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mysql - 包含 EnumSet 的 ResultSet 生成 Table 不存在

问题描述

我有一个像

enum class AmenitiesEnum {
POOL,
PETS_ALLOWED
// shortened list, for demo purpose only
}

这个枚举被包含在一个类中，然后在 MySQL 数据库中持久化。

@Entity
@Table(name = "hotels")
class Hotel (

@Id
@GeneratedValue(strategy= GenerationType.AUTO)
override val id: Int? = null,

override val name: String,

@Enumerated(EnumType.STRING)
val rating: RatingsEnum?,

@OneToMany(mappedBy = "hotel", cascade = [CascadeType.ALL])
val rooms : List<Room> = ArrayList<Room>(),

// below code seems to be the issue

@ElementCollection
@Enumerated(EnumType.STRING)
val amenities: Set<AmenitiesEnum> = EnumSet.noneOf(AmenitiesEnum::class.java)
) : Accommodation, Serializable

我用演示值初始化数据库

CREATE TABLE hotels (
id int UNSIGNED NOT NULL AUTO_INCREMENT,
name varchar(125) NOT NULL,
description TEXT NOT NULL,
type ENUM('BED_AND_BREAKFAST', 'SPA') DEFAULT NULL,
rating ENUM('HALF_STAR', 'ONE_STAR', 'TWO_STARS', 'THREE_STARS', 'FOUR_STARS', 'FIVE_STARS') DEFAULT NULL,
amenities SET('POOL','PETS_ALLOWED','PARKING_INCLUDED', 'BAR', 'DISABLED_FRIENDLY', 'FREE_DRINKS', 'WIFI') DEFAULT NULL,
PRIMARY KEY (id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO hotels VALUES (NULL, 'First ever Hotel', 'First hotel ever build. Amazing.', null, 'one_star', 'bar');
SELECT @last := LAST_INSERT_ID();
INSERT INTO rooms VALUES
(NULL, @last, 2, 'QUEEN', 'PETS_ALLOWED'),
(NULL, @last, 1, 'SINGLE', 'PETS_ALLOWED'),
(NULL, @last, 3, 'KING', 'PETS_ALLOWED');

当我启动应用程序并对数据库发起查询时，我得到

... binding parameter [1] as [INTEGER] - [1]
... SQL Error: 1146, SQLState: 42S02
... Table '_database_.hotel_amenities' doesn't exist
... Resolved [org.springframework.http.converter.HttpMessageNotWritableException: Could not write JSON: could not extract ResultSet; nested exception is com.fasterxml.jackson.databind.JsonMappingException: could not extract ResultSet (through reference chain: java.util.ArrayList[0]->com.katanox.hotelbooking.domain.room.value.Room["hotel"]->com._.hotelbooking.domain.hotel.value.Hotel["amenities"])]

为什么 MySQL 状态Table 'd035b60f.hotel_amenities' doesn't exist，我该如何解决这个问题？将类型更改为 List 而不是 EnumSet 会导致相同的错误。

更新

查看调试，Hibernate 将其转换为：

select
    amenities0_.hotel_id as hotel_id1_0_0_,
    amenities0_.amenities as amenitie2_0_0_ 
from
    hotel_amenities amenities0_ 
where
    amenities0_.hotel_id=?

我仍在寻找问题的解决方案，任何提示将不胜感激。

标签： mysqlspringhibernatekotlinjdbc

解决方案

为此，您将需要一个自定义 JPA 属性转换器，将其映射到字符串。据我了解，在 JDBC 级别上，这必须表示为逗号分隔的字符串。有关如何使用它的示例，请参阅文档：https ://docs.oracle.com/javaee/7/api/javax/persistence/Convert.html

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