How to map a map JSON column to Java Object with JPA

Question

We have a big table with a lot of columns. After we moved to MySQL Cluster, the table cannot be created because of:

ERROR 1118 (42000): Row size too large. The maximum row size for the used table type, not counting BLOBs, is 14000. This includes storage overhead, check the manual. You have to change some columns to TEXT or BLOBs

As an example:

@Entity @Table (name = "appconfigs", schema = "myproject")
public class AppConfig implements Serializable
{
    @Id @Column (name = "id", nullable = false)
    @GeneratedValue (strategy = GenerationType.IDENTITY)
    private int id;

    @OneToOne @JoinColumn (name = "app_id")
    private App app;

    @Column(name = "param_a")
    private ParamA parama;

    @Column(name = "param_b")
    private ParamB paramb;
}

It's a table for storing configuration parameters. I was thinking that we can combine some columns into one and store it as JSON object and convert it to some Java object.

For example:

@Entity @Table (name = "appconfigs", schema = "myproject")
public class AppConfig implements Serializable
{
    @Id @Column (name = "id", nullable = false)
    @GeneratedValue (strategy = GenerationType.IDENTITY)
    private int id;

    @OneToOne @JoinColumn (name = "app_id")
    private App app;

    @Column(name = "params")
    //How to specify that this should be mapped to JSON object?
    private Params params;
}

Where we have defined:

public class Params implements Serializable
{
    private ParamA parama;
    private ParamB paramb;
}

By using this we can combine all columns into one and create our table. Or we can split the whole table into several tables. Personally I prefer the first solution.

Anyway my question is how to map the Params column which is text and contains JSON string of a Java object?

Answer 1

You can use a JPA converter to map your Entity to the database. Just add an annotation similar to this one to your params field:

@Convert(converter = JpaConverterJson.class)

and then create the class in a similar way (this converts a generic Object, you may want to specialize it):

@Converter(autoApply = true)
public class JpaConverterJson implements AttributeConverter<Object, String> {

  private final static ObjectMapper objectMapper = new ObjectMapper();

  @Override
  public String convertToDatabaseColumn(Object meta) {
    try {
      return objectMapper.writeValueAsString(meta);
    } catch (JsonProcessingException ex) {
      return null;
      // or throw an error
    }
  }

  @Override
  public Object convertToEntityAttribute(String dbData) {
    try {
      return objectMapper.readValue(dbData, Object.class);
    } catch (IOException ex) {
      // logger.error("Unexpected IOEx decoding json from database: " + dbData);
      return null;
    }
  }

}

That's it: you can use this class to serialize any object to json in the table.

Answer 2

The JPA AttributeConverter is way too limited to map JSON object types, especially if you want to save them as JSON binary.

You don’t have to create a custom Hibernate Type to get JSON support, All you need to do is use the Hibernate Types OSS project.

For instance, if you're using Hibernate 5.2 or newer versions, then you need to add the following dependency in your Maven pom.xml configuration file:

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

Now, you need to declare the new type either at the entity attribute level or, even better, at the class level in a base class using @MappedSuperclass:

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

And the entity mapping will look like this:

@Type(type = "json")
@Column(columnDefinition = "json")
private Location location;

If you're using Hibernate 5.2 or later, then the JSON type is registered automatically by MySQL57Dialect.

Otherwise, you need to register it yourself:

public class MySQLJsonDialect extends MySQL55Dialect {

    public MySQLJsonDialect() {
        super();
        this.registerColumnType(Types.JAVA_OBJECT, "json");
    }
}

And, set the hibernate.dialect Hibernate property to use the fully-qualified class name of the MySQLJsonDialect class you have just created.

Answer 3

If you need to map json type property to json format when responding to the client (e.g. rest API response), add @JsonRawValue as the following:

@Column(name = "params", columnDefinition = "json")
@JsonRawValue
private String params;

This might not do the DTO mapping for server-side use, but the client will get the property properly formatted as json.

Answer 4

It is simple

@Column(name = "json_input", columnDefinition = "json")
private String field;

and in mysql database your column 'json_input' json type

Answer 5

There is a workaround for those don't want write too much code.

Frontend -> Encode your JSON Object to string base64 in POST method, decode it to json in GET method

In POST Method
data.components = btoa(JSON.stringify(data.components));

In GET
data.components = JSON.parse(atob(data.components))

Backend -> In your JPA code, change the column to String or BLOB, no need Convert.

@Column(name = "components", columnDefinition = "json")
private String components;

Answer 6

In this newer version of spring boot and MySQL below code is enough

@Column( columnDefinition = "json" )
private String string;

I was facing quotes issue so I commented below line in my project

#spring.jpa.properties.hibernate.globally_quoted_identifiers=true

Answer 7

If you are using JPA version 2.1 or higher you can go with this case. Link Persist Json Object

public class HashMapConverter implements AttributeConverter<Map<String, Object>, String> {

@Override
public String convertToDatabaseColumn(Map<String, Object> customerInfo) {

    String customerInfoJson = null;
    try {
        customerInfoJson = objectMapper.writeValueAsString(customerInfo);
    } catch (final JsonProcessingException e) {
        logger.error("JSON writing error", e);
    }

    return customerInfoJson;
}

@Override
public Map<String, Object> convertToEntityAttribute(String customerInfoJSON) {

    Map<String, Object> customerInfo = null;
    try {
        customerInfo = objectMapper.readValue(customerInfoJSON, 
            new TypeReference<HashMap<String, Object>>() {});
    } catch (final IOException e) {
        logger.error("JSON reading error", e);
    }

    return customerInfo;
}

}

A standard JSON object would represent those attributes as a HashMap:

@Convert(converter = HashMapConverter.class) private Map<String, Object> entityAttributes;

Answer 8

I had a similar problem, and solved it by using @Externalizer annotation and Jackson to serialize/deserialize data (@Externalizer is OpenJPA-specific annotation, so you have to check with your JPA implementation similar possibility).

@Persistent
@Column(name = "params")
@Externalizer("toJSON")
private Params params;

Params class implementation:

public class Params {
    private static final ObjectMapper mapper = new ObjectMapper();

    private Map<String, Object> map;

    public Params () {
        this.map = new HashMap<String, Object>();
    }

    public Params (Params another) {
        this.map = new HashMap<String, Object>();
        this.map.putAll(anotherHolder.map);
    }

    public Params(String string) {
        try {
            TypeReference<Map<String, Object>> typeRef = new TypeReference<Map<String, Object>>() {
            };
            if (string == null) {
                this.map = new HashMap<String, Object>();
            } else {
                this.map = mapper.readValue(string, typeRef);
            }
        } catch (IOException e) {
            throw new PersistenceException(e);
        }
    }

    public String toJSON() throws PersistenceException {
        try {
            return mapper.writeValueAsString(this.map);
        } catch (IOException e) {
            throw new PersistenceException(e);
        }
    }

    public boolean containsKey(String key) {
        return this.map.containsKey(key);
    }

    // Hash map methods
    public Object get(String key) {
        return this.map.get(key);
    }

    public Object put(String key, Object value) {
        return this.map.put(key, value);
    }

    public void remove(String key) {
        this.map.remove(key);
    }

    public Object size() {
        return map.size();
    }
}

HTH

Answer 9

Make use of hibernate json types with this dependency in your pom.xml

    <dependency>
        <groupId>com.vladmihalcea</groupId>
        <artifactId>hibernate-types-52</artifactId>
        <version>2.10.4</version>
    </dependency>

and defining column with json data type in database as follows:

@Type(type = "json")
@Column(columnDefinition = "json")
private JsonNode column;

Here, columnDefinition attribute in the @Column annotation to tell Hibernate that the score column should be mapped to a JSON data type column in the database.