关于Hibernate的问题

Question

是外键关联错了吧，相关表中没数据，就不能插入

Answer 1

----BlogUser 表 public class BlogUser implements java.io.Serializable {    private static final long serialVersionUID = 8850445505288137285L;    private Integer userId;    private String userName;    private String passWordOne;    private Set<BlogFile> blogFile = new HashSet<BlogFile>();
//省略set & get ....
}----BlogFile 表 
public class BlogFile implements java.io.Serializable {    private static final long serialVersionUID = -1963389377781856294L;    private Integer fileId;    private String filePath;    private String fileName;
    
    private BlogUser blogUser;    //private String createUserId;
//省略set & get ....
}
---BlogUser.hbm.xml 
<hibernate-mapping>
    <class name="model.user.BlogUser" table="BLOG_User" schema="dbo" catalog="Blog"
            dynamic-update="true"  optimistic-lock="version" >
        <id name="userId" type="java.lang.Integer">
            <column name="userId" />
            <generator class="native" />
        </id>
        <property name="userName" type="java.lang.String">
            <column name="userName" length="50"  />
        </property>
        <property name="passWordOne" type="java.lang.String">
            <column name="passWordOne" length="50"  />
        </property>
        <set name="blogFile" inverse="true" lazy="true" cascade="all-delete-orphan">
            <key>
                <column name="userId" />
            </key>
            <one-to-many class="model.user.BlogFile"/>
        </set>
    </class>
</hibernate-mapping>
---BlogFile.hbm.xml 
<hibernate-mapping>
    <class name="model.user.BlogFile" table="Blog_File" schema="dbo" catalog="Blog">
        <id name="fileId" type="java.lang.Integer">
            <column name="fileId" />
            <generator class="native" />
        </id>
        <property name="filePath" type="java.lang.String">
            <column name="filePath" length="200" />
        </property>
        <property name="fileName" type="java.lang.String">
            <column name="fileName" length="200" />
        </property>
        <many-to-one name="blogUser"
            class="model.user.BlogUser">
            <column name="createUserId" /> <!--createUserId 是blogUser的主键userId -->
        </many-to-one>
    </class>
</hibernate-mapping>---action public ActionForward insertUser(ActionMapping mapping, ActionForm form,
            HttpServletRequest request, HttpServletResponse response)
            throws Exception {
        BlogUser blogUser = new BlogUser();
        
        Set<BlogFile> file = new HashSet<BlogFile>();
        
        blogUser.setUserName("mousetsun");
        blogUser.setPassWordOne("111111");
        
        BlogFile blogFile = new BlogFile();
        blogFile.setFileName("aaa");
        blogFile.setFilePath("bbb");
        blogFile.setBlogUser(blogUser);
        file.add(blogFile);
        
        BlogFile blogFile1 = new BlogFile();
        blogFile1.setFileName("ccc");
        blogFile1.setFilePath("ddd");
        blogFile1.setBlogUser(blogUser);
        file.add(blogFile1);
        
        blogUser.setBlogFile(file);
        //spring...service 调用dao
        BlogUserService blogUserService = Locator
                .lookupService(BlogUserService.class);
        BlogUser user = blogUserService.insertUser(blogUser);
        //********
        System.out.println(user.getBlogFile().size());
        
        return mapping.findForward("index");
    }
---DAO //GenericDaoHibernateImpl 是我自己写的一些hibernate常用接口方法
public class BlogUserDaoImpl extends GenericDaoHibernateImpl<BlogUser, Integer> implements BlogUserDao{    public BlogUser insertUser(BlogUser blogUser) {
        // TODO Auto-generated method stub
        this.getHibernateTemplate().clear();
        blogUser = (BlogUser) this.getHibernateTemplate()
                .merge(blogUser);
        this.getHibernateTemplate().flush();
        return blogUser;
    }
}希望对你有用 >_<

Answer 2

谢谢楼上各位 我再仔细看看吧