[英]Hibernate can't insert UUID to VARCHAR column
I am working on setting up a MySQL database and using Hibernate and Play Framework for the backend.我正在建立一个 MySQL 数据库,并为后端使用 Hibernate 和 Play Framework。 I am having problems with the IDs of entries.
我遇到了条目 ID 的问题。 I defined my id column as
VARCHAR(36)
:我将我的 id 列定义为
VARCHAR(36)
:
CREATE TABLE `logaritmical`.`users` (
`id` VARCHAR(36) NOT NULL,
`username` TEXT NOT NULL,
`email` TEXT NOT NULL,
PRIMARY KEY (`id`),
UNIQUE INDEX `id_UNIQUE` (`id` ASC)
);
Now, the @Entity
class is like this:现在,@Entity
@Entity
是这样的:
@Entity
@Table(name = "users")
public class UserDO {
@Id
@GeneratedValue(generator = "uuid2")
@GenericGenerator(name = "uuid2", strategy = "uuid2")
@Column(name = "id", updatable = false, nullable = false, columnDefinition = "VARCHAR(36)")
private UUID id;
@Column
private String username;
@Column
private String email;
}
When doing the insert, I get the following error: Incorrect string value: '\xEF\xA5.\x89\xF3K...' for column 'id' at row 1
执行插入时,出现以下错误:
Incorrect string value: '\xEF\xA5.\x89\xF3K...' for column 'id' at row 1
If I change the column type and columnDefinition to BINARY(16)
the insert works, but it has the disadvantage that the ID is not human-readable when doing selects.如果我将列类型和 columnDefinition 更改为
BINARY(16)
插入工作,但它的缺点是 ID 在执行选择时不可读。
Additional info: persistence.xml
looks like this:附加信息:
persistence.xml
看起来像这样:
<persistence-unit name="defaultPersistenceUnit" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
<non-jta-data-source>DefaultDS</non-jta-data-source>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQL8Dialect"/>
</properties>
</persistence-unit>
DB Configuration and versions of libraries looks like this: jpa.default=defaultPersistenceUnit
"org.hibernate" % "hibernate-entitymanager" % "5.4.24.Final",
"mysql" % "mysql-connector-java" % "8.0.22",
数据库配置和库版本如下所示:
jpa.default=defaultPersistenceUnit
"org.hibernate" % "hibernate-entitymanager" % "5.4.24.Final",
"mysql" % "mysql-connector-java" % "8.0.22",
What can be done to have the UUID
working with VARCHAR
?可以做些什么来让
UUID
与VARCHAR
一起工作? Is there something I am missing?有什么我想念的吗?
Try to use the following definition:尝试使用以下定义:
import org.hibernate.annotations.Type;
@Entity
@Table(name = "users")
public class UserDO {
@Id
@GeneratedValue(generator = "uuid2")
@GenericGenerator(name = "uuid2", strategy = "uuid2")
@Column(name = "id", updatable = false, nullable = false, columnDefinition = "VARCHAR(36)")
@Type(type = "uuid-char")
private UUID id;
// ...
}
It looks like that for your dialect UUID
by default mapped to the uuid-binary
basic type .默认情况下,您的方言
UUID
看起来像映射到uuid-binary
基本类型。
PS Please note that saving UUID
PK as a string can lead to the performance issue, as it explained in this article : PS 请注意,将
UUID
PK 保存为字符串可能会导致性能问题,如本文所述:
Aside from the 9x cost in size (36 vs. 4 bytes for an int), strings don't sort as fast as numbers because they rely on collation rules.
除了 9 倍的大小成本(36 对 int 的 4 个字节)之外,字符串的排序不如数字快,因为它们依赖于排序规则。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.