Java 如何在 JPA 中使用 Postgres JSONB 数据类型？

How to use Postgres JSONB datatype with JPA?

提问by justcode

Im not finding a way to map the JSON and JSONB datatypes from PostgreSQL using JPA (EclipseLink). Is some one using this datatypes with JPA and can give me some working examples?

我没有找到使用 JPA (EclipseLink) 从 PostgreSQL 映射 JSON 和 JSONB 数据类型的方法。有人在 JPA 中使用这种数据类型，可以给我一些工作示例吗？

采纳答案by justcode

All the answers helped me to reach the final solution that is ready for JPA and not EclipseLink or Hibernate specifically.

所有的答案都帮助我找到了为 JPA 而不是 EclipseLink 或 Hibernate 准备好的最终解决方案。

import com.fasterxml.Hymanson.core.type.TypeReference;
import com.fasterxml.Hymanson.databind.ObjectMapper;
import java.io.IOException;
import javax.json.Json;
import javax.json.JsonObject;
import javax.persistence.Converter;
import org.postgresql.util.PGobject;

@Converter(autoApply = true)
public class JsonConverter implements javax.persistence.AttributeConverter<JsonObject, Object> {

  private static final long serialVersionUID = 1L;
  private static ObjectMapper mapper = new ObjectMapper();

  @Override
  public Object convertToDatabaseColumn(JsonObject objectValue) {
    try {
      PGobject out = new PGobject();
      out.setType("json");
      out.setValue(objectValue.toString());
      return out;
    } catch (Exception e) {
      throw new IllegalArgumentException("Unable to serialize to json field ", e);
    }
  }

  @Override
  public JsonObject convertToEntityAttribute(Object dataValue) {
    try {
      if (dataValue instanceof PGobject && ((PGobject) dataValue).getType().equals("json")) {
        return mapper.reader(new TypeReference<JsonObject>() {
        }).readValue(((PGobject) dataValue).getValue());
      }
      return Json.createObjectBuilder().build();
    } catch (IOException e) {
      throw new IllegalArgumentException("Unable to deserialize to json field ", e);
    }
  }
}

回答by Tobb

Edit: I see now that this is pretty much Hibernatedependant. But perhaps you can find something similar for EclipseLink..

编辑：我现在看到这非常Hibernate依赖。但也许你可以找到类似的东西EclipseLink......

I'll just add what I have as an answer, it originates from another SO answer but whatever.. This will map jsonbto JsonObjectof Google gson, but you can change it to something else if needed. To change to something else, change nullSafeGet, nullSafeSetand deepCopymethods.

我只会添加我所拥有的作为答案，它源自另一个 SO 答案，但无论如何.. 这将映射jsonb到JsonObjectGoogle gson，但如果需要，您可以将其更改为其他内容。要改变别的东西，改变nullSafeGet，nullSafeSet和deepCopy方法。

public class JsonbType implements UserType {

    @Override
    public int[] sqlTypes() {
        return new int[] { Types.JAVA_OBJECT };
    }

    @Override
    public Class<JsonObject> returnedClass() {
        return JsonObject.class;
    }

    @Override
    public boolean equals(final Object x, final Object y) {
        if (x == y) {
            return true;
        }
        if (x == null || y == null) {
            return false;
        }
        return x.equals(y);
    }

    @Override
    public int hashCode(final Object x) {
        if (x == null) {
            return 0;
        }

        return x.hashCode();
    }

    @Nullable
    @Override
    public Object nullSafeGet(final ResultSet rs,
                              final String[] names,
                              final SessionImplementor session,
                              final Object owner) throws SQLException {
        final String json = rs.getString(names[0]);
        if (json == null) {
            return null;
        }

        final JsonParser jsonParser = new JsonParser();
        return jsonParser.parse(json).getAsJsonObject();
    }

    @Override
    public void nullSafeSet(final PreparedStatement st,
                            final Object value,
                            final int index,
                            final SessionImplementor session) throws SQLException {
        if (value == null) {
            st.setNull(index, Types.OTHER);
            return;
        }

        st.setObject(index, value.toString(), Types.OTHER);
    }

    @Nullable
    @Override
    public Object deepCopy(@Nullable final Object value) {
        if (value == null) {
            return null;
        }
        final JsonParser jsonParser = new JsonParser();
        return jsonParser.parse(value.toString()).getAsJsonObject();
    }

    @Override
    public boolean isMutable() {
        return true;
    }

    @Override
    public Serializable disassemble(final Object value) {
        final Object deepCopy = deepCopy(value);

        if (!(deepCopy instanceof Serializable)) {
            throw new SerializationException(
                    String.format("deepCopy of %s is not serializable", value), null);
        }

        return (Serializable) deepCopy;
    }

    @Nullable
    @Override
    public Object assemble(final Serializable cached, final Object owner) {
        return deepCopy(cached);
    }

    @Nullable
    @Override
    public Object replace(final Object original, final Object target, final Object owner) {
        return deepCopy(original);
    }
}

To use this, do:

要使用它，请执行以下操作：

public class SomeEntity {

    @Column(name = "jsonobject")
    @Type(type = "com.myapp.JsonbType") 
    private JsonObject jsonObject;

In addition, you need to set your dialect to indicate that JAVA_OBJECT= jsonb:

此外，您需要设置您的方言以指示JAVA_OBJECT= jsonb：

registerColumnType(Types.JAVA_OBJECT, "jsonb");

回答by coladict

I think I found an analogy to Hibernate's UserType for EclipseLink.

我想我找到了一个类比 Hibernate 的 UserType for EclipseLink。

http://www.eclipse.org/eclipselink/documentation/2.6/jpa/extensions/annotations_ref.htm#CHDEHJEB

http://www.eclipse.org/eclipselink/documentation/2.6/jpa/extensions/annotations_ref.htm#CHDEHJEB

You have to make a class that implements org.eclipse.persistence.mappings.converters.Converterand does the conversion for you, then use the @Convertannotation on every field where you are using that type.

您必须创建一个实现org.eclipse.persistence.mappings.converters.Converter并为您进行转换的类，然后@Convert在您使用该类型的每个字段上使用注释。