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

How to map a MySQL JSON column to a Java entity property using JPA and Hibernate

提问by Heril Muratovic

I have MySQL column declared as type JSONand I have problems to map it with Jpa/Hibernate. I'm using Spring Boot on back-end.

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

Here is small part of my code:

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;

The program returns me an error and tells me that I can't map the column.

该程序返回一个错误并告诉我我无法映射该列。

In mysql table the column is defined as:

在 mysql 表中，列定义为：

json_value JSON NOT NULL;

json_value JSON 非空；

采纳答案by Heril Muratovic

I prefer to do this way:

我更喜欢这样做：

• Creating converter (attribute converter) from Map to String and vice versa.
• Using Map to map mysql JSON column type in domain (entity) class

• 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

The code is bellow.

代码如下。

JsonToMapConverted.java

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;
        }
    }
}

Part of domain (entity-mapping) class

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<>();

...

This solution perfectly works for me.

这个解决方案非常适合我。

回答by Songle Bao

Heril Muratovic's answeris good, but I think the JsonToMapConvertershould implement AttributeConverter<Map<String, Object>, String>, not AttributeConverter<String, Map<String, Object>>. Here is the code that works for me

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;
        }
    }
}

回答by Vlad Mihalcea

As I explained in this article, it's very easy to persist JSON objects using Hibernate.

正如我在本文中解释的那样，使用 Hibernate 保留 JSON 对象非常容易。

You don't have to create all these types manually, you can simply get them via Maven Central using the following dependency:

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

For more info, check out the hibernate-types open-source project.

您不必手动创建所有这些类型，您可以使用以下依赖项通过 Maven Central 轻松获取它们：

<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。

Now, to explain how it all works.

现在，解释这一切是如何运作的。

Assuming you have the following entity:

假设您有以下实体：

@Entity(name = "Book")
@Table(name = "book")
@TypeDef(
    name = "json", 
    typeClass = JsonStringType.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
}

Notice two things in the code snippet above:

请注意上面代码片段中的两件事：

• the @TypeDefis used to define a new custom Hibernate Type, jsonwhich is handled by the JsonStringType
• the propertiesattribute has a jsoncolumn type and it's mapped as a String

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

That's it!

而已！

Now, if you save an entity:

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 is going to generate the following SQL statement:

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
)

And you can also load it back and modify it:

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/\"" +
    "}"
);

Hibernate taking caare of the UPDATEstatement for you:

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

All code avilable on GitHub.

所有代码都可在GitHub 上使用。

回答by J. Wang

If the values inside your json array are simple strings you can do this:

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;

回答by user1686407

For anyone can't make @J. Wang answer work :

对于任何人都无法制作@J。王回答工作：

Try add this dependency(it's for hibernate 5.1 and 5.0, other version check here)

尝试添加此依赖项（适用于 hibernate 5.1 和 5.0，其他版本在此处检查）

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

And add this line to the entity

并将这一行添加到实体

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

So full version of the entity 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;
}

I test the code with spring boot 1.5.9 and hibernate-types-5 1.2.0 .

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