Question

我有部门和员工实体，我想将它们映射为一对多。如何在XML映射中使用List而不是Set？我在Internet上找到了这段代码，但是当执行"from Department"时，错误是"Repeat property department_id in Employee.department"，但是如果我删除了部门中的<index>，它就会是"Unable to read XML"。 Department.hbm.xml

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<!-- Generated Aug 2, 2015 7:14:52 AM by Hibernate Tools 4.3.1 -->
<hibernate-mapping>
    <class name="emp.entity.Department" table="department" schema="dbo" catalog="Emp" optimistic-lock="version">
        <id name="departmentId" type="long">
            <column name="department_id" />
            <generator class="assigned" />
        </id>
        <property name="deptName" type="string">
            <column name="dept_name" length="50" not-null="true" />
        </property>
        <list name="employees" table="Employee" inverse="true" cascade="all" >
            <key>
                <column name="department_id" />
            </key>
            <index column="department_id"></index>
            <one-to-many class="emp.entity.Employee" />
        </list>
    </class>
</hibernate-mapping>

Employee.hbm.xml

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<!-- Generated Aug 2, 2015 7:14:52 AM by Hibernate Tools 4.3.1 -->
<hibernate-mapping>
    <class name="emp.entity.Employee" table="Employee" schema="dbo" catalog="Emp" optimistic-lock="version">
        <id name="employeeId" type="int">
            <column name="employee_id" />
            <generator class="assigned" />
        </id>
        <many-to-one name="department"  class="emp.entity.Department" fetch="select">
            <column name="department_id" />
        </many-to-one>
        <property name="firstname" type="serializable">
            <column name="firstname" />
        </property>
    </class>
</hibernate-mapping>

Answer 1

如上所述，列表中的子项标识符和子项索引必须有两列不同，因此列表定义应为：

<list name="employees" inverse="true" cascade="all" >
        <key>
            <column name="department_id" />
        </key>
        <index column="other_column_to_keep_the_order_of_the_element_in_the_list"></index>
        <one-to-many class="emp.entity.Employee" />
</list>

如果您只需要将元素放在列表中而不需要将其订单保存在数据库中，则可以声明<bag>而不是<list>。

如何在Hibernate XML映射中使用List？

1 个答案: