Question

我有2个班，学生和班级。每个学生只能参加一个班级。

我正在尝试使用以下代码插入学生

            for (Class cls : classList) {
            if (cls.getId().getClassId() == selectedClassId.intValue()) {
                selectedStudent.setClass(cls);
            }
        }
        HibernateUtil.beginTransaction();
        StudentHome stuhome = new StudentHome();
        stuhome.persist(selectedStudent);
        HibernateUtil.commitTransaction();

但由于某些原因，Hibernate没有在Student表中插入classId字段。

以下是我从Hibernate获得的SQL输出

Hibernate: 
select
    class_.ClassId,
    class_.EntityId,
    class_.ClassName as ClassName2_ 
from
    school.class102 class_ 
where
    class_.ClassId=? 
    and class_.EntityId=?

Hibernate: 
insert 
into
    school.student102
    (StudentId, ParentId, RouteId, FirstName, LastName, MiddleName, DoB, DoJ, DoL, Status, RegistrationId, EntityId) 
values
    (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)

Asyou可以看到，insert语句中明显缺少classId。我不确定Hibernate没有在insert语句中包含classId。

EntityId和RegistrationId是学生和学生的PK EntityId和ClassId是Class的PK。两个表中都有（EntityId，ClassId）的外键关系。通过具有预定义EntityId

的视图尝试插入

可能是什么问题？

以下是映射

<hibernate-mapping>
<class name="Student" table="student104" catalog="school">
    <composite-id name="id" class="studentId">
        <key-property name="registrationId" type="int">
            <column name="RegistrationId" />
        </key-property>
        <key-property name="entityId" type="int">
            <column name="EntityId" />
        </key-property>
    </composite-id>
    <many-to-one name="class" class="class" update="false" insert="false" fetch="select">
        <column name="ClassId" />
        <column name="EntityId" not-null="true" />
    </many-to-one>
    <property name="studentId" type="java.lang.Integer">
        <column name="StudentId" />
    </property>
    <property name="parentId" type="java.lang.Integer">
        <column name="ParentId" />
    </property>
    <property name="routeId" type="java.lang.Integer">
        <column name="RouteId" />
    </property>
    <property name="firstName" type="string">
        <column name="FirstName" length="45" />
    </property>
    <property name="lastName" type="string">
        <column name="LastName" length="45" />
    </property>
    <property name="middleName" type="string">
        <column name="MiddleName" length="45" />
    </property>
    <property name="doB" type="date">
        <column name="DoB" length="10" />
    </property>
    <property name="doJ" type="date">
        <column name="DoJ" length="10" />
    </property>
    <property name="doL" type="date">
        <column name="DoL" length="10" />
    </property>
    <property name="status" type="string">
        <column name="Status" length="45" />
    </property>
</class>

<hibernate-mapping>
<class name="class" table="class102" catalog="school">
    <composite-id name="id" class="classId">
        <key-property name="classId" type="int">
            <column name="ClassId" />
        </key-property>
        <key-property name="entityId" type="int">
            <column name="EntityId" />
        </key-property>
    </composite-id>
    <property name="className" type="string">
        <column name="ClassName" length="45" not-null="true" />
    </property>
    <set name="students" table="student104" inverse="true" lazy="true" fetch="select">
        <key>
            <column name="ClassId" />
            <column name="EntityId" not-null="true" />
        </key>
        <one-to-many class="student" />
    </set>
</class>

Answer 1

如果你看看学生的映射

<many-to-one name="class" class="class" update="false" insert="false" fetch="select">
        <column name="ClassId" />
        <column name="EntityId" not-null="true" />
</many-to-one>

你说不要在学生中插入或更新关系。如果您认为一旦分配给某个班级的学生无法修改，那么设置update="false" insert="true"可以稍后修改其他设置，然后设置update="true" insert="true"。默认情况下，这两个属性值（如果您从映射中省略它们）仅{。}}。

hibernate不会在persist上插入外键id的值

1 个答案: