java JPA/Hibernate 持久化似乎不起作用
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JPA/Hibernate persist does not appear to work
提问by Bart van Heukelom
I'm using JPA (Hibernate implementation) to save objects to the database. Selecting works fine, but for some reason, saving doesn't work. I don't get any errors, but the database doesn't get changed either. This goes for both new entities and existing ones.
我正在使用 JPA（休眠实现）将对象保存到数据库。选择工作正常，但由于某种原因，保存不起作用。我没有收到任何错误，但数据库也没有更改。这适用于新实体和现有实体。
EPayment pay = new EPayment();
pay.setAmount(payment.getAmount());
...
pay.setUserByToUserId(receiver);
CompayDAO.get().save(pay);
CompayDAO.save()
CompayDAO.save()
public void save(Object ent) {
System.out.println("Persisting: " + ent + " using " + this);
this.em.persist(ent);
}
Console output:
控制台输出：
Opening DOA nl.compay.entities.CompayDAO@b124fa
Persisting: nl.compay.entities.EUser@1e2fe5d using nl.compay.entities.CompayDAO@b124fa
Persisting: nl.compay.entities.EUser@30b601 using nl.compay.entities.CompayDAO@b124fa
Persisting: nl.compay.entities.EPayment@ed3b53 using nl.compay.entities.CompayDAO@b124fa
Closing DOA nl.compay.entities.CompayDAO@b124fa
EPayment
电子支付
package nl.compay.entities;
// Generated 21-mei-2009 12:27:07 by Hibernate Tools 3.2.2.GA
import java.util.Date;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import static javax.persistence.GenerationType.IDENTITY;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;
/**
* Payment generated by hbm2java
*/
@Entity
@Table(name = "payment", catalog = "compay")
public class EPayment implements java.io.Serializable {
private static final long serialVersionUID = -2578493336948256566L;
private Integer id;
private EUser userByToUserId;
private EUser userByFromUserId;
private String description;
private float amount;
private String method;
private Date paydate;
public EPayment() {
}
public EPayment(EUser userByToUserId, EUser userByFromUserId, float amount,
Date paydate) {
this.userByToUserId = userByToUserId;
this.userByFromUserId = userByFromUserId;
this.amount = amount;
this.paydate = paydate;
}
public EPayment(EUser userByToUserId, EUser userByFromUserId,
String description, float amount, String method, Date paydate) {
this.userByToUserId = userByToUserId;
this.userByFromUserId = userByFromUserId;
this.description = description;
this.amount = amount;
this.method = method;
this.paydate = paydate;
}
@Id
@GeneratedValue(strategy = IDENTITY)
@Column(name = "id", unique = true, nullable = false)
public Integer getId() {
return this.id;
}
public void setId(Integer id) {
this.id = id;
}
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "to_user_id", nullable = false)
public EUser getUserByToUserId() {
return this.userByToUserId;
}
public void setUserByToUserId(EUser userByToUserId) {
this.userByToUserId = userByToUserId;
}
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "from_user_id", nullable = false)
public EUser getUserByFromUserId() {
return this.userByFromUserId;
}
public void setUserByFromUserId(EUser userByFromUserId) {
this.userByFromUserId = userByFromUserId;
}
@Column(name = "description", length = 1024)
public String getDescription() {
return this.description;
}
public void setDescription(String description) {
this.description = description;
}
@Column(name = "amount", nullable = false, precision = 8)
public float getAmount() {
return this.amount;
}
public void setAmount(float amount) {
this.amount = amount;
}
@Column(name = "method", length = 50)
public String getMethod() {
return this.method;
}
public void setMethod(String method) {
this.method = method;
}
@Temporal(TemporalType.TIMESTAMP)
@Column(name = "paydate", nullable = false, length = 0)
public Date getPaydate() {
return this.paydate;
}
public void setPaydate(Date paydate) {
this.paydate = paydate;
}
}
采纳答案by rudolfson
As Sherkaner mentioned, a save doesn't result in an INSERT or UPDATE directly. You have to flush the session or - better in my opinion - close the unit of work / commit the transaction. You do have transactions?
正如 Sherkaner 所提到的，保存不会直接导致 INSERT 或 UPDATE。您必须刷新会话或 - 在我看来更好 - 关闭工作单元/提交事务。你有交易吗？
回答by Ansh jain
use @Transactionalon your method.....
用@Transactional在你的方法上.....
@Transactional
public void save(Object ent){
.....
.....
}
回答by Khangharoth
Don't think this as bug in Hibernate implementation.This is desired behavior,you would like to have minimum communication with database so Hibernate(or any good ORM framework) will consolidate all your changes and will flush your changes in one go.
不要认为这是 Hibernate 实现中的错误。这是理想的行为，您希望与数据库进行最少的通信，以便 Hibernate（或任何好的 ORM 框架）将整合您的所有更改，并一次性刷新您的更改。
回答by Johan
The program doesn't have to sync with the database right away, have you tried this.em.flush();somewhere?
该程序不必立即与数据库同步，您是否尝试过this.em.flush();？
