public interface CustomerDAO
{
public Customer getCustomer(String num);
}
public class Customer implements java.io.Serializable
{
public String id;
public void setId(String id){this.id = id;}
public void getId(){return id;} public String toString()
{
return new ToStringBuilder(this).append("id",getId()).toString();
}
public boolean equals(Object other)
{
if((this == other))return true;
if(!(other instanceof Customer))return false;
Customer castOther = (Customer)other;
return new EqualsBuilder().append(this.getId(),castOther.getId()).isEquals();
}
public int hashCode()
{
return new HashCodeBuilder().append(getId()).toHashCode();
}
}
public class CustomerDAOHibernate extends HibernateDaoSupport implements CustomerDAO
{
public Customer getCustomer(String num)
{
Customer cust = //请在此填入相应的代码!才能获取最新的Customer信息!
return cust;
}}
请您帮帮忙!
小弟感激不尽!
find("from Customer order by id desc")
2 那里返回的就是当前保存好的对象
3 如果你不在乎是谁持久化的,只是泛泛的要数据库的最新的一个被持久化的记录,可以用1楼的方法find("from Customer order by id desc").setMaxNumber(1);只返回第一个就行了。