Question

我正在研究具有以下三个实体的hibernate项目，当我在实体之间添加onetomany关系时得到Field＆＃39; user_id＆＃39;没有默认值。

当我尝试插入新主题时

Hibernate：插入主题（content，date，image，image_name）值（？，？，？，？） 2016年5月8日11:46:56 org.hibernate.util.JDBCExceptionReporter logExceptions 警告：SQL错误：1364，SQLState：HY000 2016年5月8日11:46:56 org.hibernate.util.JDBCExceptionReporter logExceptions 严重：字段＆＃39; user_id＆＃39;没有默认值

并在尝试插入新记录

时获得相同的错误

用户类

public class User implements Serializable{

private static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "user_id", unique = true, nullable = false)
private long id;

@Column(name = "first_name",  nullable = false, length = 10)
private String firstName;

@Column(name = "last_name",  nullable = false, length = 10)
private String lastName;

@Column(name = "role",  nullable = false, length = 10)
private String role;


@Column(name = "email", unique = true, nullable = false, length = 10)
private String email;


@Column(name = "password", nullable = false, length = 10)
private String password;

@OneToMany(fetch = FetchType.LAZY, mappedBy = "user")
private Set<Topics> topics=new HashSet<Topics>();

@OneToMany(fetch = FetchType.LAZY, mappedBy = "user1")
private Set<Like> likes=new HashSet<Like>();

//getter and setter

喜欢上课

@Entity
@Table(name = "likes")
 public class Like implements Serializable{

private static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "likes_id", unique = true, nullable = false)
private long id;

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "user_id", nullable = false,insertable = false, updatable = false)
private User user1;

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "topics_id", nullable = false,insertable = false,   updatable = false)
private Topics topics;
 //getter and setter

主题课

 @Entity
 @Table(name = "topics")
 public class Topics implements Serializable{

/**
 * 
 */
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "topics_id", unique = true, nullable = false)
private long id;

@Column(name = "content", unique = true, nullable = false)
private String content;

@Temporal(TemporalType.DATE)
@Column(name = "date")
private Date date;

@Column(name = "image_name")
private String imageName;

@Lob
@Column(name="image", columnDefinition="longblob")
private byte[] image;


@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "user_id", nullable = false,insertable = false, updatable = false)
private User user;

@OneToMany(fetch = FetchType.LAZY, mappedBy = "topics")
private Set<Like> likes=new HashSet<Like>();
 //getter and setter

hibernate配置文件

 <hibernate-configuration>
<session-factory>
    <property name="hibernate.bytecode.use_reflection_optimizer">false</property>
    <property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property>
    <property name="hibernate.connection.password">root</property>
    <property name="hibernate.connection.url">jdbc:mysql://localhost:3306/social</property>
    <property name="hibernate.connection.username">root</property>
    <property name="hibernate.dialect">org.hibernate.dialect.MySQLDialect</property>
    <property name="show_sql">true</property>
    <property name="hbm2ddl.auto">create</property>
    <mapping class="com.pro.model.User"></mapping>
    <mapping class="com.pro.model.Topics"></mapping>
    <mapping class="com.pro.model.Like"></mapping>
</session-factory>

Answer 1

可能您没有设置试图插入主题表的主题用户。

还有一件事，就是使用单数类名。主题 - ＆gt;主题

Answer 2

我在这里看到很多问题：

你确定主题不是ManyToMany而不是OneToMany（似乎不是有效的东西）......（一个人可以对很多主题感兴趣，主题可以包含很多人）< / LI>
如果您想要使用新用户保留新主题，并且您应该使用@OneToMany（级联 = {CascadeType.PERSIST，CascadeType.MERGE}），如果您在字段中正确设置，如果你不这样做，你可能会受到循环引用的影响......
我宁愿使用 GenerationType.SEQUENCE ，因为AUTO就像一个黑盒子。你不会确切地知道发生了什么......（例如oracle会创建一个序列来提供唯一的主键）

hibernate Field＆＃39; user_id＆＃39;没有默认值

2 个答案: