Jpa 对简单连接长异常的奇怪值 table

Question

我有 2 个实体项目和用户：

@Entity
public class Project implements Serializable {


    private static final long serialVersionUID = 1L;


    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id", updatable = false, nullable = false)
    private Long id;


    @Version
    @Column(name = "version")
    private int version;

    @Column(unique = true)
    private String name;


    @Column
    private String description;
}

和

@Entity
public class User implements Serializable {

    private static final long serialVersionUID = 1L;


    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id", updatable = false, nullable = false)
    private Long id;


    @Version
    @Column(name = "version")
    private int version;


    @Column(unique = true)
    private String name;

}

我想创建一个加入 table 实体（分配给用户的项目）并想在预订实体中使用它。我使用了@IdClass 方法（我得到了与可嵌入相同的结果）：

@Entity
@IdClass(UserProjectKey.class)
public class UserProjects implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @ManyToOne
    private User user;

    @Id
    @ManyToOne
    private Project project;

    public UserProjects() {

    }

    public UserProjects(final User user, final Project project){
        this.user = user;
        this.project = project;

    }
}

public class UserProjectKey implements Serializable {

    private static final long serialVersionUID = 1L;

    private Long user;

    private Long project;

    public UserProjectKey() {

    }

    public UserProjectKey(final long userId, final long projectId){
        user = userId;
        project = projectId;
    }

}

在预订中使用：

@Entity
public class Booking implements Serializable {


    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id", updatable = false, nullable = false)
    private Long id;

    @Version
    @Column(name = "version")
    private int version;


    @Column
    @NotNull
    private UserProjects userProject;


    @Column
    @NotNull
    private Timestamp start;

    @Column
    @NotNull
    private Timestamp end;


}

我省略了所有的 getter 和 setter。问题出在持久预订上，我得到一个奇怪的异常：

Value too long for column "USERPROJECT BINARY(255) NOT NULL": "X'aced00057372002d636f6d2e70726f64796e612e7061632e74696d747261636b65722e6d6f64656c2e5573657250726f6a6563747300000000000000010200... (572)"; SQL statement:
insert into Booking (end, start, userProject, version, id) values (?, ?, ?, ?, ?) [22001-173]

因此出于某种原因，jpa 尝试将 UserProject 存储为二进制文件而不是 UserProject table 的外键。那么如何修复它，如何为这个用例正确设计 jpa 类？

Answer 1

您与 UserProjects 的关系未定义为外键。您需要使用 @ManyToOne（我假设许多预订可能涉及一个 UserProjects 实例？）以及 @JoinColumn（而不是您的 @Column）才能触发一个外键。我相信您看到的是试图插入的序列化值。

Jpa 对简单连接长异常的奇怪值 table

Jpa strange value to long exception on simple join table

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jointable