Question

Employee.java

public class Employee
{
String id;
String name;
List<String> designations;
List<String> qualifications;
}

Employee.hbm.xml

<hibernate-mapping>
  <class name="com.novelty.Employee" table="Employee">
    <id column="ID" name="id" type="long">
      <generator class="increment"/>
    </id>
    <property column="name" name="name"/>

    <list lazy="false" name="designations" table="designations">
      <key column="ID"/>
      <list-index column="idx"/>
      <element column="designation" type="string"/>
    </list>

   <list lazy="false" name="qualifications" table="qualifications">
      <key column="ID"/>
      <list-index column="idx"/>
      <element column="qualification" type="string"/>
    </list>
</hibernate-mapping>

我需要获取特定员工的名称列表（我有名字或ID）。我不想将Employee对象作为一个整体获取并获取列表。我尝试了预测，但徒劳无功。

Criteria criteria = session.createCriteria(c);
ProjectionList proList = Projections.projectionList();
proList.add(Projections.property("designations"));
criteria.setProjection(proList);
List list = criteria.list();

我得到了以下异常堆栈跟踪。

Caused by: java.lang.ArrayIndexOutOfBoundsException: 0
    at org.hibernate.loader.criteria.CriteriaLoader.getResultColumnOrRow(CriteriaLoader.java:148)
    at org.hibernate.loader.Loader.getRowFromResultSet(Loader.java:639)
    at org.hibernate.loader.Loader.doQuery(Loader.java:829)
    at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:274)
    at org.hibernate.loader.Loader.doList(Loader.java:2533)
    at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2276)
    at org.hibernate.loader.Loader.list(Loader.java:2271)
    at org.hibernate.loader.criteria.CriteriaLoader.list(CriteriaLoader.java:119)
    at org.hibernate.impl.SessionImpl.list(SessionImpl.java:1716)
    at org.hibernate.impl.CriteriaImpl.list(CriteriaImpl.java:347)

Answer 1

由于designations是一个列表，您应该join

    try {
            Criteria criteria = session.createCriteria(Employee.class);
            criteria.setFetchMode("designations", FetchMode.JOIN);
             List list = criteria.list();

        } catch (Exception e) {
             //print e 
        }

Answer 2

无法获得符合条件的元素集合（阅读here）。您必须使用本机SQL

使用投影获取集合（休眠）

2 个答案: