Question

我想将另一个表导入到我的结果中，但是我有一个小问题。以下是我的控制器，对重要的地方进行了评论。问题在于NULL值where: { id: transactions.id_sender }

我认为我必须使用.map或for循环，但是我想知道，什么是解决此问题以及可能答案的好方法。

exports.getRecipientdata = (req, res) => {
  const userId = req.params.recipientId;
  Transaction.findAll({
    where: {
      id_recipient: userId,
    },
  }).then(transactions => {
    console.log('\n transactions', `${JSON.stringify(transactions)}\n`);

    Transaction.findAll({
      where: { id_sender: userId },
      attributes: [
        'amount_money',
        'date_time',
        'transfer_title',
        'id_recipient',
        'id_sender',
      ],
      include: [
        {
          model: User,
          where: { id: transactions.id_sender }, // <- this is need a results with: '2', '2', '3';
          attributes: ['name', 'surname'],
        },
      ],
    })
      .then(transactionsWithName => {
        console.log(
          '\n transactionsWithName',
          `${JSON.stringify(transactionsWithName)}\n`,
        );
        res.send(transactionsWithName);
      })
      .catch(err => {
        console.log(err);
      });
  });
};

交易JSON：

[
  {
    "id": 1,
    "id_sender": 2,
    "id_recipient": 1,
    "date_time": "2019-01-19T20:09:30.000Z",
    "amount_money": 7,
    "transfer_title": "payment1"
  },
  {
    "id": 4,
    "id_sender": 2,
    "id_recipient": 1,
    "date_time": "2019-01-19T20:23:33.000Z",
    "amount_money": 8.4,
    "transfer_title": "payment2"
  },
  {
    "id": 11,
    "id_sender": 3,
    "id_recipient": 1,
    "date_time": "2019-01-20T10:34:29.000Z",
    "amount_money": 16.5,
    "transfer_title": "payment3"
  }
]

transactionsWithName JSON：

[]

在Sequelize，NodeJS中联接表，返回null

0 个答案: