无法同步mongoose操作以返回数组

const getData = () => {
  return new Promise((resolve, reject) => {
    Model.findOne({}).exec()
      .then(result1 => {
        return Model.find({}).exec();
      })
      .then(result2 => {
        resolve([result1, result2]);
      })
      .catch(error => {
        reject(error);
      });
  });
};

getData()
  .then(results => {
    console.log(results); // 返回包含两个结果的数组
  })
  .catch(error => {
    console.error(error);
  });

const getData = async () => {
  try {
    const result1 = await Model.findOne({}).exec();
    const result2 = await Model.find({}).exec();
    return [result1, result2];
  } catch (error) {
    throw error;
  }
};

getData()
  .then(results => {
    console.log(results); // 返回包含两个结果的数组
  })
  .catch(error => {
    console.error(error);
  });