SpringBoot：持久化嵌套JSON [使用spring-boot-starter-data-rest + sql]

huangapple
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英文:

SpringBoot: Persist nested JSON [using spring-boot-starter-data-rest + sql]

问题

以下是您提供的内容的翻译部分：

我有以下的POST请求：

  1. {
  2.     "name": "Peter",
  3.     "lastName": "Smith",
  4.     "contact": {
  5.         "phone": "12345679",
  6.         "email": "peter@smith.com"
  7.     }
  8. }

我想要将其存储在SQL数据库中，如下所示：

| id (int) | name (varchar) | lastName (varchar) | contact (JSON) |

我正在使用spring-boot-starter-data-rest，因此我只有UserRepository和User实体，其中有一个嵌入式属性contact。

User.java

  1. @Entity
  2. @Table(name="user")
  3. public class User {
  5.     @Id
  6.     @GeneratedValue(strategy = GenerationType.IDENTITY)
  7.     private Long id;
  9.     @Column(name = "name")
  10.     private String title;
  12.     @Column(name = "lastName")
  13.     private String lastName;
  15.     @Embedded
  16.     @Column(name = "contact")
  17.     private Contact contact;
  19. }

Contact.java

  1. @Embeddable
  2. public class Contact {
  3.    private String phone;
  4.    private String email;
  5. }

UserRepository.java

  1. public interface UserRepository extends JpaRepository<User, Integer> {
  2.     //     
  3. }

如果我进行POST请求，会出现错误，因为（猜测）我没有将Contact转换为JSON。

我已经尝试添加了@Convert(converter = HashMapConverter.class)，但是出现了错误。

HashMapConverter

  1. public class HashMapConverter implements AttributeConverter<Object, String> {
  3.     private static final ObjectMapper om = new ObjectMapper();
  5.     @Override
  6.     public String convertToDatabaseColumn(Object attribute) {
  7.         try {
  8.             return om.writeValueAsString(attribute);
  9.         } catch (JsonProcessingException ex) {
  10.             //log.error("Error while transforming Object to a text datatable column as json string", ex);
  11.             return null;
  12.         }
  13.     }
  15.     @Override
  16.     public Object convertToEntityAttribute(String dbData) {
  17.         try {
  18.             return om.readValue(dbData, Object.class);
  19.         } catch (IOException ex) {
  20.             //log.error("IO exception while transforming json text column in Object property", ex);
  21.             return null;
  22.         }
  23.     }
  25. }
英文:

I have the following POST request:

  1. {
  2. 	&quot;name&quot;: &quot;Peter&quot;,
  3.     &quot;lastName&quot;: &quot;Smith&quot;, 
  4. 	&quot;contact&quot;: {
  5. 		&quot;phone&quot;:&quot;12345679&quot;,
  6. 		&quot;email&quot;: &quot;peter@smith.com&quot;
  7. 	}
  8. }

And I would like to store that in a SQL DB as follow:

| id (int) | name (varchar) | lastName (varchar) | contact (JSON) |

I'm using spring-boot-starter-data-rest so I only have the UserRepository and User Entity, which has an Embedded property contact

User.java

  1. @Entity
  2. @Table(name=&quot;user&quot;)
  3. public class User {
  5.     @Id
  6.     @GeneratedValue(strategy = GenerationType.IDENTITY)
  7.     private Long id;
  9.     @Column(name = &quot;name&quot;)
  10.     private String title;
  12.     @Column(name = &quot;lastName&quot;)
  13.     private String lastName;
  15.     @Embedded
  16.     @Column(name = &quot;contact&quot;)
  17.     private Contact contact;
  19.    
  20.  }

Contact.java

  1. @Embeddable
  2. public class Contact {
  3.    private String phone;
  4.    private String email;
  5. }

UserRepository.java

  1. public interface UserRepository extends JpaRepository&lt;User, Integer&gt; {
  2.     //     
  3. }

If I make a POST request I get an error, because (guess) I'm not converting Contact to JSON.

I've already tried adding a @Convert(converter = HashMapConverter.class) but I get an error.

HashMapConverter

  1. public class HashMapConverter implements AttributeConverter&lt;Object, String&gt; {
  3.     private static final ObjectMapper om = new ObjectMapper();
  5.     @Override
  6.     public String convertToDatabaseColumn(Object attribute) {
  7.         try {
  8.             return om.writeValueAsString(attribute);
  9.         } catch (JsonProcessingException ex) {
  10.             //log.error(&quot;Error while transforming Object to a text datatable column as json string&quot;, ex);
  11.             return null;
  12.         }
  13.     }
  15.     @Override
  16.     public Object convertToEntityAttribute(String dbData) {
  17.         try {
  18.             return om.readValue(dbData, Object.class);
  19.         } catch (IOException ex) {
  20.             //log.error(&quot;IO exception while transforming json text column in Object property&quot;, ex);
  21.             return null;
  22.         }
  23.     }
  25. }

答案1

得分: 0

你需要为Contact创建一个实体，然后在两者之间创建一对一的关系。查看这个示例。

英文:

You need to create an entity for Contact and then create a one to one relationship between the two. Check out this example.

答案2

得分: 0

  1. 我已经有一个用于存储 JSON 字段的相同情况，它完美地工作。请尝试：
  3. 在 pom.xml 中添加依赖：
  5.     <dependency>
  6.         <groupId>com.vladmihalcea</groupId>
  7.         <artifactId>hibernate-types-52</artifactId>
  8.         <version>2.9.7</version>
  9.     </dependency>
  11. 在 User 类中进行编辑：
  13.     @Entity
  14.     @TypeDef(name = "json", typeClass = JsonStringType.class)
  15.     public class User {
  16.         // 其他字段在这里
  17.         @Type(type = "json")
  18.         @Column(columnDefinition = "json")
  19.         private Contact contact;
  20.         // getter 和 setter 方法
  21.     }
  23. 当然，您的数据库应支持 json 类型。例如，对于 MariaDB，您可以参考 https://mariadb.com/kb/en/json-data-type/
英文:

i've got the same case for storing json field which is working perfectly. Please try :

Add dependency to pom.xml :

  1. &lt;dependency&gt;
  2.     &lt;groupId&gt;com.vladmihalcea&lt;/groupId&gt;
  3.     &lt;artifactId&gt;hibernate-types-52&lt;/artifactId&gt;
  4.     &lt;version&gt;2.9.7&lt;/version&gt;
  5. &lt;/dependency&gt;

Edit User class within :

  1. @Entity
  2. @TypeDef(name = &quot;json&quot;, typeClass = JsonStringType.class)
  3. public class User {
  4.     // other field here
  5.     @Type(type = &quot;json&quot;)
  6.     @Column(columnDefinition = &quot;json&quot;)
  7.     private Contact contact;
  8.     // getters, setters
  9. }

Of course your database should support json type. For MariaDB for example you can refer to https://mariadb.com/kb/en/json-data-type/

huangapple
  • 本文由 发表于 2020年4月5日 03:22:51
  • 转载请务必保留本文链接：https://java.coder-hub.com/61033575.html
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