Java JPA/HibernatePersist似乎不起作用

我正在使用JPA（Hibernate实现）将对象保存到数据库中。选择很好，但由于某些原因，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（）

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

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;
    }

}

---

该程序不必立即与数据库同步，您是否尝试过

this.em.flush（）某处？
正如Sherkaner提到的，保存不会直接导致插入或更新。您必须刷新会话，或者——在我看来更好——关闭工作单元/提交事务。您确实有事务吗？
不要认为这是Hibernate实现中的错误。这是需要的行为，您希望与数据库的通信最少，这样Hibernate（或任何好的ORM框架）将整合所有更改并一次性刷新您的更改。
在您的方法上使用
@Transactional

@Transactional
public void save(Object ent){
.....
.....
}

没有，但我有。关闭（）。这是否也意味着刷新，还是意味着我要扔掉东西？em.close（）不会自动将东西刷新到数据库中。更好地使用事务（我同意rudolfson）或em.flush（）嗯，我不是JPA专家（只是使用Hibernate+JPA Annotations+Spring），但根据EntityManager的javadoc of the close（）方法，事务仍然是活动的。看见因此，您仍然必须提交事务（例如，通过调用em.getTransaction（）.commit（），但我建议使用声明式事务管理。我从来没有这样做过。这是我的一个错误；）我面临相同的问题：Hibernate+JPA+Spring。在日志中生成Insert查询（指示flush（）等已就位）。尝试了一切，包括在web上提交事务和其他建议。我不知道如何，但一个表删除和重新创建在数据库级别的工作对我来说。
@Transactional
public void save(Object ent){
.....
.....
}