Hibernate沒有找到一個屬性（org.hibernate.PropertyNotFoundException）二傳手

Question

我試圖讓我的事務類的SQL查詢（春季註釋）的結果：

String query = "SELECT DAT, DROP_RADIO FROM My_TABLE"; 
SQLQuery q = getCurrentSession().createSQLQuery(query); 

q.setResultTransformer(Transformers.aliasToBean(CallDrop.class)); 
List result = q.list(); 

而且CallDrop：

public class CallDrop implements java.io.Serializable { 

    private int id; 
    private Date dat; 
    private BigDecimal onRadio; 
    public BigDecimal getOnRadio() { 
     return onRadio; 
    } 
    public void setOnRadio(BigDecimal onRadio) { 
     this.onRadio = onRadio; 
    } 
    public Date getDat() { 
     return dat; 
    } 
    public void setDat(Date dat) { 
     this.dat = dat; 
    } 
} 

的的.hbm.xml是：

<?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.company.beans.CallDrop" table="My_TABLE"> 
     <id name="id" type="int" /> 

     <property name="dat" type="date"> 
      <column name="DAT" length="10" /> 
     </property> 
     <property name="onRadio" type="big_decimal"> 
      <column name="DROP_RADIO" precision="22" scale="0" /> 
     </property> 
</class> 
</hibernate-mapping> 

我敢肯定，我已經添加了的hbm.xml到ApplicationContext的文件中，會話工廠：

<bean id="sessionFactoryETL" class="org.springframework.orm.hibernate3.LocalSessionFactoryBean"> 
     <property name="dataSource"><ref local="dataSourceETL"/></property> 
     <property name="mappingResources"> 
     <list> 
        <value>com/company/beans/CallDrop.hbm.xml</value> 
     </list> 
     </property> 
     <property name="hibernateProperties"> 
      <props> 
       <prop key="hibernate.dialect">${hibernate.dialect}</prop> 
       <prop key="hibernate.show_sql">true</prop> 
      </props> 
     </property> 
    </bean> 

所以，沒有類型，但它不起作用。有沒有人知道問題是什麼？ 感謝

Answer 1

更改查詢"SELECT DAT as \"dat\", DROP_RADIO as \"onRadio\" FROM My_TABLE"

Answer 2

您可以使用addScalar（字符串columnAlias，類型類型）顯式聲明的本地SQL的列別名：

getCurrentSession().createSQLQuery("select e.id AS id, e.first_name AS firstName, e.password AS password FROM table") 
       .addScalar("id", StandardBasicTypes.INTEGER) 
       .addScalar("firstName", StandardBasicTypes.STRING) 
       .addScalar("password", StandardBasicTypes.STRING) 
       .setResultTransformer(Transformers.aliasToBean(Employee.class)) 
       .list();