如何使用 JPA 和 Hibernate 將 MySQL JSON 列映射到 Java 實體屬性

Question

我將 MySQL 列聲明為JSON類型，並且在將其映射到 Jpa/Hibernate 時遇到問題。 我在后端使用 Spring Boot。

這是我的代碼的一小部分：

@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;

該程序返回一個錯誤並告訴我我無法映射該列。

在 mysql 表中，列定義為：

json_value JSON 非空；

Answer 1

我更喜歡這樣做：

• 創建從 Map 到 String 的轉換器（屬性轉換器），反之亦然。
• 使用Map映射域（實體）類中的mysql JSON列類型

代碼如下。

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（實體映射）類的一部分

...

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

...

這個解決方案非常適合我。

Answer 2

您不必手動創建所有這些類型，您只需使用以下依賴項通過 Maven Central 獲取它們：

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

有關更多信息，請查看Hibernate Types 開源項目。

現在，解釋這一切是如何運作的。

假設您有以下實體：

@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
}

請注意上面代碼片段中的兩件事：

• 所述@TypeDef用於定義一個新的自定義Hibernate類型， json這是由處理JsonType
• properties屬性有一個json列類型，它被映射為一個String

就是這樣！

現在，如果您保存實體：

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 將生成以下 SQL 語句：

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

您還可以重新加載並修改它：

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 為您處理UPDATE語句：

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 上提供的所有代碼。

Answer 3

如果您的 json 數組中的值是簡單的字符串，您可以這樣做：

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

Answer 4

Heril Muratovic 的回答很好，但我認為JsonToMapConverter應該實現AttributeConverter<Map<String, Object>, String> ，而不是AttributeConverter<String, Map<String, Object>> 。 這是對我有用的代碼

@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;
        }
    }
}

Answer 5

對於任何人都無法制作@J。 王回答工作：

嘗試添加此依賴項（適用於 hibernate 5.1 和 5.0，其他版本在此處檢查）

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

並將這一行添加到實體

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

實體類的完整版本：

@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 測試代碼。

Answer 6

在 Kotlin 中，上述建議的以下變化/組合對我有用：

@Entity
@Table(name = "product_menu")
@TypeDef(name = "json", typeClass = JsonStringType::class)
data class ProductMenu(

    @Type(type = "json")
    @Column(name = "menu_json", columnDefinition = "json")
    @Convert(attributeName = "menuJson", converter = JsonToMapConverter::class)
    val menuJson: HashMap<String, Any> = HashMap()

) : Serializable



import com.fasterxml.jackson.core.JsonProcessingException
import com.fasterxml.jackson.databind.ObjectMapper
import org.slf4j.LoggerFactory
import java.io.IOException
import javax.persistence.AttributeConverter

class JsonToMapConverter : AttributeConverter<String, HashMap<String, Any>> {

    companion object {
        private val LOGGER = LoggerFactory.getLogger(JsonToMapConverter::class.java)
    }

    override fun convertToDatabaseColumn(attribute: String?): HashMap<String, Any> {
        if(attribute == null) {
            return HashMap()
        }
        try {
            val objectMapper = ObjectMapper()
            @Suppress("UNCHECKED_CAST")
            return objectMapper.readValue(attribute, HashMap::class.java) as HashMap<String, Any>
        } catch (e: IOException) {
            LOGGER.error("Convert error while trying to convert string(JSON) to map data structure.")
        }
        return HashMap()
    }

    override fun convertToEntityAttribute(dbData: HashMap<String, Any>?): String? {
        return try {
            val objectMapper = ObjectMapper()
            objectMapper.writeValueAsString(dbData)
        } catch (e: JsonProcessingException) {
            LOGGER.error("Could not convert map to json string.")
            return null
        }
    }
}