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Hibernate不会保存到数据库中

save mapping hibernate database java

AdrianS · 技术社区 · 16 年前

实体

public class Profesor implements Comparable<Profesor> {
private int id;
private String nume;
private String prenume;
private int departament_id;
private Set<Disciplina> listaDiscipline; //the teacher gives some courses}

public class Disciplina implements Comparable<Disciplina>{ //the course class

private int id;
private String denumire;
private String syllabus;
private String schNotare;
private Set<Lectie> lectii;
private Set<Tema> listaTeme;
private Set<Grup> listaGrupuri; // the course gets teached/assigmened to some groups of students
private Set<Assignment> listAssignments;}

映射

<hibernate-mapping default-cascade="all">
<class name="model.Profesor" table="devgar_scoala.profesori">
<id name="id" column="id">
 <generator class="increment"/>
</id>
<set name="listaDiscipline" table="devgar_scoala.`profesori-discipline`">
<key column="Profesori_id" />
<many-to-many class="model.Disciplina" column="Discipline_id"  />
</set>
<property name="nume" column="Nume" type="string" />
<property name="prenume" column="Prenume" type="string" />
<property name="departament_id" column="Departamente_id" type="integer" />
</class>

<class name="model.Grup" table="devgar_scoala.grupe">
<id name="id" unsaved-value="0">
<generator class="increment"/>
</id>
<set name="listaStudenti" table="devgar_scoala.`studenti-grupe`">
<key column="Grupe_id" />
<many-to-many column="Studenti_nrMatricol" class="model.Student" />
</set>

<property name="nume" column="Grupa" type="string"/>
<property name="programStudiu" column="progStudii_id" type="integer" />
</class>
<class name="model.Disciplina" table="devgar_scoala.discipline" >
<id name="id"  >
<generator class="increment"/>
</id> 
<property name="denumire" column="Denumire" type="string"/>
<property name="syllabus" type="string" column="Syllabus"/>
<property name="schNotare" type="string" column="SchemaNotare"/>

<set name="listaGrupuri" table="devgar_scoala.`Discipline-Grupe`">
<key column="Discipline_id" />
<many-to-many column="Grupe_id" class="model.Grup" />
</set>

<set name="lectii" table="devgar_scoala.lectii">
<key column="Discipline_id" not-null="true"/>
<one-to-many class="model.Lectie"  />
</set>
</class>

p-手动加载的Profesor对象

Profesor p2 = (Profesor) session.merge(p);
session.saveOrUpdate(p2);


//flush session of course

Hibernate: insert into devgar_scoala.grupe (Grupa, progStudii_id, id) values (?, ?, ?)

但是当我查看数据库时,Grupe表(Groups表)中没有新行

4 回复 | 直到 16 年前

Barthelemy 16 年前

正如在评论中提到的,我的猜测是您没有提交更改。你可以试试这样的方法:

Transaction tx = null;
try {
    tx = session.beginTransaction()
    Profesor p2 = (Profesor) session.merge(p);
    session.saveOrUpdate(p2);
    tx.commit();
} catch(Exception e) {
    tx.rollback();
}

您还可以在配置文件中使用自动提交模式,以避免手动提交更改。在中查找“hibernate.connection.autocommit”属性 Hibernate reference . 我不认为所有的数据库都支持autocommit。

Poso Lingofe 15 年前

将此代码段添加到hibernate配置中:

<property name = "current_session_context_class">thread</property>

使用以下代码:

Session session = factory.getCurrentSession();
Transaction tx = session.beginTransaction();

Profesor p2 = (Profesor) session.merge(p);
session.saveOrUpdate(p2);

tx.commit();

Pascal Thivent 16 年前

确保在事务中运行代码(最后提交):

Session session = factory.openSession();
Transaction tx = session.beginTransaction();

Profesor p2 = (Profesor) session.merge(p);
session.saveOrUpdate(p2);

tx.commit();
session.close();

Sebastien Lorber 16 年前

如果它不能处理事务,那是因为事务是在mysql(innoDB引擎)上激活的,所以您只需在mysql服务器上提交或禁用事务。。。

但不管怎样,保持交易不是个坏主意不是吗???