如何使用JPA和Hibernate将MySQL JSON列映射到Java实体属性-

Question

我将MySQL列声明为JSON类型，并且在使用Jpa / Hibernate映射时遇到问题。我在后端使用Spring Boot。

这是我的代码的一小部分：

@Entity
@Table(name = "some_table_name")
public class MyCustomEntity implements Serializable {

private static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;

@Column(name = "json_value")
private JSONArray jsonValue;

程序返回一个错误，并告诉我无法映射该列。

在mysql表中，该列定义为：

json_value JSON NOT NULL;

Answer 1

因为任何人都无法成为@J。王功回答：

尝试添加此依赖项（适用于 hibernate 5.1 和 5.0，其他版本请在此处查看）

<dependency>
    <groupId>com.vladmihalcea</groupId>
    <artifactId>hibernate-types-5</artifactId>
    <version>1.2.0</version>
</dependency>

并将这一行添加到实体中

@TypeDef(name = "json", typeClass = JsonStringType.class)

实体类的完整版本：

@Entity
@Table(name = "some_table_name")
@TypeDef(name = "json", typeClass = JsonStringType.class)
public class MyCustomEntity implements Serializable {

   private static final long serialVersionUID = 1L;

   @Id
   @GeneratedValue(strategy = GenerationType.AUTO)
   private Long id;

   @Type( type = "json" )
   @Column( columnDefinition = "json" )
   private List<String> jsonValue;
}

我使用 spring boot 1.5.9 和 hibernate-types-5 1.2.0 测试代码。

Answer 2

您不必手动创建所有这些类型，您只需使用以下依赖项通过 Maven Central 获取它们：

<dependency>
    <groupId>com.vladmihalcea</groupId>
    <artifactId>hibernate-types-52</artifactId>
    <version>${hibernate-types.version}</version> 
</dependency> 

有关更多信息，请查看Hibernate Types 开源项目。

现在，解释这一切是如何运作的。

假设您有以下实体：

@Entity(name = "Book")
@Table(name = "book")
@TypeDef(
    name = "json", 
    typeClass = JsonType.class
)
public class Book {
 
    @Id
    @GeneratedValue
    private Long id;
 
    @NaturalId
    private String isbn;
 
    @Type(type = "json")
    @Column(columnDefinition = "json")
    private String properties;
 
    //Getters and setters omitted for brevity
}

请注意上面代码片段中的两件事：

• 的@TypeDef用于定义新的自定义休眠类型，json这是由处理JsonType
• 该properties属性有一个json列类型，它被映射为String

就是这样！

现在，如果您保存实体：

Book book = new Book();
book.setIsbn("978-9730228236");
book.setProperties(
    "{" +
    "   \"title\": \"High-Performance Java Persistence\"," +
    "   \"author\": \"Vlad Mihalcea\"," +
    "   \"publisher\": \"Amazon\"," +
    "   \"price\": 44.99" +
    "}"
);
 
entityManager.persist(book);

Hibernate 将生成以下 SQL 语句：

INSERT INTO
    book 
(
    isbn, 
    properties, 
    id
) 
VALUES
(
    '978-9730228236', 
    '{"title":"High-Performance Java Persistence","author":"Vlad Mihalcea","publisher":"Amazon","price":44.99}',  
    1
)

您还可以重新加载并修改它：

Book book = entityManager
    .unwrap(Session.class)
    .bySimpleNaturalId(Book.class)
    .load("978-9730228236");
     
book.setProperties(
    "{" +
    "   \"title\": \"High-Performance Java Persistence\"," +
    "   \"author\": \"Vlad Mihalcea\"," +
    "   \"publisher\": \"Amazon\"," +
    "   \"price\": 44.99," +
    "   \"url\": \"https://www.amazon.com/High-Performance-Java-Persistence-Vlad-Mihalcea/dp/973022823X/\"" +
    "}"
);

HibernateUPDATE为您处理声明：

SELECT  b.id AS id1_0_
FROM    book b
WHERE   b.isbn = '978-9730228236'
 
SELECT  b.id AS id1_0_0_ ,
        b.isbn AS isbn2_0_0_ ,
        b.properties AS properti3_0_0_
FROM    book b
WHERE   b.id = 1    
 
UPDATE
    book 
SET
    properties = '{"title":"High-Performance Java Persistence","author":"Vlad Mihalcea","publisher":"Amazon","price":44.99,"url":"https://www.amazon.com/High-Performance-Java-Persistence-Vlad-Mihalcea/dp/973022823X/"}'
WHERE
    id = 1

GitHub 上提供的所有代码。

Answer 3

我更喜欢这样：

• 创建从Map到String的转换器（属性转换器），反之亦然。
• 使用Map映射域（实体）类中的mysql JSON列类型

代码在下面。

JsonToMapConverted.java

@Converter
public class JsonToMapConverter 
                    implements AttributeConverter<String, Map<String, Object>> 
{
    private static final Logger LOGGER = LoggerFactory.getLogger(JsonToMapConverter.class);

    @Override
    @SuppressWarnings("unchecked")
    public Map<String, Object> convertToDatabaseColumn(String attribute)
    {
        if (attribute == null) {
           return new HashMap<>();
        }
        try
        {
            ObjectMapper objectMapper = new ObjectMapper();
            return objectMapper.readValue(attribute, HashMap.class);
        }
        catch (IOException e) {
            LOGGER.error("Convert error while trying to convert string(JSON) to map data structure.");
        }
        return new HashMap<>();
    }

    @Override
    public String convertToEntityAttribute(Map<String, Object> dbData)
    {
        try
        {
            ObjectMapper objectMapper = new ObjectMapper();
            return objectMapper.writeValueAsString(dbData);
        }
        catch (JsonProcessingException e)
        {
            LOGGER.error("Could not convert map to json string.");
            return null;
        }
    }
}

域（实体映射）类的一部分

...

@Column(name = "meta_data", columnDefinition = "json")
@Convert(attributeName = "data", converter = JsonToMapConverter.class)
private Map<String, Object> metaData = new HashMap<>();

...

该解决方案非常适合我。

Answer 4

如果您的 json 数组中的值是简单的字符串，您可以这样做：

@Type( type = "json" )
@Column( columnDefinition = "json" )
private String[] jsonValue;

Answer 5

Heril Muratovic 的回答很好，但我认为JsonToMapConverter应该实现AttributeConverter<Map<String, Object>, String>，而不是AttributeConverter<String, Map<String, Object>>。这是对我有用的代码

@Slf4j
@Converter
public class JsonToMapConverter implements AttributeConverter<Map<String, Object>, String> {
    @Override
    @SuppressWarnings("unchecked")
    public Map<String, Object> convertToEntityAttribute(String attribute) {
        if (attribute == null) {
            return new HashMap<>();
        }
        try {
            ObjectMapper objectMapper = new ObjectMapper();
            return objectMapper.readValue(attribute, HashMap.class);
        } catch (IOException e) {
            log.error("Convert error while trying to convert string(JSON) to map data structure.", e);
        }
        return new HashMap<>();
    }

    @Override
    public String convertToDatabaseColumn(Map<String, Object> dbData) {
        try {
            ObjectMapper objectMapper = new ObjectMapper();
            return objectMapper.writeValueAsString(dbData);
        } catch (JsonProcessingException e) {
            log.error("Could not convert map to json string.", e);
            return null;
        }
    }
}