Question

我是Hibernate的新手。编写CompositeUserType。当我运行代码时，我收到错误。财产

映射列数错误：请帮帮我，我错过了什么？

我的CompositeUserType如下

 public  class EncryptedAsStringType implements CompositeUserType {
 @Override
 public String[] getPropertyNames() {
    return new String[] { "stockId", "stockCode", "stockName","stockDescription" };
}


@Override
public Type[] getPropertyTypes() {
    //stockId, stockCode,stockName,modifiedDate
    return new Type[] { 
            Hibernate.INTEGER, Hibernate.STRING, Hibernate.STRING,Hibernate.STRING

    };
}
@Override
public Object getPropertyValue(final Object component, final int property)
        throws HibernateException {
    Object returnValue = null;
    final Stock auditData = (Stock) component;

    if (0 == property) {
        returnValue = auditData.getStockId();
    } else if (1 == property) {
        returnValue = auditData.getStockCode();
    } else if (2 == property) {
        returnValue = auditData.getStockName();
    }   return returnValue; 
}

@Override
public void setPropertyValue(final Object component, final int property,
        final Object setValue) throws HibernateException {
    final Stock auditData = (Stock) component;


}


 @Override
    public Object nullSafeGet(final ResultSet resultSet,
            final String[] names,
            final SessionImplementor paramSessionImplementor, final Object paramObject)
            throws HibernateException, SQLException {
        //owner here is of type TestUser or the actual owning Object
     Stock auditData = null;
        final Integer createdBy = resultSet.getInt(names[0]);
        //Deferred check after first read
        if (!resultSet.wasNull()) {
            auditData = new Stock();

            System.out.println(">>>>>>>>>>>>"+resultSet.getInt(names[1]));
            System.out.println(">>>>>>>>>>>>"+resultSet.getString(names[2]));
            System.out.println(">>>>>>>>>>>>"+resultSet.getString(names[3]));
            System.out.println(">>>>>>>>>>>>"+resultSet.getString(names[4]));
                 }
        return auditData;
    }


    @Override
    public void nullSafeSet(final PreparedStatement preparedStatement,
            final Object value, final int property,
            final SessionImplementor sessionImplementor)
            throws HibernateException, SQLException {
        if (null == value) {


        } else {
            final Stock auditData = (Stock) value;
            System.out.println("::::::::::::::::::::::::::::::::"+auditData.getStockCode());
            System.out.println("::::::::::::::::::::::::::::::::"+auditData.getStockDescription());
            System.out.println("::::::::::::::::::::::::::::::::"+auditData.getStockId());
            System.out.println("::::::::::::::::::::::::::::::::"+auditData.getStatus());


        }
    }

我的域类库存有五个属性。（stockId，stockCode，StockName，Status，Stock 描述）

我需要将字段库存描述声明为复合字段类型。

 private Integer stockId;
private String stockCode;
private String stockName;
private String status;
private String stockDescription;

//Constructors  


@Column(name = "STOCK_CC", unique = true, nullable = false, length = 20)
@Type(type="com.mycheck.EncryptedAsStringType")
@Columns(columns = { @Column(name="STOCK_ID"),
    @Column(name="STOCK_CODE"),
    @Column(name="STOCK_NAME")

   })
public String getStockDescription() {
    return stockDescription;
}

}

当我尝试执行Stock插入时。我收到错误错误创建名称为

的bean

＆＃39;的sessionFactory＆＃39;在类路径资源[spring / config /../ database / Hibernate.xml]中定义：

调用init方法失败。嵌套异常是org.hibernate.MappingException：

属性映射列数错误：com.stock.model.Stock.stockDescription类型：

com.mycheck.EncryptedAsStringType

我哪里错了？

Answer 1

可以从代码示例和原始问题的评论中提取答案，但为了节省每个人的阅读，我编写了一个快速摘要。

如果声明将{1}映射到n列的CompositeUserType，则必须在@Columns注释之外的@Type中声明n列。例如：

public class EncryptedAsStringType implements CompositeUserType {
    @Override
    public String[] getPropertyNames() {
       return new String[] { "stockId", "stockCode", "stockName","stockDescription" };
    }

    // ...
}

此CompositeUserType映射到4个单独的列，因此必须声明4个单独的@Column注释：

@Type(type="com.mycheck.EncryptedAsStringType")
@Columns(columns = {
    @Column(name="STOCK_ID"),
    @Column(name="STOCK_CODE"),
    @Column(name="STOCK_NAME"),
    @Column(name="STOCK_DESCRIPTION")
})
public String getStockDescription() {
    return stockDescription;
}

就是这样，Hibernate很高兴。

Hibernate CompositeUserType映射的列数错误

1 个答案: