Spring MVC Hibernate Many to One 为外键提供空值

Question

我需要一些有关 Spring MVC 4 和 hibernate 5（目前都是最新版本）的帮助。

我有两个班级——会员和家庭。 Member(Staff) 可以有多个家庭成员 我想与 Family 类建立多对一关系。

Member 类的快照：

@Entity
@Table(name = "member")
public class Member {

    @Id
    @Column(name = "member_id")
    @GeneratedValue
    private int id;


    private String firstName;

    private String middleName;

    @OneToMany(cascade = CascadeType.ALL, mappedBy="member")
    private List<Family> families;

    .....

    }

Family 类的快照：

@Entity
@Table(name="family")
public class Family {

    @Id
    @GeneratedValue
    private int id;
    private String firstName;
    private String email;
    private String mobile;

    @ManyToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "member_id")
    private Member member;

    ----

    }

控制器：

`@RequestMapping("/admin/memberManagement/addMember")
    public ModelAndView addMember(@ModelAttribute("member") Member member) {
        ModelAndView view = new ModelAndView("addMember");
        return view;
    }


      @RequestMapping(value = "/admin/memberManagement/addMember", method =     RequestMethod.POST)

       public ModelAndView addMemberPost(@ModelAttribute("member") Member member, Model model) {    
        memberService.addMember(member);
            ModelAndView view = new ModelAndView("redirect:/admin/memberManagement");
        return view;
    }`

表单有一个成员，对应的家庭列表（动态添加的字段），当我运行程序时，它保存了成员没有问题，它也保存了家庭列表，但 member_id 的值为空在家庭表中。

我在这里错过了什么？我感谢您的帮助。我正在学习 Spring MVC。

谢谢，

马迪斯。

**其他信息/问题**

按照 Alexander Staroselsky 的回答，我尝试了几个选项，但家庭表中的 member_id 列仍然为空。在服务器启动时，它会生成以下 sql 表和约束：

Hibernate: 

    alter table family 
       drop 
       foreign key FKpv3nain50fsmms2j4xdig4f8u
Hibernate: 

    drop table if exists family
Hibernate: 

    drop table if exists hibernate_sequence
Hibernate: 

    drop table if exists member
Hibernate: 

    create table family (
       id integer not null,
        email varchar(255),
        firstName varchar(255),
        lastName varchar(255),
        mobile varchar(255),
        relation varchar(255),
        member_id integer,
        primary key (id)
    )
Hibernate: 

    create table hibernate_sequence (
       next_val bigint
    )
Hibernate: 

    insert into hibernate_sequence values ( 1 )
Hibernate: 

    insert into hibernate_sequence values ( 1 )
Hibernate: 

    create table member (
       member_id integer not null,
        discharge datetime,
        email varchar(255),
        firstname varchar(255),
        lastname varchar(255),
        middlename varchar(255),
        mobile varchar(255),
        number varchar(255),
        rank varchar(255),
        regiment varchar(255),
        telephone varchar(255),
        website varchar(255),
        primary key (member_id)
    )
Hibernate: 

    alter table family 
       add constraint FKpv3nain50fsmms2j4xdig4f8u 
       foreign key (member_id) 
       references member (member_id)

它已正确创建表 + 添加了必要的约束。与hibernate和spring mvc版本有关吗？有没有人遇到过这个问题？我使用的版本是：

<properties>
        <spring.version>4.3.5.RELEASE</spring.version>
        <security.version>4.2.1.RELEASE</security.version>
        <jdk.version>1.8</jdk.version>
        <hibernate.version>5.2.6.Final</hibernate.version>
        <org.aspectj-version>1.8.0</org.aspectj-version>
        <mysql.connector-version>6.0.5</mysql.connector-version>
        <jackson.databind-version>2.8.5</jackson.databind-version>
        <tiles.version>3.0.7</tiles.version>
    </properties>

