如何使用JPA和Hibernate将JSON列映射到MySQL实体属性
如何使用JPA和Hibernate将JSON列映射到MySQL实体属性
提问于 2017-06-01 20:56:44
我将MySQL列声明为JSON类型，但在将其与Jpa/Hibernate进行映射时遇到了问题。我在后端使用Spring Boot。
下面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不为空；
回答 6
Stack Overflow用户
回答已采纳
发布于 2018-10-11 04:07:57
我更喜欢这样做：
- 创建从映射到字符串的转换器(属性转换器)，反之亦然。
- 使用映射在域(实体)类
中映射mysql JSON列类型
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
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。
Stack Overflow用户
发布于 2019-09-15 01:58:05
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：
com.vladmihalcea hibernate-types-52 ${hibernate-Typees.Version}
有关更多信息，请查看Hibernate Types open-source project。
现在，来解释一下它是如何工作的。
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
}注意上面代码片段中的两件事：
- the
String
属性处理的
-
@TypeDef用于定义新的自定义Hibernate类型，它具有一个映射为json的json列类型
就这样!
现在，如果您保存一个实体：
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所有代码都可以在上找到。
Stack Overflow用户
发布于 2017-12-14 03:58:38
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;页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/44308167
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