Question

我有manyToMany关系映射，无法正常工作。我已经阅读了许多帖子和文章，无法理解这一点。如果有人有想法，请分享。

我已尽力简化图表和代码。

我的数据库设计如下：

我的实体看起来像这样（至少是在询问前的最后尝试）：客户

：

@Entity
@Table(name = "client")
public class Client implements Serializable {
    @Id
    @Column(name = "client_id")
    private int id;

    ... other fields
}

项目：

@Entity
@Table(name = "project")
public class Project implements Serializable {
    @EmbeddedId
    private ProjectId id;

    ... other fields

    @Embeddable
    class ProjectId implements Serializable {
        @ManyToOne(fetch = FetchType.LAZY)
        @JoinColumn(name = "client_id", insertable = false, updatable = false)
        private Client client;

        @Column(name = "project_id")
        private int projectId;
    }
}

用户：

@Entity
@Table(name = "user")
public class User implements Serializable {
    @EmbeddedId
    private UserId id;

    ... other fields

    @Embeddable
    class UserId implements Serializable {
        @ManyToOne(fetch = FetchType.LAZY)
        @JoinColumn(name = "client_id", insertable = false, updatable = false)
        private Client client;

        @Column(name = "user_id")
        private int userId;
    }
}

ProjectUser：

@Entity
@Table(name = "project_user")
public class ProjectUser implements Serializable {
    @EmbeddedId
    private ProjectUserId id;

    ... other fields

    @Embeddable
    class ProjectUserId implements Serializable {
        @ManyToOne(fetch = FetchType.LAZY)
        @JoinColumn(name = "client_id", insertable = false, updatable = false)
        private Client client;

        @ManyToOne(fetch = FetchType.LAZY)
        @JoinColumns({
            @JoinColumn(name = "client_id", referencedColumnName = "client_id", insertable = false, updatable = false),
            @JoinColumn(name = "project_id", referencedColumnName = "project_id", insertable = false, updatable = false) })
        private Project project;

        @ManyToOne(fetch = FetchType.LAZY)
        @JoinColumns({
            @JoinColumn(name = "client_id", referencedColumnName = "client_id", insertable = false, updatable = false),
            @JoinColumn(name = "user_id", referencedColumnName = "user_id", insertable = false, updatable = false) })
        private User user;
    }
}

在添加ProjectUser实体之前，一切正常。
现在，当我启动服务器时，它说：

实体的映射中的重复列：ProjectUser列：client_id （应使用insert = \“ false \” update = \“ false \”映射）“}}

所以，问题是我该如何进行这项工作？

编辑： Java应用程序将主要是提供数据的REST服务。数据库设计保持原样。它具有逻辑意义，大多数业务逻辑将在数据库中。我们拥有非常丰富的DB知识的人员，因此，由于JPA / Hibernate的限制，更改数据库设计没有太大意义。

Answer 1

下面的代码将让休眠状态为您的实体创建一个结构：

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "client_id", insertable = false, updatable = false)
    private Client client;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumns({
            @JoinColumn(name = "project_client_id", referencedColumnName = "client_id", insertable = false, updatable = false),
            @JoinColumn(name = "project_id", referencedColumnName = "project_id", insertable = false, updatable = false) })
    private Project project;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumns({
            @JoinColumn(name = "user_client_id", referencedColumnName = "client_id", insertable = false, updatable = false),
            @JoinColumn(name = "user_id", referencedColumnName = "user_id", insertable = false, updatable = false) })
    private User user;

但是要像您问题中的图片一样制定一个方案，请摆脱所有这些EmbeddedId。如果您希望ProjectUser中的项目和用户具有相同的client_id，请使用简单的id并在代码中添加验证。

JPA / Hibernate manyToMany关系映射以及其他字段

1 个答案: