在hibernate中使用HQL直接对数据库进行update、save、delete操作时，日志中会输出：

org.hibernate.hql.ast.tree.FromElementType -164767 [http-8080-Processor72] WARN   - Using non-qualified column reference 

 的警告信息。该警告实乃Hibernate-3.2.0版本的BUG，原因是hibernate执行HQL时没有细分操作类型，日志记录类型有误，该BUG已在3.2.1中修复。

2、deleted object would be re-saved by cascade (remove deleted object from associations) 
 

    在hibernate 删除关联时会出现eleted object would be re-saved by cascade (remove deleted object from associations)的异常，结合别人的和自己的经验通常有三种解决的方案：

        方法1 删除Set方的cascade：
        方法2 解决关联关系后，再删除

    onside.getManys().remove(thisMany);   //在所关联的一方的set中移走当前要删除的对象
    thisMany.setOne(null);                          //设置所对应的一方为空，解除它们之间的关系
    manyDao.delete(thisMany);

        方法3 在many-to-one方增加cascade 但值不能是none

import java.util.HashSet;

/**
 * 系统角色信息
 *
 */
public class SysRole {

 private String roleNo ;  //角色编码
 
 private String roleName ; //角色名称
 
 private String roleDesc ; //角色描述
 
 private java.util.Set<Role_Menu_Perm> rmpSet = new java.util.HashSet<Role_Menu_Perm>();
 
 public java.util.Set<Role_Menu_Perm> getRmpSet() {
  return rmpSet;
 }

 public void setRmpSet(java.util.Set<Role_Menu_Perm> rmpSet) {
  this.rmpSet = rmpSet;
 }

 public String getRoleNo() {
  return roleNo;
 }

 public void setRoleNo(String roleNo) {
  this.roleNo = roleNo;
 }

 public String getRoleName() {
  return roleName;
 }

 public void setRoleName(String roleName) {
  this.roleName = roleName;
 }

 public String getRoleDesc() {
  return roleDesc;
 }

 public void setRoleDesc(String roleDesc) {
  this.roleDesc = roleDesc;
 }
}

<?xml version="1.0" encoding="UTF-8"?>
<!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.permissions.model.SysRole" table="SysRole">
  <id column="roleNo" name="roleNo" type="string">
   <generator class="assigned"/>
  </id>
  <property column="roleName" name="roleName" type="string" length="20" not-null="true" unique="true"/>
  <property column="roleDesc" name="roleDesc" type="string" length="200"/>
  <!-- 如果这里的注释信息打开，Role_Menu_Perm类中的注释部分也要打开
   <set name="rmpSet" table="Role_Menu_Perm" inverse="false" lazy="false">
     <key column="roleNo" not-null="true"></key>
     <composite-element class="com.permissions.model.Role_Menu_Perm">
   <property name="menuNo"></property>
   <property name="permNo"></property>
     </composite-element>
   </set>
  -->
  <set name="rmpSet" table="Role_Menu_Perm" inverse="false" lazy="false">
     <key column="roleNo" not-null="true"></key>
     <composite-element class="com.permissions.model.Role_Menu_Perm">
   <many-to-one name="menu"  column="menuNo"></many-to-one>
   <many-to-one name="perm" column="permNo"></many-to-one>
     </composite-element>
   </set>
 </class>
</hibernate-mapping>
/**
*角色-资源-权限
*
**/

public class Role_Menu_Perm {

// private String roleNo ;
// 
// private String menuNo ;
// 
// private String permNo ;
// 
 private SysRole role ;
 
 private SysMenu menu ;
 
 private SysPermissions perm ;

 public SysRole getRole() {
  return role;
 }

 public void setRole(SysRole role) {
  this.role = role;
 }

 public SysMenu getMenu() {
  return menu;
 }

 public void setMenu(SysMenu menu) {
  this.menu = menu;
 }

 public SysPermissions getPerm() {
  return perm;
 }

 public void setPerm(SysPermissions perm) {
  this.perm = perm;
 }

// public String getRoleNo() {
//  return roleNo;
// }
//
// public void setRoleNo(String roleNo) {
//  this.roleNo = roleNo;
// }
//
// public String getMenuNo() {
//  return menuNo;
// }
//
// public void setMenuNo(String menuNo) {
//  this.menuNo = menuNo;
// }
//
// public String getPermNo() {
//  return permNo;
// }
//
// public void setPermNo(String permNo) {
//  this.permNo = permNo;
// }
}

借助于Hibernate强大的O/R Mapping能力，我们能够通过discriminator轻易地将一颗继承树映射到一个表中，通过discriminator确定具体映射的子类。

在设置@hibernate.discriminator column="type" type="integer"后，启动Hibernate报错：

Could not format discriminator value to SQL string

 http://dev.csdn.net/article/82/82734.shtm

http://forum.hibernate.org/viewtopic.php?t=974225

搜索Hibernate官方文档后发现问题，原来Hibernate默认的discriminator的type是String，当设置discriminator的type为integer后，需要为父类也设置@hibernate.class table="TABLE_NAME" discriminator-value="not null"，否则，Hibernate默认的discriminator-value是完整的类名，在转换String到int时造成NumberFormatException。

最后运行XDoclet，生成hbm文件：

<?xml version="1.0" encoding="iso-8859-1"?>

<!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.crackj2ee.example.AbstractClass" table="TABLE_NAME" discriminator-value="not null">

        <id name="id" column="id" type="java.lang.Long" unsaved-value="null">

            <generator class="increment"/>

        </id>

        <discriminator column="type" not-null="true" type="integer"/>

        <subclass name="com.crackj2ee.example.SubClass1" discriminator-value="1">

        <subclass name="com.crackj2ee.example.SubClass2" discriminator-value="2">

    </class>

</hibernate-mapping>