Question

我对hibernate很新，我正在尝试在PersonDAO中设置一个新方法。

我的hibernate映射文件如下所示：

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">

<hibernate-mapping package="com.foo.bar.domain">
<class name="Person" table="person">
    <meta attribute="class-description">A Person</meta>
    <id name="id" type="java.lang.Long" column="rid" unsaved-value="null">
        <generator class="native" />
    </id>


    <version name="version" type="integer" column="rversion" unsaved-value="null" />
    <property name="UID"           type="string"  column="UID" length="16" not-null="true" unique="true"/>
    <property name="lastName"         type="string"  column="last_name"  not-null="true" />
    <property name="firstName"        type="string"  column="first_name" not-null="true" />
    <property name="ownDepartment"    type="string"  column="own_department"/>

    <!-- a person has many responsibilities and a responsibility can can assigned to many person -->
    <set name="responsibilities" table="person_responsibility">
        <key column="person_id"/>
        <many-to-many column="responsibility_id" class="Responsibility"/>
    </set>

    <set name="additionalDepartments" table="PERSON_TO_ADDL_DEPARTMENT">    
        <key column="person_id"/>               
        <element column="ADDITIONAL_DEPARTMENT" type="string"/>         
    </set>                                              
</class>

我在java中编写了一个这样的方法来获取给定部门的所有管理器：

public List<Person> getManagerByDepartment(final String givenDepartment){
    List<Person> l = (List<Person>) this.getHibernateTemplate().executeFind(new HibernateCallback<List<Person>>() {

    public List<Person> doInHibernate(Session session) throws HibernateException, SQLException {
        String query = "select p from Person p join p.responsibilities responsibilities join p.additionalDepartments additionalDepartments where responsibilities.name = 'manager' and (p.ownDepartment = :givenDepartment or additionalDepartments = :givenDepartment)";
        List<Person> result = (List<Person>) session.createQuery(query).setString("givenDepartment", givenDepartment).list();
        return result;  
        }
    });

    return l;
}

现在我在SQL中进行手动查询，我可以看到，对于给定的部门，肯定不止一个人有额外的责任'经理'...为什么我的方法只返回一个人，而不是全部？

我强烈怀疑我的方法，特别是我的查询，而不是映射，是问题，但我看不出它有什么问题......

我已经跳到了这里的深处，所以任何帮助都会非常感激。

编辑：注意，我正在处理数百条记录，而不是数百万条记录，这不是一个瓶颈操作，所以我不太担心性能...如果我正在做的话一些毫无意义的浪费，指出来

Answer 1

如果没有示例数据，很难确定，但是当您在HQL中执行join时，它会在SQL中转换为inner join。因此，如果您知道应该有多个具有给定责任的结果，则问题可能是join p.additionalDepartments。

对于其他部门，请使用left join尝试此查询并查看其是否有效

String query = "select p from Person p join p.responsibilities responsibilities left join p.additionalDepartments additionalDepartments where responsibilities.name = 'manager' and (p.ownDepartment = :givenDepartment or additionalDepartments = :givenDepartment)";

Answer 2

您可以通过启用showsql选项打印hibernate查询并检查创建的查询，然后针对数据库对其进行测试。

初学者休眠：为什么我的hibernate查询只返回一个结果，而不是列表？

2 个答案: