Spring Hibernate CompositeUserType映射的列数错误

我刚开始冬眠。写一个复合字体。当我运行代码时，我得到一个错误。 财产

映射的列数错误： 请帮帮我，我错过了什么

我的组合类型如下

 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());


        }
    }

我的域类股票有五个属性。（股票ID、股票代码、股票名称、状态、股票 （说明）

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;
}

}

当我尝试执行股票插入时。我在创建名为的bean时出错

在类路径资源[spring/config/./database/Hibernate.xml]中定义的“sessionFactory”：

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

属性映射的列数错误：com.stock.model.stock.stockDescription类型：

com.mycheck.EncryptedAsStringType

---

我哪里出错了？

可以从代码示例和原始问题的注释中提取答案，但为了节省大家的阅读时间，我编写了一个快速摘要

如果声明一个将类型映射到n列的

CompositeUserType

，则除了

@type

注释外，还必须在

@columns

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。

您错过了@Column（name=“STOCK\u DESCRIPTION”）对吗？STOCK\u CC指的是描述。你看到有什么问题吗？@Column（name=“STOCK_CC”，unique=true，nullable=false，length=20）当我在用户类型类中的返回新字符串[]{“stockDescription”}中声明一个one属性时，代码是有效的。如果我添加了多个列，则返回新字符串[]{“stockDescription”，“stockName”}；它说属性映射的列数不正确：因为您试图将3列拆分为1个类字段。将3列分隔为3个字段。这将解决部分问题Custom NamingStrategy应该能够为具有2列或更多列的CompositeUserType定义唯一的列名：