[英]How to map a MySQL JSON column to a Java entity property using JPA and Hibernate
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。
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;
该程序返回一个错误并告诉我我无法映射该列。
在 mysql 表中,列定义为:
json_value JSON 非空;
我更喜欢这样做:
代码如下。
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(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<>();
...
这个解决方案非常适合我。
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:
@TypeDef
用于定义一个新的自定义Hibernate类型, json
这是由处理JsonType
properties
属性有一个json
列类型,它被映射为一个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/\"" +
"}"
);
Hibernate 为您处理UPDATE
语句:
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 上提供的所有代码。
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;
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;
}
}
}
对于任何人都无法制作@J。 王回答工作:
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>
并将这一行添加到实体
@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。
在 Kotlin 中,上述建议的以下变化/组合对我有用:
@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
}
}
}
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