使用Spring启动将数据插入关系表(多对多)

时间:2018-05-12 12:11:23

标签: java mysql spring-boot jpa many-to-many

我制作了一个包含一些用户的Spring Boot应用程序。这些用户可以属于0,一个或多个组(为了更好的可视化,我省略了一些代码行):

@Entity
public class User {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    private String name;

    @ManyToMany
    @JoinTable(
            name = "group_user",
            joinColumns = {@JoinColumn(name = "user_id")},
            inverseJoinColumns = {@JoinColumn(name = "group_id")}
    )
    private List<Group> groups = new ArrayList<>();

    public User(String name, List<Group> groups) {
        this.name = name;
        this.groups = groups;
    }
}

一个组可以包含0个,一个或多个用户。

@Entity
public class Group {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    private String name;

    @ManyToMany(mappedBy = "groups")
    private List<User> users = new ArrayList<>();

    public Group(String name, List<User> users) {
        this.name = name;
        this.users = users;
    }
}

我正在使用MySQL,我创建了3个表:

CREATE TABLE user (
    id integer NOT NULL AUTO_INCREMENT, 
    name varchar(255), 
    PRIMARY KEY (id)
);

CREATE TABLE group (
    id integer NOT NULL AUTO_INCREMENT,
    name varchar(255),
    PRIMARY KEY (id)
);

CREATE TABLE group_user (
    user_id int NOT NULL,
    group_id int NOT NULL,
    PRIMARY KEY (user_id, group_id),
    KEY group_id (group_id),
    CONSTRAINT group_user_ibfk_1
    FOREIGN KEY (user_id) REFERENCES user (id),
    CONSTRAINT group_user_ibfk_2
    FOREIGN KEY (group_id) REFERENCES group (id)
);

我设法通过创建这个新用户,将一个用户链接到一个组,将一个组传递给它的构造函数,并调用我的userDao.save()方法:

userDao.save(new User(name, groups));

现在我想编辑我创建的用户并让他属于另一个组。 如何在不创建新用户的情况下执行此操作?

例如,我有这个用户,他没有任何组:

INSERT INTO user VALUES(1, 'Jordan');

这些团体:

INSERT INTO group VALUES(1, 'Group 1');
INSERT INTO group VALUES(2, 'Group 2');

现在,我如何(在Java中)将我的用户链接到第1组和第2组?

2 个答案:

答案 0 :(得分:2)

您必须使用您选择的 id 从数据库中选择实体。您可以使用 findById 方法为该实体使用JPA 界面,也可以在界面中创建自己的自定义方法。

如果您拥有的实体,则将其添加到用户实体字段 。之后,您保存用户实体

同一过程以双向关系进入另一方

推荐阅读:

  1. Difference between unidirectional and bidirectional associations
  2. A bidirectional association has two sides - owner side and an inverse side
  3. Hibernate ManyToMany persistance tutorial

答案 1 :(得分:0)

创建如:

的表格
mysql> CREATE TABLE users (
    ->     id integer NOT NULL AUTO_INCREMENT, 
    ->     name varchar(255), 
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.24 sec)

mysql> CREATE TABLE groups (
    ->     id integer NOT NULL AUTO_INCREMENT,
    ->     name varchar(255),
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.17 sec)

mysql> CREATE TABLE group_users (
    ->     user_id int NOT NULL,
    ->     group_id int NOT NULL,
    ->     PRIMARY KEY (user_id, group_id),
    ->     CONSTRAINT group_user_ibfk_1
    ->     FOREIGN KEY (user_id) REFERENCES users(id),
    ->     CONSTRAINT group_user_ibfk_2
    ->     FOREIGN KEY (group_id) REFERENCES groups(id)
    -> );
Query OK, 0 rows affected (0.23 sec)

将虚拟数据插入其中:

mysql> INSERT INTO user VALUES(1, 'Jordan');
Query OK, 1 row affected (0.09 sec)

mysql> INSERT INTO users VALUES(1, 'Jordan');
Query OK, 1 row affected (0.08 sec)

mysql> 
mysql> INSERT INTO groups VALUES(1, 'Group 1');
Query OK, 1 row affected (0.02 sec)

mysql> INSERT INTO groups VALUES(2, 'Group 2');
Query OK, 1 row affected (0.02 sec)

mysql> insert into group_users values(1,1);
Query OK, 1 row affected (0.05 sec)

mysql> insert into group_users values(1,2);
Query OK, 1 row affected (0.08 sec)

表中的实际数据如下所示:

mysql> select * from users;
+----+--------+
| id | name   |
+----+--------+
|  1 | Jordan |
+----+--------+
1 row in set (0.00 sec)

mysql> select * from groups;
+----+---------+
| id | name    |
+----+---------+
|  1 | Group 1 |
|  2 | Group 2 |
+----+---------+
2 rows in set (0.00 sec)

mysql> select * from group_users;
+---------+----------+
| user_id | group_id |
+---------+----------+
|       1 |        1 |
|       1 |        2 |
+---------+----------+
2 rows in set (0.00 sec)

加入所有表后的数据将如下所示。

mysql> select u.id user_id, u.name user_name,g.id group_id, g.name group_name from users u, groups g, group_users gu where u.id = gu.user_id and g.id = gu.group_id;
+---------+-----------+----------+------------+
| user_id | user_name | group_id | group_name |
+---------+-----------+----------+------------+
|       1 | Jordan    |        1 | Group 1    |
|       1 | Jordan    |        2 | Group 2    |
+---------+-----------+----------+------------+
2 rows in set (0.05 sec)

希望,你清除了设计,以及它如何在你的情况下工作。