Hibernate CompositeUserType映射的列数错误 [英] Hibernate CompositeUserType mapping has wrong number of columns
问题描述
财产
映射的列数不正确：
请帮助我我缺少什么？
My CompositeUserType如下所示：
public class EncryptedAsStringType implements CompositeUserType {
@Override
public String [] getPropertyNames（）{
return new String [] {stockId，stockCode，stockName，stockDescription};
$ b @Override
public Type [] getPropertyTypes（）{
// stockId，stockCode，stockName，modifiedDate
return new键入[] {
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）组件;
if（0 == property）{
returnValue = auditData.getStockId（）;
} else if（1 == property）{
returnValue = auditData.getStockCode（）;
} else if（2 == property）{
returnValue = auditData.getStockName（）;
} return returnValue;
$ b @Override
public void setPropertyValue（final Object component，final int property，
final Object setValue）throws HibernateException {
final Stock auditData = （股票）组件;
$ b @Override
public Object nullSafeGet（final ResultSet resultSet，
final String [] names，
final SessionImplementor paramSessionImplementor，final Object paramObject）
抛出HibernateException，SQLException {
//这里的owner是类型为TestUser或者实际拥有的对象
Stock auditData = null;
final Integer createdBy = resultSet.getInt（names [0]）;
//首次读取
后的延期检查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]））;
}
返回auditData;
$ b @Override
public void nullSafeSet（final PreparedStatement preparedStatement，
final Object value，final int property，
final SessionImplementor sessionImplementor ）
抛出HibernateException，SQLException {
if（null == value）{
$ b} 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。 （stockId，stockCode，StockName，状态，股票
描述）
我需要将字段股票描述声明为复合字段类型。
private Integer stockId;
私人字符串stockCode;
私人字符串stockName;
私人字符串状态;
private String stockDescription;
//构造函数
@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。我收到错误错误创建bean名称
'sessionFactory'在类路径资源[spring / config /../ database / Hibernate.xml]中定义： p>
调用init方法失败。嵌套异常是org.hibernate.MappingException：
属性映射的列数不正确：com.stock.model.Stock.stockDescription类型：
com.mycheck.EncryptedAsStringType
我在哪里错了？
可以从代码示例中提取答案并将注释提取为原始问题，但为了节省大家的一些阅读时间，我编写了一个快速概要。
如果声明将类型映射到n列的 CompositeUserType ，则必须在 @Columns 中声明n列c $ c>除 @Type 注释外。示例：
public class EncryptedAsStringType implements CompositeUserType {
@Override
public String [] getPropertyNames（）{
return new String [] {stockId，stockCode，stockName，stockDescription};
}
// ...
}
<这个 CompositeUserType 映射到4个独立的列，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很开心。
I am new to Hibernate. Writing a CompositeUserType. When I run the code I am getting error. property
mapping has wrong number of columns: Please help me what am I missing?
My CompositeUserType goes as follows
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());
}
}
My Domain class Stock has five attributes. (stockId,stockCode,StockName,Status , Stock Description)
I need to declare the field Stock description as Composite field Type.
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;
}
}
When I try to execute a insert for Stock. I am getting the error Error creating bean with name
'sessionFactory' defined in class path resource [spring/config/../database/Hibernate.xml]:
Invocation of init method failed. nested exception is org.hibernate.MappingException:
property mapping has wrong number of columns: com.stock.model.Stock.stockDescription type:
com.mycheck.EncryptedAsStringType
Where am I going wrong ?
One can extract the answer from the code samples and the comments to the original question, but to save everyone some reading, I've compiled a quick summary.
If you declare a CompositeUserType that maps a type to n columns, you have to declare n columns in @Columns besides the @Type annotation. Example:
public class EncryptedAsStringType implements CompositeUserType {
@Override
public String[] getPropertyNames() {
return new String[] { "stockId", "stockCode", "stockName","stockDescription" };
}
// ...
}
This CompositeUserType maps to 4 separate columns, therefore 4 separate @Column annotations have to be declared:
@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;
}
That's it and Hibernate is happy.
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