即时类型和日期类型

Question

public class UserEntity {
...
    @Basic(optional = false)
    @Column(name = "created", nullable = false, updatable = false)
    @CreatedDate
    private Date created;
...
}

[EL Fine]: sql: 2018-05-02 21:54:25.578--ClientSession(210949780)--Connection(1559972721)--INSERT INTO users (activation_token, avatar_id_in_cloud, avatar_provider, created, email, email_change_token, enabled, entity_version, modified_date, new_email, password, unique_id, username) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
bind => [null, null, null, [here created ==> 2018-05-02 21:54:25.279 <== here created, [e-mail], null, true, 1, B@6f8060ac, null, $2a$12$/MwdBoXwBgHlxnsicVW.w.UWdYo6ulLH87aZtSxhdbBYA.9QacUXy, 74a45e29-c1f8-40dc-ac25-a7196d7870b4, JonkiPro]

但是，根据Java8，我将date类型更改为instant。

public class UserEntity {
{
    @Basic(optional = false)
    @Column(name = "created", nullable = false, updatable = false)
    @CreatedDate
    private Instant created;
...
}

现在它自动创建一个表，其中包含longvarbinary创建的列，生成的命令如下所示

[EL Fine]: sql: 2018-05-02 21:54:25.578--ClientSession(210949780)--Connection(1559972721)--INSERT INTO users (activation_token, avatar_id_in_cloud, avatar_provider, created, email, email_change_token, enabled, entity_version, modified_date, new_email, password, unique_id, username) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)
bind => [null, null, null, [here created ==> B@6f8060ac <== here created, [e-mail], null, true, 1, B@6f8060ac, null, $2a$12$/MwdBoXwBgHlxnsicVW.w.UWdYo6ulLH87aZtSxhdbBYA.9QacUXy, 74a45e29-c1f8-40dc-ac25-a7196d7870b4, JonkiPro]

代替日期的是字符串B@6f8060ac，在数据库中它已经被写入ACED00057372000D6A6176612E74696D652E536572955D84BA1B2248B20C00007870770D02000000005AEA177110A133C078。日期是怎么回事？我使用的是H2 base和Eclipselink。

Answer 1

JPA似乎不支持instant类型，甚至支持大多数新时间类型的最新版本2.2也不支持。

Hibernate和DataNucleus支持instant类型，它们自动将其映射到时间戳。但对于EclipseLink却不是这样，它将其视为任何其他对象，并使用序列化将其存储到数据库中。

JPA2.2规范只列出了以特殊方式处理的以下类型：