@NotFound(action=NotFoundAction.IGNORE)

最新推荐文章于 2020-11-10 10:31:35 发布

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最新推荐文章于 2020-11-10 10:31:35 发布

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本文介绍如何在Hibernate中配置实体类关联关系，当引用的数据在数据库中不存在时，通过设置@NotFound(action=NotFoundAction.IGNORE)注解来避免抛出异常。并提供了实例代码，包括一对多关系的查询条件编写，以及处理对象属性为null的情况。

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引用

转载：http://blog.sina.com.cn/s/blog_7a8f609501014j34.html

使用hibernate 注解配置实体类的关联关系，在many-to-one,one-to-one关联中，一边引用自另一边的属性，如果属性值为某某的数据在数据库不存在了，hibernate默认会抛出异常。解决此问题，加上如下注解就可以了：
@NotFound(action=NotFoundAction.IGNORE)，意思是找不到引用的外键数据时忽略，NotFound默认是exception
下面贴出hibernate 注解的实例代码
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@Entity
@Table(name = "ICT_COMPUTER_LOCATION")
public class IctComputerLocation {
    private static final long serialVersionUID = 1L;

    private Integer id;

    private String idcNum;

    private Integer ictBaseId;


    private IctBase ictBase;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "ID")
    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    @Column(name = "IDC_NUM")
    public String getIdcNum() {
        return idcNum;
    }

    public void setIdcNum(String idcNum) {
        this.idcNum = idcNum;
    }

    @Column(name = "ICT_BASE_ID")
    public Integer getIctBaseId() {
        return ictBaseId;
    }

    public void setIctBaseId(Integer ictBaseId) {
        this.ictBaseId = ictBaseId;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "ICT_BASE_ID", referencedColumnName = "ID", unique = false, nullable = false, insertable = false, updatable = false)
    @NotFound(action=NotFoundAction.IGNORE)
    public IctBase getIctBase() {
        return ictBase;
    }

    public void setIctBase(IctBase ictBase) {
        this.ictBase = ictBase;
    }

本文来自优快云博客，转载请标明出处：http://blog.youkuaiyun.com/hardwin/archive/2009/07/08/4330779.aspx

对象A一对多对象B，现在要查询对象A，条件中包含对象B的属性，但对象A的属性B可能为Null，这时候怎么写条件？
比如对象A是用户：
Java code
@Entity
@Table(name = "base_staff")
public class User {
@Id
@GeneratedValue
@Column(name = "id")
private int id;
@Column(name = "name")
private String name;
@Column(name = "logname")
private String logname;
@Column(name = "logpass")
private String logpass;
@ManyToOne
@JoinColumn(name = "gender")
@NotFound(action=NotFoundAction.IGNORE)
private Gender gender;
@Column(name = "birthdate")
private Date birthdate;
@ManyToOne
@JoinColumn(name = "user_status")
@NotFound(action=NotFoundAction.IGNORE)
private UserStatus status;
@Column(name = "note")
private String note;
@ManyToMany(fetch = FetchType.EAGER)
@JoinTable(name = "sys_user_role",
joinColumns = { @JoinColumn(name = "user_id") },
inverseJoinColumns = { @JoinColumn(name = "role_id") })
@NotFound(action=NotFoundAction.IGNORE)
private Set roles;
……
}
对象B是性别：
Java code
@Entity
@Table(name = "code_gender")
public class Gender {
@Id
@Column(name = "id")
private int id;
@Column(name = "name")
private String name;
}
UserDao:
Java code
Query query = session.createQuery("from User u" +
" where u.name like :key" +
" or u.gender.name like :key" +
" or u.logname like :key" +
" or u.status.name like :key" +
" or u.note like :key" +
" order by u.id");
query.setParameter("key", "%" + key + "%");
query.setFirstResult(offset);
query.setMaxResults(size);
List list = query.list();
这种情况查询不到User的gender为null的数据。想要得到那些数据应该怎么些HQL呢？