return false;
Location other = (Location) object;
if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id)))
return false;
return true;
@Override
public String toString()
return "com.fgsecure.geoloc.entities.Location[ id=" + id + " ]";
商家
import java.io.Serializable;
import javax.persistence.*;
import javax.validation.constraints.NotNull;
import javax.validation.constraints.Size;
import javax.xml.bind.annotation.XmlRootElement;
import org.codehaus.jackson.annotate.JsonAutoDetect;
import org.codehaus.jackson.annotate.JsonIgnore;
@JsonAutoDetect
@Entity
@Table(name = "retailer")
@XmlRootElement
public class Merchant implements Serializable
private static final long serialVersionUID = 1L;
@JsonIgnore
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Basic(optional = false)
@Column(name = "id", nullable = false)
private Long id;
@Basic(optional = false)
@NotNull
@Size(min = 1, max = 50)
@Column(name = "public_id", nullable = false, length = 50)
private String publicId;
@Basic(optional = false)
@NotNull
@Size(min = 1, max = 50)
@Column(name = "name", nullable = false, length = 50)
private String name;
@Size(max = 50)
@Column(name = "organization", length = 50)
private String organization;
@Size(max = 128)
@Column(name = "address", length = 128)
private String address;
@Size(max = 10)
@Column(name = "zip_code", length = 10)
private String zipCode;
@Size(max = 50)
@Column(name = "city", length = 50)
private String city;
@Size(max = 50)
@Column(name = "country", length = 50)
private String country;
@JoinColumn(name = "location_id", referencedColumnName = "id", nullable = false)
@ManyToOne(optional = false)
private Location location;
@JoinColumn(name = "application_id", referencedColumnName = "id", nullable = false)
@ManyToOne(optional = false)
private Application application;
public Merchant()
public Merchant(Long id)
this.id = id;
public Merchant(Long id, String publicId, String name)
this.id = id;
this.publicId = publicId;
this.name = name;
//getters and setters
@Override
public int hashCode()
int hash = 0;
hash += (id != null ? id.hashCode() : 0);
return hash;
@Override
public boolean equals(Object object)
// TODO: Warning - this method won't work in the case the id fields are not set
if (!(object instanceof Merchant))
return false;
Merchant other = (Merchant) object;
if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id)))
return false;
return true;
@Override
public String toString()
return "com.fgsecure.geoloc.entities.Retailer[ id=" + id + " ]";
每次我调用我的函数时,我都会得到:
org.hibernate.id.IdentifierGenerationException: ids for this class must be manually assigned before calling save():
对于这一行:
location = this.locationDao.save(location);
我不明白为什么。 因为在这种情况下,无需手动分配位置 id,因为它是自动生成的。 我一定是做错了什么,但看了 3 天后我没有找到。
我正在使用 PostGreSQL,并且 id 是具有以下序列的自动生成的 ID:
nextval('location_id_seq'::regclass)`
编辑求解:
好的,谢谢您的回答。
如果以后可以帮助到其他人,只需要修改id的参数:
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Basic(optional = false)
@Column(name = "id",unique=true, nullable = false)
【问题讨论】:
你使用的是哪个数据库 商家为您正常工作 尝试将id算法名称更改为另一个进行测试 尝试将类型从 IDENTITY 更改为另一种仅用于测试目的 不应将您的问题标记为已解决并将解决方案添加到您的问题中,而应将您的解决方案添加为答案并接受该答案。 【参考方案1】:Assign primary key in hibernate
确保该属性是数据库中的主键和Auto Incrementable。然后将其映射到数据类中,并使用 IDENTITY 使用带有 @GeneratedValue
注释的注释。
@Entity
@Table(name = "client")
data class Client(
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id")
private val id: Int? = null
)
GL
Source
【讨论】:
【参考方案2】:您的 id 属性未设置。这可能是因为 DB 字段未设置为自动递增?你用的是什么数据库? mysql?您的字段是否设置为 AUTO INCREMENT?
【讨论】:
感谢您的回答。我正在使用带有自动生成 ID 的 PostGre。我刚刚编辑了我的帖子。 对于 MySQL 和 AI,这是可行的。您可能想检查 postgresql 是否也是这种情况。我的 pojo 几乎总是这样:@Id @GeneratedValue(strategy = GenerationType.IDENTITY) @Column(name="id", unique=true, nullable=false) public Long getId() return this.id; 在 oracle 中这可能是由于序列名称引起的以上是关于Hibernate:必须在调用 save() 之前手动分配此类的 id的主要内容,如果未能解决你的问题,请参考以下文章
Hibernate Session中的save(),update(),delete(),saveOrUpdate() 细粒度分析
HIbernate - 对象引用未保存的瞬态实例 - 在刷新之前保存瞬态实例
hibernate 对象的持久化状态。 在session.save()之后 commit()之前 数据库有对应的记录吗?
来自其他包的 Api 出现错误“必须在调用 save() 之前手动分配此类的 ID”