Hibernate这样的问题怎么处理

Question

[code="java"]
/*==============================================================*/
/* DBMS name: MySQL 5.0 /
/ Created on: 2011-11-3 19:37:33 /
/==============================================================*/

drop database if exists db_hi;

create database db_hi;

use db_hi;

/*==============================================================*/
/* Table: tb_bbs /
/==============================================================*/

drop table if exists tb_bbs;
create table tb_bbs
(
bbs_id int(4) not null auto_increment,
bbs_boardid smallint(2),
bbs_title varchar(70),
bbs_content varchar(2000),
bbs_sender varchar(20),
bbs_sendtime datetime,
bbs_face varchar(8),
bbs_optime datetime,
bbs_istop varchar(1),
bbs_totoptime datetime,
bbs_isgood varchar(1),
bbs_togoodtime datetime,
primary key (bbs_id)
);

/*==============================================================*/
/* Index: index_1 /
/==============================================================*/

create index index_1 on tb_bbs
(
bbs_sender
);

/*==============================================================*/
/* Table: tb_user /
/==============================================================*/

drop table if exists tb_user;
create table tb_user
(
id int(4) not null auto_increment,
user_password varchar(15),
user_name varchar(20),
user_face varchar(8),
user_sex varchar(2),
user_phone varchar(15),
user_qicq varchar(20),
user_email varchar(20),
user_from varchar(20),
user_able varchar(4),
primary key (id),
unique key ak_key_2 (user_name)
);

/*==============================================================*/
/* Index: un_username [code="java"][/code] /
/==============================================================*/
create unique index un_username on tb_user
(
user_name
);

alter table tb_bbs add constraint fk_reference_1 foreign key (bbs_sender)
references tb_user (user_name) on delete cascade on update cascade;

[/code]
以上是mysql 的脚本

然后使用hibernate处理

[code="java"]
hibernate中配置文件

BbsForm对应的xml
<?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">

UserForm对应的xml

<?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">

[/code]
[img]http://dl.iteye.com/upload/attachment/581843/34ac6a45-0016-3819-9a2e-6ede08130b3c.jpg[/img]

如果在数据库操作层使用

[code="java"]Session session=SessFactory.newInstance().openSession();
BbsForm bbsForm=(BbsForm) session.get(BbsForm.class, id);
[/code]

java.sql.sqlexception Invalid value for getInt() "xxxx"

虽然知道hibernate内部一定是以user表中的id作为主键,但mysql既然支持这种不正确的主键关系,为什么hibernate就不能支持这种不通过只能使用id来标识对象,不知道大牛们对这种有什么看法

Answer 1

hibernate＋不合范式的DB绝对是折腾，虽然hiberante支持～～～～
[code="xml"]


[/code]