Вставьте и получите JSON DATA IN Sequelize и postgres - PullRequest
Новая
       21

Вставьте и получите JSON DATA IN Sequelize и postgres

0 голосов
/ 16 марта 2020

В модели 

 {
  data:{
  type: Sequelize.JSONB
  }
}

Я новичок, чтобы продолжить. Пожалуйста, кто-нибудь подскажет, как вставить и получить данные в sequelize Заранее спасибо ...

1 Ответ

0 голосов
/ 17 марта 2020

Вот пример: 

import { sequelize } from '../../db';
import Sequelize, { Model } from 'sequelize';

class SomeModel extends Model {}
SomeModel.init(
  {
    data: {
      type: Sequelize.JSONB,
    },
  },
  { sequelize, modelName: 'somemodels' },
);

(async function test() {
  try {
    await sequelize.sync({ force: true });
    // insert
    const data = { name: 'orm', age: 23 };
    await SomeModel.create({ data });

    // find
    const row = await SomeModel.findByPk(1, { raw: true });
    console.log(row);
  } catch (error) {
    console.log(error);
  } finally {
    await sequelize.close();
  }
})();

Результат выполнения: 

Executing (default): DROP TABLE IF EXISTS "somemodels" CASCADE;
Executing (default): DROP TABLE IF EXISTS "somemodels" CASCADE;
Executing (default): CREATE TABLE IF NOT EXISTS "somemodels" ("id"   SERIAL , "data" JSONB, PRIMARY KEY ("id"));
Executing (default): SELECT i.relname AS name, ix.indisprimary AS primary, ix.indisunique AS unique, ix.indkey AS indkey, array_agg(a.attnum) as column_indexes, array_agg(a.attname) AS column_names, pg_get_indexdef(ix.indexrelid) AS definition FROM pg_class t, pg_class i, pg_index ix, pg_attribute a WHERE t.oid = ix.indrelid AND i.oid = ix.indexrelid AND a.attrelid = t.oid AND t.relkind = 'r' and t.relname = 'somemodels' GROUP BY i.relname, ix.indexrelid, ix.indisprimary, ix.indisunique, ix.indkey ORDER BY i.relname;
Executing (default): INSERT INTO "somemodels" ("id","data") VALUES (DEFAULT,$1) RETURNING *;
Executing (default): SELECT "id", "data" FROM "somemodels" AS "somemodels" WHERE "somemodels"."id" = 1;
{ id: 1, data: { age: 23, name: 'orm' } }

записей данных в базе данных: 

=# select * from "somemodels";
 id |            data
----+----------------------------
  1 | {"age": 23, "name": "orm"}
(1 row)
...