谢谢，

感谢 Alexander，您抽出宝贵时间 - 非常感谢。更改为 @GeneratedValue(strategy = GenerationType.IDENTITY) 不起作用。将成员属性设置为 'nullable=false' 也不起作用，这个值实际上是给 null。插入sql语句如下：

Hibernate: 
    select
        member0_.member_id as member_i1_1_,
        member0_.email as email2_1_,
        member0_.firstname as firstnam3_1_,
        member0_.lastname as lastname4_1_,
        member0_.website as website5_1_ 
    from
        member member0_
Hibernate: 
    select
        next_val as id_val 
    from
        hibernate_sequence for update

Hibernate: 
    update
        hibernate_sequence 
    set
        next_val= ? 
    where
        next_val=?
Hibernate: 
    insert 
    into
        member
        (email, firstname, lastname, website, member_id) 
    values
        (?, ?, ?, ?, ?)
Hibernate: 
    insert 
    into
        family
        (email, firstName, lastName, member_id, mobile) 
    values
        (?, ?, ?, ?, ?)
Hibernate: 
    insert 
    into
        family
        (email, firstName, lastName, member_id, mobile) 
    values
        (?, ?, ?, ?, ?)
Hibernate: 
    select
        member0_.member_id as member_i1_1_,
        member0_.email as email2_1_,
        member0_.firstname as firstnam3_1_,
        member0_.lastname as lastname4_1_,
        member0_.website as website5_1_ 
    from
        member member0_

sql 插入看起来很正常——除了family 表中的member_id 之外，所有其他值都在那里。

两个表上插入的测试值：

mysql> select * from member;
+-----------+----------------+-----------+----------+---------+
| member_id | email          | firstname | lastname | website |
+-----------+----------------+-----------+----------+---------+
|         1 | test@email.com | firstname | lastname | dot.com |
+-----------+----------------+-----------+----------+---------+
1 row in set (0.00 sec)

mysql> select * from family;
+----+----------------+-----------+----------+--------+-----------+
| id | email          | firstName | lastName | mobile | member_id |
+----+----------------+-----------+----------+--------+-----------+
|  1 | test@email.com | test1     | test`    | i3i3   |      NULL |
|  2 | test@gmail.com | test2     | test2    | 7373   |      NULL |
+----+----------------+-----------+----------+--------+-----------+
2 rows in set (0.00 sec)

谢谢

再测试一次

我刚刚删除了 mappedBy="member" 以测试它是否在单独的表中工作。它工作得很好。正如预期的那样，它创建了第三个表——member_family——并正确插入了 id 值。以下是测试值：

mysql> select * from member;
+-----------+----------------+-----------+----------+---------+
| member_id | email          | firstname | lastname | website |
+-----------+----------------+-----------+----------+---------+
|         1 | test@email.com | tre       | jjj      | dot.com |
+-----------+----------------+-----------+----------+---------+
1 row in set (0.00 sec)

mysql> select * from family;
+----+-----------------+-----------+----------+----------+-----------+
| id | email           | firstName | lastName | mobile   | member_id |
+----+-----------------+-----------+----------+----------+-----------+
|  1 | dkdke@email.com | test      | teke     | d8i38303 |      NULL |
|  2 | e@gmai.com      | test1     | testie   | 38383    |      NULL |
+----+-----------------+-----------+----------+----------+-----------+
2 rows in set (0.00 sec)

mysql> select * from member_family;
+------------------+-------------+
| Member_member_id | families_id |
+------------------+-------------+
|                1 |           1 |
|                1 |           2 |
+------------------+-------------+

现在，问题似乎只出在 mappedBy 上（我可能错过了）。任何指针将不胜感激。 谢谢

Answer 1

尝试将@GeneratedValue 替换为@GeneratedValue(strategy = GenerationType.AUTO) 以确保为每个对象生成一个ID：

...
public class Member {

    @Id
    @Column(name = "member_id")
    @GeneratedValue(GenerationType.AUTO)
    private int id;
...

...
public class Family {

    @Id
    @GeneratedValue(GenerationType.AUTO)
    private int id;
    private String firstName;
    private String email;
    private String mobile;
...