我在PostgreSQL中有一个名为"transaction_response"的简单表.在此表中,只有两列:transaction_id和response_id.
我在那里的数据有两行.两者都具有相同的transaction_id,但响应值不同.
我创建了以下Java类来保存这些数据:
public class TransactionResponseDAO implements java.io.Serializable {
    private Integer transactionId;
    private Integer responseId;
    public Integer getTransactionId() {
        return transactionId;
    }
    public void setTransactionId(Integer transactionId) {
        this.transactionId = transactionId;
    }
    public Integer getResponseId() {
        return responseId;
    }
    public void setResponseId(Integer responseId) {
        this.responseId = responseId;
    }
    public String toString() {
        return "Transaction Id: " + transactionId + "\nResponse Id: " + responseId;
    }
}
然后我创建了名为"TransactionResponse.hbm.xml"的Hibernate配置文件:
<?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>
    <class name="com.data.TransactionResponseDAO" table="transaction_response">
        <id name="transactionId" column="transaction_id" type="java.lang.Integer">
            <generator class="increment"/>
        </id>
        <property name="responseId" type="java.lang.Integer">
            <column name="responses_id" />
        </property>
    </class>
    <query name="findTransactionResponseByTransactionId">
        <![CDATA[from com.data.TransactionResponseDAO where transaction_id = :transactionId]]>
    </query>
    <query name="findTransactionResponseByBothIds">
        <![CDATA[from com.data.TransactionResponseDAO where transaction_id = :transactionId 
                 and responses_id = :responseId]]>
    </query>
</hibernate-mapping>
在我正在测试的Java类中,我有以下代码:
Query q = null;
        SessionFactory sFactory = new Configuration().configure("/conf/hibernate.cfg.xml").buildSessionFactory();
        Session session = null;
        try {
            session = sFactory.openSession();
            q = session.getNamedQuery("findTransactionResponseByTransactionId");
            q.setInteger("transactionId", transactionId);
        }catch (Exception e) {
            System.err.println("Error running getStatusTab");
            e.printStackTrace();
            session.close();
            sFactory.close();
        }
        ArrayList<TransactionResponseDAO> results = (ArrayList<TransactionResponseDAO>) q.list();
        session.close();
        sFactory.close();
        System.out.println(">>> Size: " + results.size());
        for (TransactionResponseDAO tr : results) {
            System.out.println(tr);
            System.out.println();
        }
当我运行此代码时,它会返回两行,就像它应该的那样.但是,在打印出值时...两行的response_id相同.因此,出于某种原因,它似乎没有引入第二个response_id值.
这是什么原因?一如既往,我感谢您的帮助!
Dio*_*ana 12
问题是hibernate的缓存机制.当hibernate获取第一行并创建一个对象时,它会通过指定的id列(transactionId)缓存该对象.当hibernate获取第二行时,它认为该行与第一行相同,因为这些行具有相同的id,从缓存中获取对象(同一对象)并将其再次放在结果上.
如果colum transactionId不是唯一的,请不要使用id标记映射它.
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